## Data: ChEDNT filename: chedamsLD1hindbrainsimple (SAP 12)

Parameter: Hindbrain dams Lactation Day 1

Model: R progr am – Single Sex

BMR: 10%

Power parameter set to 1

Female Dose-Response Curve for Chlorpyrifos

## Thu Mar 26 09:16:05 2009

Redundancy Analysis

1 2 3 4 5 6 CondIndex 14.379 14.379 1.447 1.447 1.000 1.000 mu 0.099 0.099 0.983 0.983 1.422 1.422 lA.sexF 0.000 0.400 0.589 0.000 0.011 0.000 lA.sexM 0.400 0.000 0.000 0.589 0.000 0.011 lD.sexF 0.000 0.995 0.000 0.000 0.005 0.000 lD.sexM 0.995 0.000 0.000 0.000 0.000 0.005 lg.sexF 0.000 0.995 0.001 0.000 0.005 0.000 lg.sexM 0.995 0.000 0.000 0.001 0.000 0.005

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.580 0.070 lA.sexM 1.580 0.070 lD.sexF 0.370 0.928 lD.sexM 0.370 0.928 lg.sexF 0.788 0.743 lg.sexM 0.788 0.743

---- Mean Absolute Value of Gradient Val lA.sexF 1.9126817 lA.sexM 1.9126817 lD.sexF 0.4933735 lD.sexM 0.4933735 lg.sexF 0.5612878 lg.sexM 0.5612878

## Variance is heterogeneous ::: Power parameter fit

Intercept estimate (standard error): 4.86148 (0.232147) BMD10 estimate (standard error) : 1.451653 (1.674269)

Power parameter estimate (s tandard error): 2.203495 (1.759592) The confidence intervals for the variables are given by lower est. upper Intercept 4.4376629 4.861480 5.325774 BMD 0.4390024 1.451653 4.800192 Power Parameter 0.8484474 2.203495 5.722679

BMDL 0.5366617

## Data: ChEDNT filename: chedamsLD1RBCsimple (SAP5)

Parameter: RBC dams Lactation Day 1

Model: R progr am – Single Sex

BMR: 10%

Power parameter set to 1

Female Dose-Response Curve for Chlorpyrifos

Wed Mar 25 09:22:59 2009

Redundancy Analysis

1 2 3 4 5 6 CondIndex 8.758 8.060 2.160 1.956 1.023 1.000

## mu 0.178 0.194 0.723 0.798 1.527 1.562 lA.sexF 0.000 0.493 0.469 0.000 0.000 0.038 lA.sexM 0.206 0.000 0.000 0.737 0.057 0.000 lD.sexF 0.000 0.986 0.004 0.000 0.000 0.010 lD.sexM 0.984 0.000 0.000 0.006 0.010 0.000 lg.sexF 0.000 0.936 0.050 0.000 0.000 0.013 lg.sexM 0.972 0.000 0.000 0.018 0.011 0.000

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.610 0.089 lA.sexM 1.750 0.077 lD.sexF -1.470 2.090 lD.sexM -2.520 0.826 lg.sexF -0.740 0.660 lg.sexM -0.138 0.332

---- Mean Absolute Value of Gradient Val lA.sexF 2.0816586 lA.sexM 1.5309286 lD.sexF 0.1692696 lD.sexM 0.4306021 lg.sexF 0.3497506 lg.sexM 0.9885340

## Variance is heterogeneous ::: Power parameter set to 1

Intercept estimate (standard error): 5.099891 (0.2995380) BMD10 estimate (standard error) : 0.05452521 (0.006826087)

The confidence intervals for the variables are given by lower est. upper Intercept 4.56378961 5.09989061 5.69896653 BMD 0.04360681 0.05452521 0.06817738

BMDL 0.0452703

>

### Chlorpyrifos 91107 Repeat oral (Corn Oil) Brain F (ADLT) 8 time (hrs)



## BMDS Exponential MODEL variance Bmr = 10%

!!! Warning: optimum may not have been found for Model 3 !!! !!! Bad completion code in maximum likelihood optimization routine !!! !!! Try choosing different initial values !!!

Warning: optimum may not have been found form model 2 (BMDL). Bad process completion in Optimization routine.

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d) Gnuplot Plotting File: Fri Aug 20 10:46:28 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) The variance is to be modeled as Var(i) = exp(lalpha + log(mean(i)) * rho)

Total number of dose groups = 6 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 19.6141 19.6141 19.6141 19.6141 rho -0.42086 -0.42086 -0.42086 - 0.42086 a 31357.8 366.014 54589.5 54589.5 b 0.341823 -0.412747 0.329655 0.315193 c -- -- 0.000294793 0.280755 d -- 2 -- 14

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 148.694 19.6141 148.694 17.0871 rho -12.2946 -0.42086 -12.2946 - 0.175616 a 52527.4 366.014 52527.4 51964.8 b 0.106739 -0.412747 0.106739 0.622646 c -- -- 0 0.309675 d -- 2 -- 4.18895

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 8 5.198e+004 1803 0.05 8 5.193e+004 1031 0.1 8 5.199e+004 3291 0.5 8 5.169e+004 1862 1 8 4.736e+004 1938 3.5 8 1.609e+004 2362

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 5.253e+004 1865 -0.8333 0.05 5.225e+004 1927 -0.46780.1 5.197e+004 1992 0.02887 0.5 4.98e+004 2590 2.071 1 4.721e+004 3595 0.1163 3.5 3.615e+004 1.854e+004 -3.06 3 0 0 1.397e+069 1.053e-064 0.05 0 1.397e+069 1.052e-064 0.1 0 1.397e+069 1.053e-064 0.5 0 1.397e+069 1.047e-064 1 0 1.397e+069 9.591e-065 3.5 0 1.397e+069 3.259e-065 4 0 5.253e+004 1865 -0.8333 0.05 5.225e+004 1927 -0.4678 0.1 5.197e+004 1992 0.02887 0.5 4.98e+004 2590 2.071 1 4.721e+004 3595 0.1163 3.5 3.615e+004 1.854e+004 -3.06 5 0 5.196e+004 1979 0.01852 0.05 5.196e+004 1979 -0.05122 0.1 5.196e+004 1979 0.03642 0.5 5.17e+004 1979 -0.002227 1 4.736e+004 1995 -0.001818 3.5 1.609e+004 2193 0.0005003

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -389.2882 7 792.5764 A2 -383.9951 12 791.9902 A3 -389.2158 8 794.4316 R -479.6601 2 963.3202 2 -412.5291 4 833.0582 3 -2271.155 5 4552.31 4 -412.5291 4 833.0582 5 -389.2173 6 790.4345

Additive constant for all log-likelihoods = -44.11. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R)Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 191.3 10 < 0.0001 Test 2 10.59 5 0.06023 Test 3 10.44 4 0.03361 Test 4 46.63 4 < 0.0001 Test 5a 3764 3 < 0.0001 Test 5b -3717 1 N/A Test 6a 46.63 4 < 0.0001 Test 6b -1.137e-013 0 N/A Test 7a 0.002912 2 0.9985 Test 7b 3764 1 < 0.0001 Test 7c 46.62 2 < 0.0001

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. A non-homogeneous variance model appears to be appropriate.

The p-value for Test 3 is less than .1. You may want to consider a different variance model.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is less than .1. Model 3 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5b is less than .05. Model 3 appears to fit the data better than Model 2.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 6b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7a is greater than .1. Model 5 seems to adequately describe the data.

The p-value for Test 7b is less than .05. Model 5 appears to fit the data better than Model 3.The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.987087 0.686322 3 -0.786421 Bad completion 4 0.987087 0.686322 5 1.03149 0.952164

15000 20000 25000 30000 35000 40000 45000 50000 55000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

10:46 08/20 2010 BMD BMDL Exponential0 10000 20000 30000 40000 50000 60000

-1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

10:46 08/20 2010 BMD BMDL Exponential

15000 20000 25000 30000 35000 40000 45000 50000 55000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

10:46 08/20 2010 BMD BMDL Exponential15000 20000 25000 30000 35000 40000 45000 50000 55000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

10:46 08/20 2010 BMD BMDL Exponential

### Chlorpyrifos 91107 Repeat oral (Corn Oil) RBC M (PND) 8 time (hrs)



## BMDS Exponential MODEL variance Bmr = 10%

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d) Gnuplot Plotting File: Mon Aug 23 10:41:12 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) The variance is to be modeled as Var(i) = exp(lalpha + log(mean(i)) * rho)

Total number of dose groups = 6 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

-----lnalpha 3.5618 3.5618 3.5618 3.5618 rho 1.24098 1.24098 1.24098 1.24098 a 1763.69 1763.69 6730.5 6730.5 b 0.688802 0.688802 0.924147 0.924147 c -- -- 0.040101 0.040101 d -- 1 -- 1

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 4.58344 4.58344 3.37202 3.37202 rho 1.11897 1.11897 1.25812 1.25812 a 5900.69 5900.69 6128.48 6128.48 b 0.733957 0.733957 1.00583 1.00583 c -- -- 0.0626966 0.0626966 d -- 1 -- 1

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 8 6410 1605 0.05 8 5517 1180 0.1 8 5450 1294 0.5 8 4054 1112 1 8 2478 532.5 3.5 8 539.8 332.2

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 5901 1274 1.131 0.05 5688 1248 -0.3889 0.1 5483 1222 -0.0771 0.5 4088 1037 -0.09311 1 2832 844.7 -1.187 3.5 452.1 302.6 0.8195 3 0 5901 1274 1.131 0.05 5688 1248 -0.3889 0.1 5483 1222 -0.0771 0.5 4088 1037 -0.09311 1 2832 844.7 -1.187 3.5 452.1 302.6 0.81954 0 6128 1302 0.6115 0.05 5847 1264 -0.7388 0.1 5579 1227 -0.2973 0.5 3858 973.3 0.5691 1 2485 738.1 -0.02734 3.5 554.2 287.2 -0.1417 5 0 6128 1302 0.6115 0.05 5847 1264 -0.7388 0.1 5579 1227 -0.2973 0.5 3858 973.3 0.5691 1 2485 738.1 -0.02734 3.5 554.2 287.2 -0.1417

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -356.9792 7 727.9583 A2 -346.5243 12 717.0486 A3 -347.6798 8 711.3596 R -394.8545 2 793.709 2 -350.2632 4 708.5264 3 -350.2632 4 708.5264 4 -348.573 5 707.146 5 -348.573 5 707.146

Additive constant for all log-likelihoods = -44.11. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 96.66 10 < 0.0001 Test 2 20.91 5 0.0008425 Test 3 2.311 4 0.6788 Test 4 5.167 4 0.2706 Test 5a 5.167 4 0.2706 Test 5b 0 0 N/A Test 6a 1.786 3 0.6179 Test 6b 3.38 1 0.06598 Test 7a 1.786 3 0.6179 Test 7b 3.38 1 0.06598 Test 7c 0 0 N/A

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. A non-homogeneous variance model appears to be appropriate.

The p-value for Test 3 is greater than .1. The modeled variance appears to be appropriate here.

The p-value for Test 4 is greater than .1. Model 2 seems to adequately describe the data.

The p-value for Test 5a is greater than .1. Model 3 seems to adequately describe the data.

Degrees of freedom for Test 5b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 6a is greater than .1. Model 4 seems to adequately describe the data.

The p-value for Test 6b is greater than .05. Model 4 does not seem to fit the data better than Model 2.

The p-value for Test 7a is greater than .1. Model 5 seems to adequately describe the data.

The p-value for Test 7b is greater than .05. Model 5 does not seem to fit the data better than Model 3.

Degrees of freedom for Test 7c are less than or equal to 0. The Chi-Square test for fit is not valid.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviationConfidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.143551 0.119435 3 0.143551 0.119435 4 0.112166 0.0912869 5 0.112166 0.0912869

0 1000 2000 3000 4000 5000 6000 7000 8000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

10:41 08/23 2010 BMD BMDL Exponential0 1000 2000 3000 4000 5000 6000 7000 8000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

10:41 08/23 2010 BMD BMDL Exponential

0 1000 2000 3000 4000 5000 6000 7000 8000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

10:41 08/23 2010 BMD BMDL Exponential0 1000 2000 3000 4000 5000 6000 7000 8000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

10:41 08/23 2010 BMD BMDL Exponential

### Chlorpyrifos 91107 acute oral (Diet) Brain F (ADLT) 8 time (hrs) 

## BMDS Exponential MODEL variance Bmr = 10%

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d) Gnuplot Plotting File: Fri Aug 20 09:01:08 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) The variance is to be modeled as Var(i) = exp(lalpha + log(mean(i)) * rho)

Total number of dose groups = 6 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 143.444 143.444 143.444 143.444 rho -11.8212 -11.8212 -11.8212 - 11.8212a 48011.3 48011.3 56582 56582 b 0.0263921 0.0263921 0.217692 0.217692 c -- -- 0.692218 0.692218 d -- 1 -- 1

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 147.106 129.648 147.106 131.002 rho -12.1486 -10.5465 -12.1486 - 10.6729 a 53554.6 53371.1 53554.6 53338.2 b 0.0235669 0.0374083 0.0235669 0.427433 c -- -- 0 0.764479 d -- 1.32525 -- 11.6354

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 8 5.389e+004 1242 0.05 8 5.27e+004 2265 0.1 8 5.302e+004 1698 0.5 8 5.384e+004 1737 2 8 5.129e+004 1639 10 8 4.113e+004 7650

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 5.355e+004 1658 0.5683 0.05 5.349e+004 1670 -1.341 0.1 5.343e+004 1681 -0.6866 0.5 5.293e+004 1781 1.457 2 5.109e+004 2207 0.2623 10 4.231e+004 6937 -0.4831 3 0 5.337e+004 1677 0.8712 0.05 5.336e+004 1679 -1.109 0.1 5.334e+004 1682 -0.5354 0.5 5.31e+004 1723 1.225 2 5.168e+004 1987 -0.5504 10 4.067e+004 7026 0.1819 4 0 5.355e+004 1658 0.5683 0.05 5.349e+004 1670 -1.341 0.1 5.343e+004 1681 -0.6866 0.5 5.293e+004 1781 1.4572 5.109e+004 2207 0.2623 10 4.231e+004 6937 -0.4831 5 0 5.334e+004 1664 0.9338 0.05 5.334e+004 1664 -1.085 0.1 5.334e+004 1664 -0.5404 0.5 5.334e+004 1664 0.8603 2 5.147e+004 2013 -0.2437 10 4.078e+004 6975 0.1417

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -412.5905 7 839.181 A2 -389.5215 12 803.043 A3 -390.8265 8 797.653 R -437.9293 2 879.8586 2 -394.317 4 796.634 3 -393.454 5 796.9079 4 -394.317 4 796.634 5 -393.0053 6 798.0106

Additive constant for all log-likelihoods = -44.11. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 96.82 10 < 0.0001 Test 2 46.14 5 < 0.0001 Test 3 2.61 4 0.625 Test 4 6.981 4 0.1369 Test 5a 5.255 3 0.1541 Test 5b 1.726 1 0.1889 Test 6a 6.981 4 0.1369 Test 6b -2.274e-013 0 N/A Test 7a 4.358 2 0.1132 Test 7b 0.8973 1 0.3435 Test 7c 2.623 2 0.2694

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. A non-homogeneous variance model appears to be appropriate.

The p-value for Test 3 is greater than .1. The modeled variance appears to be appropriate here.

The p-value for Test 4 is greater than .1. Model 2 seems to adequately describe the data.

The p-value for Test 5a is greater than .1. Model 3 seems to adequately describe the data.

The p-value for Test 5b is greater than .05. Model 3 does not seem to fit the data better than Model 2.

The p-value for Test 6a is greater than .1. Model 4 seems to adequately describe the data.

Degrees of freedom for Test 6b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7a is greater than .1. Model 5 seems to adequately describe the data.

The p-value for Test 7b is greater than .05. Model 5 does not seem to fit the data better than Model 3.

The p-value for Test 7c is greater than .05. Model 5 does not seem to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000

BMD and BMDL by ModelModel BMD BMDL

------- ------------ ---------- 2 4.4707 3.3015 3 4.89291 3.61336 4 4.4707 3.27823 5 2.2233 2.07086

35000 40000 45000 50000 55000

0 2 4 6 8 10Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

09:01 08/20 2010 BMD BMDL Exponential35000 40000 45000 50000 55000

0 2 4 6 8 10Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

09:01 08/20 2010 BMD BMDL Exponential

35000 40000 45000 50000 55000

0 2 4 6 8 10Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

09:01 08/20 2010 BMD BMDL Exponential35000 40000 45000 50000 55000

0 2 4 6 8 10Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

09:01 08/20 2010 BMD BMDL Exponential

### Chlorpyrifos 91107 acute oral (Diet) RBC F (ADLT) 8 time (hrs) 

## BMDS Exponential MODEL variance Bmr = 10%

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d) Gnuplot Plotting File: Fri Aug 20 08:36:34 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) The variance is to be modeled as Var(i) = exp(lalpha + log(mean(i)) * rho)

Total number of dose groups = 6 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 6.5256 6.5256 6.5256 6.5256 rho 0.685737 0.685737 0.685737 0.685737a 1255.83 1255.83 5648.48 5648.48 b 0.350191 0.350191 0.374436 0.374436 c -- -- 0.0059131 0.0059131 d -- 1 -- 1

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 5.70484 6.81983 5.67735 6.1726 rho 0.823671 0.683928 0.826993 0.726332 a 5529.09 5460.69 5531.07 5291.21 b 0.354011 0.346782 0.355537 0.456066 c -- -- 0.000761254 0.0311856 d -- 1.09341 -- 2.92331

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 8 5337 304.8 0.05 8 5164 575 0.1 8 5380 524.6 0.5 8 5219 575.3 2 8 2548 445.8 10 8 167 146

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 5529 602.8 -0.9014 0.05 5432 598.4 -1.267 0.1 5337 594 0.2034 0.5 4632 560.4 2.961 2 2724 450.3 -1.105 10 160.4 140.3 0.1331 3 0 5461 574 -0.6095 0.05 5396 571.7 -1.149 0.1 5324 569 0.2754 0.5 4713 545.8 2.62 2 2794 456.4 -1.523 10 111.1 151.5 1.044 4 0 5531 603.2 -0.91 0.05 5434 598.8 -1.274 0.1 5338 594.4 0.1974 0.5 4631 560.5 2.9672 2719 449.7 -1.074 10 162.1 140.1 0.09854 5 0 5291 492.8 0.2629 0.05 5291 492.8 -0.7297 0.1 5291 492.7 0.5104 0.5 5224 490.5 -0.02753 2 2552 378.1 -0.03281 10 165 139.9 0.04026

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -314.7161 7 643.4321 A2 -306.9955 12 637.991 A3 -308.9629 8 633.9257 R -389.2592 2 782.5184 2 -316.5189 4 641.0378 3 -315.9325 5 641.865 4 -316.5177 5 643.0355 5 -309.3693 6 630.7387

Additive constant for all log-likelihoods = -44.11. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 164.5 10 < 0.0001 Test 2 15.44 5 0.008635 Test 3 3.935 4 0.4149 Test 4 15.11 4 0.004474 Test 5a 13.94 3 0.002989 Test 5b 1.173 1 0.2788 Test 6a 15.11 3 0.001725 Test 6b 0.00237 1 0.9612 Test 7a 0.8129 2 0.666 Test 7b 13.13 1 0.0002912 Test 7c 14.3 1 0.0001561

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. A non-homogeneous variance model appears to be appropriate.

The p-value for Test 3 is greater than .1. The modeled variance appears to be appropriate here.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is less than .1. Model 3 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5b is greater than .05. Model 3 does not seem to fit the data better than Model 2.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

The p-value for Test 6b is greater than .05. Model 4 does not seem to fit the data better than Model 2.

The p-value for Test 7a is greater than .1. Model 5 seems to adequately describe the data.

The p-value for Test 7b is less than .05. Model 5 appears to fit the data better than Model 3.

The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000

BMD and BMDL by ModelModel BMD BMDL

------- ------------ ---------- 2 0.297619 0.266265 3 0.368228 0.274132 4 0.29658 0.25531 5 1.02712 0.598015

0 1000 2000 3000 4000 5000 6000

0 2 4 6 8 10Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

08:36 08/20 2010 BMD BMDL Exponential0 1000 2000 3000 4000 5000 6000

0 2 4 6 8 10Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

08:36 08/20 2010 BMD BMDL Exponential

0 1000 2000 3000 4000 5000 6000

0 2 4 6 8 10Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

08:36 08/20 2010 BMD BMDL Exponential0 1000 2000 3000 4000 5000 6000

0 2 4 6 8 10Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

08:36 08/20 2010 BMD BMDL Exponentialmridstudytypesextime(hrs)mg/kgmeanChEChEunitSDnspecies48139301acute oral (gavage) BrainF(ADLT)8054727.8U/L1283.28RATExponential modelConstant varianceBmr 10Risk – relative====================================================================Exponential Model. (Version: 1.7; Date: 12/10/2009)Input Data File: C:/Usepa/BMDS212/Data/exp_Testrunscontinuous_Setting.(d)Gnuplot Plotting File:Tue Jan 25 13:25:18 2011====================================================================BMDS Model Run~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~The form of the response function by Model:Model 2: Y[dose] = a * exp{sign * b * dose}Model 3: Y[dose] = a * exp{sign * (b * dose)^d}Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}]Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]Note: Y[dose] is the median response for exposure = dose;sign = +1 for increasing trend in data;sign = -1 for decreasing trend.Model 2 is nested within Models 3 and 4.Model 3 is nested within Model 5.Model 4 is nested within Model 5.Dependent variable = MeanIndependent variable = DoseData are assumed to be distributed: normallyVariance Model: exp(lnalpha +rho *ln(Y[dose]))rho is set to 0.A constant variance model is fit.Total number of dose groups = 6Total number of records with missing values = 0Maximum number of iterations = 250Relative Function Convergence has been set to: 1e-008Parameter Convergence has been set to: 1e-008MLE solution provided: ExactInitial Parameter ValuesVariable Model 2 Model 3 Model 4 Model 5-------- ------- ------- ------- -------lnalpha 16.9406 16.9406 16.9406 16.9406rho(S) 0 0 0 0a 38663.1 38663.1 57464.2 57464.2b 0.0840269 0.0840269 0.0891764 0.0891764c -- -- 0.000405057 0.000405057d -- 1 -- 1(S) = SpecifiedParameter Estimates by ModelVariable Model 2 Model 3 Model 4 Model 5-------- ------- ------- ------- -------lnalpha 17.2111 17.0051 17.2111 17.0051rho 0 0 0 0a 54754.3 53165.5 54754.3 53165.5b 0.077807 0.0920755 0.077807 0.0920755c -- -- 0 7.90968e-024d -- 2.31692 -- 2.31692Table of Stats From Input DataDose N Obs Mean Obs Std Dev----- --- ---------- -------------0 8 5.473e+004 12830.05 8 5.446e+004 23010.1 8 5.112e+004 24840.5 8 5.224e+004 15862 8 5.219e+004 258010 8 2.328e+004 1.157e+004Estimated Values of InterestModel Dose Est Mean Est Std Scaled Residual------- ------ ---------- --------- ----------------2 0 5.475e+004 5462 -0.013720.05 5.454e+004 5462 -0.041580.1 5.433e+004 5462 -1.6640.5 5.267e+004 5462 -0.21882 4.686e+004 5462 2.7610 2.515e+004 5462 -0.96933 0 5.317e+004 4927 0.89680.05 5.317e+004 4927 0.7440.1 5.316e+004 4927 -1.1760.5 5.312e+004 4927 -0.50552 5.212e+004 4927 0.0414610 2.328e+004 4927 -0.0010594 0 5.475e+004 5462 -0.013720.05 5.454e+004 5462 -0.041580.1 5.433e+004 5462 -1.6640.5 5.267e+004 5462 -0.21882 4.686e+004 5462 2.7610 2.515e+004 5462 -0.96935 0 5.317e+004 4927 0.89680.05 5.317e+004 4927 0.7440.1 5.316e+004 4927 -1.1760.5 5.312e+004 4927 -0.50552 5.212e+004 4927 0.0414610 2.328e+004 4927 -0.001059Other models for which likelihoods are calculated:Model A1: Yij = Mu(i) + e(ij)Var{e(ij)} = Sigma^2Model A2: Yij = Mu(i) + e(ij)Var{e(ij)} = Sigma(i)^2Model A3: Yij = Mu(i) + e(ij)Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)Model R: Yij = Mu + e(i)Var{e(ij)} = Sigma^2Likelihoods of InterestModel Log(likelihood) DF AIC------- ----------------- ---- ------------A1 -430.5745 7 875.1491A2 -399.161 12 822.322A3 -430.5745 7 875.1491R -475.292 2 954.58392 -437.0672 3 880.13453 -432.1223 4 872.24474 -437.0672 3 880.13455 -432.1223 5 874.2447Additive constant for all log-likelihoods = -44.11. This constant added to theabove values gives the log-likelihood including the term that does notdepend on the model parameters.Explanation of TestsTest 1: Does response and/or variances differ among Dose levels? (A2 vs. R)Test 2: Are Variances Homogeneous? (A2 vs. A1)Test 3: Are variances adequately modeled? (A2 vs. A3)Test 4: Does Model 2 fit the data? (A3 vs. 2)Test 5a: Does Model 3 fit the data? (A3 vs 3)Test 5b: Is Model 3 better than Model 2? (3 vs. 2)Test 6a: Does Model 4 fit the data? (A3 vs 4)Test 6b: Is Model 4 better than Model 2? (4 vs. 2)Test 7a: Does Model 5 fit the data? (A3 vs 5)Test 7b: Is Model 5 better than Model 3? (5 vs. 3)Test 7c: Is Model 5 better than Model 4? (5 vs. 4)Tests of InterestTest -2*log(Likelihood Ratio) D. F. p-value-------- ------------------------ ------ --------------Test 1 152.3 10 < 0.0001Test 2 62.83 5 < 0.0001Test 3 62.83 5 < 0.0001Test 4 12.99 4 0.01135Test 5a 3.096 3 0.3771Test 5b 9.89 1 0.001662Test 6a 12.99 4 0.01135Test 6b -1.137e-013 0 N/ATest 7a 3.096 2 0.2127Test 7b -2.274e-013 1 N/ATest 7c 9.89 2 0.00712The p-value for Test 1 is less than .05. There appears to be adifference between response and/or variances among the doselevels, it seems appropriate to model the data.The p-value for Test 2 is less than .1. Consider runninga non-homogeneous variance model.The p-value for Test 3 is less than .1. You may want toconsider a different variance model.The p-value for Test 4 is less than .1. Model 2 may not adequatelydescribe the data; you may want to consider another model.The p-value for Test 5a is greater than .1. Model 3 seemsto adequately describe the data.The p-value for Test 5b is less than .05. Model 3 appearsto fit the data better than Model 2.The p-value for Test 6a is less than .1. Model 4 may not adequatelydescribe the data; you may want to consider another model.Degrees of freedom for Test 6b are less than or equal to 0.The Chi-Square test for fit is not valid.The p-value for Test 7a is greater than .1. Model 5 seemsto adequately describe the data.The p-value for Test 7b is less than .05. Model 5 appearsto fit the data better than Model 3.The p-value for Test 7c is less than .05. Model 5 appearsto fit the data better than Model 4.Benchmark Dose Computations:Specified Effect = 0.100000Risk Type = Relative deviationConfidence Level = 0.950000BMD and BMDL by ModelModel BMD BMDL------- ------------ ----------2 1.35413 1.157583 4.11184 2.261154 1.35413 1.137795 4.11184 2.261151000020000300004000050000600000246810

Mean ResponsedoseExponential Model 2 with 0.95 Confidence Level13:25 01/25 2011BMDBMDLExponential1000020000300004000050000600000246810

Mean ResponsedoseExponential Model 3 with 0.95 Confidence Level13:25 01/25 2011BMDBMDLExponential1000020000300004000050000600000246810

Mean ResponsedoseExponential Model 4 with 0.95 Confidence Level13:25 01/25 2011BMDBMDLExponential1000020000300004000050000600000246810

Mean ResponsedoseExponential Model 5 with 0.95 Confidence Level13:25 01/25 2011BMDBMDLExponential48139301acute oral (gavage) RBCF(ADLT)805409.3U/L350.58RATExponential modelConstant varianceBmr 10Risk – relative====================================================================Exponential Model. (Version: 1.7; Date: 12/10/2009)Input Data File: C:/Usepa/BMDS212/Data/exp_Testrunscontinuous_Setting.(d)Gnuplot Plotting File:Tue Jan 25 14:08:23 2011====================================================================BMDS Model Run~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~The form of the response function by Model:Model 2: Y[dose] = a * exp{sign * b * dose}Model 3: Y[dose] = a * exp{sign * (b * dose)^d}Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}]Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]Note: Y[dose] is the median response for exposure = dose;sign = +1 for increasing trend in data;sign = -1 for decreasing trend.Model 2 is nested within Models 3 and 4.Model 3 is nested within Model 5.Model 4 is nested within Model 5.Dependent variable = MeanIndependent variable = DoseData are assumed to be distributed: normallyVariance Model: exp(lnalpha +rho *ln(Y[dose]))rho is set to 0.A constant variance model is fit.Total number of dose groups = 6Total number of records with missing values = 0Maximum number of iterations = 250Relative Function Convergence has been set to: 1e-008Parameter Convergence has been set to: 1e-008MLE solution provided: ExactInitial Parameter ValuesVariable Model 2 Model 3 Model 4 Model 5-------- ------- ------- ------- -------lnalpha 12.8192 12.8192 12.8192 12.8192rho(S) 0 0 0 0a 2602.7 2602.7 6273.23 6273.23b 0.188905 0.188905 0.199083 0.199083c -- -- 0.000135704 0.000135704d -- 1 -- 1(S) = SpecifiedParameter Estimates by ModelVariable Model 2 Model 3 Model 4 Model 5-------- ------- ------- ------- -------lnalpha 13.1907 13.0394 13.1907 12.9692rho 0 0 0 0a 5659.79 5508.33 5659.79 5505.95b 0.157864 0.158148 0.157864 0.455332c -- -- 0 0.154615d -- 1.42637 -- 13.5011Table of Stats From Input DataDose N Obs Mean Obs Std Dev----- --- ---------- -------------0 8 5409 350.50.05 8 5146 533.30.1 8 5494 356.90.5 8 5975 11502 8 4360 536.210 8 851.3 622.2Estimated Values of InterestModel Dose Est Mean Est Std Scaled Residual------- ------ ---------- --------- ----------------2 0 5660 731.7 -0.96830.05 5615 731.7 -1.8140.1 5571 731.7 -0.29820.5 5230 731.7 2.8772 4127 731.7 0.898210 1167 731.7 -1.2223 0 5508 678.4 -0.41290.05 5503 678.4 -1.4880.1 5493 678.4 0.0021370.5 5363 678.4 2.5512 4539 678.4 -0.74610 805.3 678.4 0.19184 0 5660 731.7 -0.96830.05 5615 731.7 -1.8140.1 5571 731.7 -0.29820.5 5230 731.7 2.8772 4127 731.7 0.898210 1167 731.7 -1.2225 0 5506 655 -0.41740.05 5506 655 -1.5540.1 5506 655 -0.05160.5 5506 655 2.0232 4360 655 -1.746e-00710 851.3 655 3.65e-008Other models for which likelihoods are calculated:Model A1: Yij = Mu(i) + e(ij)Var{e(ij)} = Sigma^2Model A2: Yij = Mu(i) + e(ij)Var{e(ij)} = Sigma(i)^2Model A3: Yij = Mu(i) + e(ij)Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)Model R: Yij = Mu + e(i)Var{e(ij)} = Sigma^2Likelihoods of InterestModel Log(likelihood) DF AIC------- ----------------- ---- ------------A1 -331.6616 7 677.3232A2 -323.0454 12 670.0909A3 -331.6616 7 677.3232R -384.4019 2 772.80382 -340.5775 3 687.15493 -336.9452 4 681.89044 -340.5775 3 687.15495 -335.262 5 680.524Additive constant for all log-likelihoods = -44.11. This constant added to theabove values gives the log-likelihood including the term that does notdepend on the model parameters.Explanation of TestsTest 1: Does response and/or variances differ among Dose levels? (A2 vs. R)Test 2: Are Variances Homogeneous? (A2 vs. A1)Test 3: Are variances adequately modeled? (A2 vs. A3)Test 4: Does Model 2 fit the data? (A3 vs. 2)Test 5a: Does Model 3 fit the data? (A3 vs 3)Test 5b: Is Model 3 better than Model 2? (3 vs. 2)Test 6a: Does Model 4 fit the data? (A3 vs 4)Test 6b: Is Model 4 better than Model 2? (4 vs. 2)Test 7a: Does Model 5 fit the data? (A3 vs 5)Test 7b: Is Model 5 better than Model 3? (5 vs. 3)Test 7c: Is Model 5 better than Model 4? (5 vs. 4)Tests of InterestTest -2*log(Likelihood Ratio) D. F. p-value-------- ------------------------ ------ --------------Test 1 122.7 10 < 0.0001Test 2 17.23 5 0.00408Test 3 17.23 5 0.00408Test 4 17.83 4 0.001331Test 5a 10.57 3 0.01431Test 5b 7.265 1 0.007033Test 6a 17.83 4 0.001331Test 6b 1.137e-013 0 N/ATest 7a 7.201 2 0.02731Test 7b 3.366 1 0.06654Test 7c 10.63 2 0.004915The p-value for Test 1 is less than .05. There appears to be adifference between response and/or variances among the doselevels, it seems appropriate to model the data.The p-value for Test 2 is less than .1. Consider runninga non-homogeneous variance model.The p-value for Test 3 is less than .1. You may want toconsider a different variance model.The p-value for Test 4 is less than .1. Model 2 may not adequatelydescribe the data; you may want to consider another model.The p-value for Test 5a is less than .1. Model 3 may not adequatelydescribe the data; you may want to consider another model.The p-value for Test 5b is less than .05. Model 3 appearsto fit the data better than Model 2.The p-value for Test 6a is less than .1. Model 4 may not adequatelydescribe the data; you may want to consider another model.Degrees of freedom for Test 6b are less than or equal to 0.The Chi-Square test for fit is not valid.The p-value for Test 7a is less than .1. Model 5 may not adequatelydescribe the data; you may want to consider another model.The p-value for Test 7b is greater than .05. Model 5 doesnot seem to fit the data better than Model 3.The p-value for Test 7c is less than .05. Model 5 appearsto fit the data better than Model 4.Benchmark Dose Computations:Specified Effect = 0.100000Risk Type = Relative deviationConfidence Level = 0.950000BMD and BMDL by ModelModel BMD BMDL------- ------------ ----------2 0.667414 0.5531073 1.30544 0.8612544 0.667414 0.5528155 1.88369 1.16012010002000300040005000600070000246810

Mean ResponsedoseExponential Model 2 with 0.95 Confidence Level14:08 01/25 2011BMDBMDLExponential010002000300040005000600070000246810

Mean ResponsedoseExponential Model 3 with 0.95 Confidence Level14:08 01/25 2011BMDBMDLExponential010002000300040005000600070000246810

Mean ResponsedoseExponential Model 4 with 0.95 Confidence Level14:08 01/25 2011BMDBMDLExponential010002000300040005000600070000246810

Mean ResponsedoseExponential Model 5 with 0.95 Confidence Level14:08 01/25 2011BMDBMDLExponential

### 91107 acute oral (gavage) brain F (PND) 8 

## Exponential model Constant variance Bmr 10 Risk – relative

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d) Gnuplot Plotting File: Tue Sep 07 15:07:15 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) rho is set to 0. A constant variance model is fit.

Total number of dose groups = 6 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5-------- ------- ------- ------- --

----- lnalpha 15.7893 15.7893 15.7893 15.7893 rho(S) 0 0 0 0 a 17507.3 17507.3 26641 26641 b 0.160447 0.160447 0.163349 0.163349 c -- -- 0.000411223 0.000411223 d -- 1 -- 1

(S) = Specified

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 16.149 15.7974 16.149 15.7962 rho 0 0 0 0 a 25897.2 25018.4 25897.2 25009.1 b 0.136582 0.185163 0.136582 0.42834 c -- -- 0 0.438057 d -- 2.4765 -- 11.8526

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 8 2.463e+004 3768 0.05 8 2.512e+004 2987 0.1 8 2.491e+004 3080 0.5 8 2.537e+004 2069 2 8 2.293e+004 3188 5 8 1.096e+004 1517

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 2.59e+004 3212 -1.117 0.05 2.572e+004 3212 -0.525 0.1 2.555e+004 3212 -0.5596 0.5 2.419e+004 3212 1.043 2 1.971e+004 3212 2.841 5 1.308e+004 3212 -1.873 3 0 2.502e+004 2694 -0.4091 0.05 2.502e+004 2694 0.11190.1 2.502e+004 2694 -0.1121 0.5 2.495e+004 2694 0.444 2 2.297e+004 2694 -0.03888 5 1.095e+004 2694 0.005068 4 0 2.59e+004 3212 -1.117 0.05 2.572e+004 3212 -0.525 0.1 2.555e+004 3212 -0.5596 0.5 2.419e+004 3212 1.043 2 1.971e+004 3212 2.841 5 1.308e+004 3212 -1.873 5 0 2.501e+004 2692 -0.3996 0.05 2.501e+004 2692 0.1216 0.1 2.501e+004 2692 -0.1037 0.5 2.501e+004 2692 0.3817 2 2.293e+004 2692 -3.368e-008 5 1.096e+004 2692 9.54e-009

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -402.9422 7 819.8844 A2 -399.1602 12 822.3204 A3 -402.9422 7 819.8844 R -439.9613 2 883.9227 2 -411.577 3 829.154 3 -403.1386 4 814.2772 4 -411.577 3 829.154 5 -403.1082 5 816.2164

Additive constant for all log-likelihoods = -44.11. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 81.6 10 < 0.0001 Test 2 7.564 5 0.182 Test 3 7.564 5 0.182 Test 4 17.27 4 0.001713 Test 5a 0.3927 3 0.9417 Test 5b 16.88 1 < 0.0001 Test 6a 17.27 4 0.001713 Test 6b 0 0 N/A Test 7a 0.332 2 0.847 Test 7b 0.06072 1 0.8054 Test 7c 16.94 2 0.0002099

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is greater than .1. A homogeneous variance model appears to be appropriate here.

The p-value for Test 3 is greater than .1. The modeled variance appears to be appropriate here.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is greater than .1. Model 3 seems to adequately describe the data.

The p-value for Test 5b is less than .05. Model 3 appears to fit the data better than Model 2.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 6b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7a is greater than .1. Model 5 seems to adequately describe the data.

The p-value for Test 7b is greater than .05. Model 5 does not seem to fit the data better than Model 3.

The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.771408 0.641655 3 2.17675 1.5596 4 0.771408 0.641603 5 2.03466 1.56599

10000 15000 20000 25000

0 1 2 3 4 5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

15:07 09/07 2010 BMD BMDL Exponential10000 15000 20000 25000

0 1 2 3 4 5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

15:07 09/07 2010 BMD BMDL Exponential

10000 15000 20000 25000

0 1 2 3 4 5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

15:07 09/07 2010 BMD BMDL Exponential10000 15000 20000 25000

0 1 2 3 4 5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

15:07 09/07 2010 BMD BMDL Exponential

## Chlorpyrifos MRID 48139301 Acute Oral – Gavage Female PND 11 Brain ChE R-based single sex program

Fri Jan 21 09:52:10 2011

Redundancy Analysis

1 2 3 4 5 6 CondIndex 9.633 9.633 1.512 1.512 1.000 1.000 mu 0.149 0.149 0.952 0.952 1.439 1.439 lA.sexF 0.204 0.000 0.000 0.765 0.031 0.000 lA.sexM 0.000 0.204 0.765 0.000 0.000 0.031 lD.sexF 0.989 0.000 0.000 0.001 0.010 0.000 lD.sexM 0.000 0.989 0.001 0.000 0.000 0.010 lg.sexF 0.987 0.000 0.000 0.003 0.010 0.000 lg.sexM 0.000 0.987 0.003 0.000 0.000 0.010 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 0.001 0.000 -0.006 0.000 -0.007 0.000 lA.sexM 0.000 0.001 0.000 -0.006 0.000 -0.007 lD.sexF -0.006 0.000 0.141 0.000 0.166 0.000 lD.sexM 0.000 -0.006 0.000 0.141 0.000 0.166 lg.sexF -0.007 0.000 0.166 0.000 0.204 0.000 lg.sexM 0.000 -0.007 0.000 0.166 0.000 0.204 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 1.000 0.000 -0.460 0.000 -0.415 0.000 lA.sexM 0.000 1.000 0.000 -0.460 0.000 -0.415 lD.sexF -0.460 0.000 1.000 0.000 0.976 0.000 lD.sexM 0.000 -0.460 0.000 1.000 0.000 0.976 lg.sexF -0.415 0.000 0.976 0.000 1.000 0.000 lg.sexM 0.000 -0.415 0.000 0.976 0.000 1.000

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.595 0.037 lA.sexM 1.595 0.037 lD.sexF 0.778 0.376 lD.sexM 0.778 0.376 lg.sexF 0.907 0.451

## lg.sexM 0.907 0.451

---- Mean Absolute Value of Gradient Val lA.sexF 2.1993035 lA.sexM 2.1993035 lD.sexF 0.4505348 lD.sexM 0.4505348 lg.sexF 0.3170473 lg.sexM 0.3170473

Variance is heterogeneous ::: Power parameter fit

Intercept estimate (standard error): 4.930366 (0.1090588) BMD10 estimate (standard error) : 2.174100 (0.5658673)

## Power parameter estimate (s tandard error): 2.473317 (0.790127) The confidence intervals for the variables are given by lower est. upper Intercept 4.721489 4.930366 5.148483 BMD 1.429808 2.174100 3.305834 Power Parameter 1.509113 2.473317 4.053571

BMDL 1.531176

## Chlorpyrifos MRID 48139301 Acute Oral – Gavage Female PND 11 RBC ChE R-based single sex program

Fri Jan 21 09:23:46 2011

Redundancy Analysis

1 2 3 4 5 6 CondIndex 5.752 5.752 1.500 1.500 1.000 1.000 mu 0.248 0.248 0.951 0.951 1.426 1.426 lA.sexF 0.000 0.185 0.000 0.778 0.037 0.000 lA.sexM 0.185 0.000 0.778 0.000 0.000 0.037 lD.sexF 0.000 0.971 0.000 0.002 0.027 0.000 lD.sexM 0.971 0.000 0.002 0.000 0.000 0.027 lg.sexF 0.000 0.960 0.000 0.013 0.027 0.000 lg.sexM 0.960 0.000 0.013 0.000 0.000 0.027 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 0.001 0.000 -0.005 0.000 -0.003 0.000 lA.sexM 0.000 0.001 0.000 -0.005 0.000 -0.003 lD.sexF -0.005 0.000 0.097 0.000 0.069 0.000 lD.sexM 0.000 -0.005 0.000 0.097 0.000 0.069 lg.sexF -0.003 0.000 0.069 0.000 0.056 0.000 lg.sexM 0.000 -0.003 0.000 0.069 0.000 0.056 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 1.000 0.000 -0.434 0.000 -0.353 0.000 lA.sexM 0.000 1.000 0.000 -0.434 0.000 -0.353 lD.sexF -0.434 0.000 1.000 0.000 0.933 0.000 lD.sexM 0.000 -0.434 0.000 1.000 0.000 0.933 lg.sexF -0.353 0.000 0.933 0.000 1.000 0.000 lg.sexM 0.000 -0.353 0.000 0.933 0.000 1.000

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.590 0.039 lA.sexM 1.590 0.039 lD.sexF -0.031 0.312 lD.sexM -0.031 0.312

## lg.sexF 0.588 0.236 lg.sexM 0.588 0.236

---- Mean Absolute Value of Gradient Val lA.sexF 1.9530866 lA.sexM 1.9530866 lD.sexF 0.4152111 lD.sexM 0.4152111 lg.sexF 0.4827111 lg.sexM 0.4827111

pnd female rbc che

## Variance is heterogeneous ::: Power parameter fit

Intercept estimate (standard error): 4.905236 (0.1169381) BMD10 estimate (standard error) : 0.9689255 (0.1676837)

Power parameter estimate (s tandard error): 1.799136 (0.1841666) The confidence intervals for the variables are given by lower est. upper Intercept 4.6820031 4.9052358 5.139112 BMD 0.7197897 0.9689255 1.304293 Power Parameter 1.4933959 1.7991357 2.167469

BMDL 0.755621

## Chlorpyrifos MRID 48139301 Acute Oral – Gavage Male PND 11 Brain ChE R-based single sex program

Fri Jan 21 08:37:26 2011

Redundancy Analysis

1 2 3 4 5 6 CondIndex 9.307 9.307 1.521 1.521 1.000 1.000 mu 0.155 0.155 0.948 0.948 1.441 1.441 lA.sexF 0.207 0.000 0.760 0.000 0.000 0.033 lA.sexM 0.000 0.207 0.000 0.760 0.033 0.000 lD.sexF 0.988 0.000 0.001 0.000 0.000 0.010 lD.sexM 0.000 0.988 0.000 0.001 0.010 0.000 lg.sexF 0.986 0.000 0.004 0.000 0.000 0.010 lg.sexM 0.000 0.986 0.000 0.004 0.010 0.000 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 0.001 0.000 -0.007 0.000 -0.007 0.000 lA.sexM 0.000 0.001 0.000 -0.007 0.000 -0.007 lD.sexF -0.007 0.000 0.159 0.000 0.181 0.000 lD.sexM 0.000 -0.007 0.000 0.159 0.000 0.181 lg.sexF -0.007 0.000 0.181 0.000 0.218 0.000 lg.sexM 0.000 -0.007 0.000 0.181 0.000 0.218 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 1.000 0.000 -0.465 0.000 -0.415 0.000 lA.sexM 0.000 1.000 0.000 -0.465 0.000 -0.415 lD.sexF -0.465 0.000 1.000 0.000 0.974 0.000 lD.sexM 0.000 -0.465 0.000 1.000 0.000 0.974 lg.sexF -0.415 0.000 0.974 0.000 1.000 0.000 lg.sexM 0.000 -0.415 0.000 0.974 0.000 1.000

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.556 0.038 lA.sexM 1.556 0.038 lD.sexF 0.756 0.399 lD.sexM 0.756 0.399 lg.sexF 0.859 0.467 lg.sexM 0.859 0.467

## ---- Mean Absolute Value of Gradient Val lA.sexF 2.1195136 lA.sexM 2.1195137 lD.sexF 0.4147902 lD.sexM 0.4147902 lg.sexF 0.2949989 lg.sexM 0.2949989

Variance is homogeneous ::: Power parameter fit

Intercept estimate (standard error): 4.741785 (0.09557595) BMD10 estimate (standard error) : 2.129423 (0.5433625)

## Power parameter estimate (s tandard error): 2.359922 (0.7317253) The confidence intervals for the variables are given by lower est. upper Intercept 4.558040 4.741785 4.932937 BMD 1.409899 2.129423 3.216148 Power Parameter 1.456628 2.359922 3.823372

BMDL 1.508193

PND 11 Male Brain ChE Dose-Response Curve for Chlorpyrifos

## Chlorpyrifos MRID 48139301 Acute Oral – Gavage Male PND 11 RBC ChE R-based single sex program

Fri Jan 21 09:02:30 2011

Redundancy Analysis

1 2 3 4 5 6 CondIndex 6.080 6.080 1.513 1.513 1.000 1.000 mu 0.235 0.235 0.946 0.946 1.431 1.431 lA.sexF 0.201 0.000 0.761 0.000 0.000 0.038 lA.sexM 0.000 0.201 0.000 0.761 0.038 0.000 lD.sexF 0.974 0.000 0.002 0.000 0.000 0.024 lD.sexM 0.000 0.974 0.000 0.002 0.024 0.000 lg.sexF 0.964 0.000 0.012 0.000 0.000 0.024 lg.sexM 0.000 0.964 0.000 0.012 0.024 0.000 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 0.002 0.000 -0.006 0.000 -0.004 0.000 lA.sexM 0.000 0.002 0.000 -0.006 0.000 -0.004 lD.sexF -0.006 0.000 0.121 0.000 0.080 0.000 lD.sexM 0.000 -0.006 0.000 0.121 0.000 0.080 lg.sexF -0.004 0.000 0.080 0.000 0.060 0.000 lg.sexM 0.000 -0.004 0.000 0.080 0.000 0.060 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 1.000 0.000 -0.454 0.000 -0.374 0.000 lA.sexM 0.000 1.000 0.000 -0.454 0.000 -0.374 lD.sexF -0.454 0.000 1.000 0.000 0.940 0.000 lD.sexM 0.000 -0.454 0.000 1.000 0.000 0.940 lg.sexF -0.374 0.000 0.940 0.000 1.000 0.000 lg.sexM 0.000 -0.374 0.000 0.940 0.000 1.000

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.584 0.040 lA.sexM 1.584 0.040 lD.sexF -0.211 0.348 lD.sexM -0.211 0.348

## lg.sexF 0.502 0.245 lg.sexM 0.502 0.245

---- Mean Absolute Value of Gradient Val lA.sexF 1.9059388 lA.sexM 1.9059388 lD.sexF 0.3995886 lD.sexM 0.3995886 lg.sexF 0.5079132 lg.sexM 0.5079132

Variance is heterogeneous ::: Power parameter fit

## Intercept estimate (standard error): 4.868777 (0.1039630) BMD10 estimate (standard error) : 0.825921 (0.1277144)

Power parameter estimate (s tandard error): 1.678496 (0.1418870) The confidence intervals for the variables are given by lower est. upper Intercept 4.6693674 4.868777 5.076703 BMD 0.6308277 0.825921 1.081350 Power Parameter 1.4357601 1.678496 1.962270

BMDL 0.6592323

### 91107 acute oral (gavage) RBC F (PND) 8 

## Exponential model Model variance Bmr 10 Risk – relative

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d) Gnuplot Plotting File: Tue Sep 07 14:30:28 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) The variance is to be modeled as Var(i) = exp(lalpha + log(mean(i)) * rho)

Total number of dose groups = 6 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 4.09606 4.09606 4.09606 4.09606rho 1.03903 1.03903 1.03903 1.03903 a 2627.12 2627.12 7007.18 7007.18 b 0.397597 0.397597 0.38917 0.38917 c -- -- 0.000124629 0.000124629 d -- 1 -- 1

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 11.3065 3.28542 11.3065 3.22766 rho 0.255862 1.14546 0.255862 1.152 a 6853.22 6553.13 6853.22 6525.17 b 0.300822 0.296078 0.300822 0.359141 c -- -- 0 0.117855 d -- 1.78248 -- 2.35354

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 8 6432 776 0.05 8 6674 633.6 0.1 8 6518 677.1 0.5 8 6345 1308 2 8 4441 392.2 5 8 873.3 281.5

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 6853 882.9 -1.351 0.05 6751 881.2 -0.2485 0.1 6650 879.5 -0.4249 0.5 5896 866.1 1.466 2 3755 817.5 2.373 5 1523 728.3 -2.523 3 0 6553 792.9 -0.4339 0.05 6550 792.6 0.4423 0.1 6541 792 -0.08138 0.5 6339 777.9 0.02145 2 4424 633.1 0.07671 5 876 250.4 -0.03085 4 0 6853 882.9 -1.351 0.05 6751 881.2 -0.24850.1 6650 879.5 -0.4249 0.5 5896 866.1 1.466 2 3755 817.5 2.373 5 1523 728.3 -2.523 5 0 6525 790.8 -0.335 0.05 6525 790.8 0.5321 0.1 6523 790.7 -0.01746 0.5 6425 783.8 -0.2884 2 4407 630.8 0.1537 5 878.1 249.1 -0.05447

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -338.7709 7 691.5419 A2 -328.0852 12 680.1704 A3 -333.0133 8 682.0267 R -393.3941 2 790.7883 2 -347.2382 4 702.4764 3 -333.2483 5 676.4965 4 -347.2382 4 702.4764 5 -333.1838 6 678.3676

Additive constant for all log-likelihoods = -44.11. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3)Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 130.6 10 < 0.0001 Test 2 21.37 5 0.000689 Test 3 9.856 4 0.04292 Test 4 28.45 4 < 0.0001 Test 5a 0.4698 3 0.9255 Test 5b 27.98 1 < 0.0001 Test 6a 28.45 4 < 0.0001 Test 6b -2.274e-013 0 N/A Test 7a 0.3409 2 0.8433 Test 7b 0.1289 1 0.7196 Test 7c 28.11 2 < 0.0001

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. A non-homogeneous variance model appears to be appropriate.

The p-value for Test 3 is less than .1. You may want to consider a different variance model.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is greater than .1. Model 3 seems to adequately describe the data.

The p-value for Test 5b is less than .05. Model 3 appears to fit the data better than Model 2.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 6b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7a is greater than .1. Model 5 seems to adequately describe the data.

The p-value for Test 7b is greater than .05. Model 5 does not seem to fit the data better than Model 3.

The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.350243 0.293888 3 0.955651 0.745873 4 0.350243 0.293888 5 1.13233 0.750633

0 1000 2000 3000 4000 5000 6000 7000 8000

0 1 2 3 4 5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

14:30 09/07 2010 BMD BMDL Exponential0 1000 2000 3000 4000 5000 6000 7000 8000

0 1 2 3 4 5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

14:30 09/07 2010 BMD BMDL Exponential

0 1000 2000 3000 4000 5000 6000 7000 8000

0 1 2 3 4 5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

14:30 09/07 2010 BMD BMDL Exponential0 1000 2000 3000 4000 5000 6000 7000 8000

0 1 2 3 4 5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

14:30 09/07 2010 BMD BMDL Exponential

### 91107 acute oral (gavage) RBC M (PND) 8



## Exponential model Model variance Bmr 10 Risk – relative

!!! Warning: optimum may not have been found for Model 3 !!! !!! Bad completion code in maximum likelihood optimization routine !!! !!! Try choosing different initial values !!!

Warning: optimum may not have been found form model 2 (BMDL). Bad process completion in Optimization routine.

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d) Gnuplot Plotting File: Tue Sep 07 14:43:34 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) The variance is to be modeled as Var(i) = exp(lalpha + log(mean(i)) * rho)

Total number of dose groups = 6 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 3.94441 3.94441 3.94441 3.94441 rho 1.05495 1.05495 1.05495 1.05495 a 2597.2 2597.2 7554.54 7554.54 b 0.428325 0.428325 0.422074 0.422074 c -- -- 0.000106466 0.000106466 d -- 1 -- 1

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 11.6629 3.94441 11.6629 3.5187 rho 0.19443 1.05495 0.19443 1.09804 a 7312.89 2597.2 7312.89 7009.08 b 0.324311 0.428325 0.324311 0.317003 c -- -- 0 3.77349e-025 d -- 1 -- 1.67104

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 8 6891 782.3 0.05 8 7034 923.3 0.1 8 7195 860.7 0.5 8 6538 544.3 2 8 4434 590.1 5 8 804.3 248.8

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 7313 809.5 -1.474 0.05 7195 808.2 -0.56440.1 7080 806.9 0.404 0.5 6218 796.8 1.133 2 3823 760 2.272 5 1445 691.4 -2.621 3 0 0 0 1.#IO 0.05 0 0 1.#IO 0.1 0 0 1.#IO 0.5 0 0 1.#IO 2 0 0 1.#IO 5 0 0 1.#IO 4 0 7313 809.5 -1.474 0.05 7195 808.2 -0.5644 0.1 7080 806.9 0.404 0.5 6218 796.8 1.133 2 3823 760 2.272 5 1445 691.4 -2.621 5 0 7009 750.6 -0.4449 0.05 7002 750.2 0.1199 0.1 6987 749.3 0.7836 0.5 6694 731.9 -0.6034 2 4394 580.9 0.1924 5 809.1 229.4 -0.05904

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -335.0147 7 684.0293 A2 -328.3503 12 680.7006 A3 -329.3416 8 674.6833 R -397.1813 2 798.3625 2 -343.4978 4 694.9957 3 -2609.113 4 5226.226 4 -343.4978 4 694.9957 5 -330.0495 6 672.099

Additive constant for all log-likelihoods = -44.11. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R)Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 137.7 10 < 0.0001 Test 2 13.33 5 0.02049 Test 3 1.983 4 0.7389 Test 4 28.31 4 < 0.0001 Test 5a 4560 4 < 0.0001 Test 5b -4531 0 N/A Test 6a 28.31 4 < 0.0001 Test 6b 0 0 N/A Test 7a 1.416 2 0.4927 Test 7b 4558 2 < 0.0001 Test 7c 26.9 2 < 0.0001

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. A non-homogeneous variance model appears to be appropriate.

The p-value for Test 3 is greater than .1. The modeled variance appears to be appropriate here.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is less than .1. Model 3 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 5b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 6b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7a is greater than .1. Model 5 seems to adequately describe the data.

The p-value for Test 7b is less than .05. Model 5 appears to fit the data better than Model 3.The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.324875 0.279152 3 0.245983 Bad completion 4 0.324875 0.284467 5 0.820499 0.652608

0 1000 2000 3000 4000 5000 6000 7000 8000

0 1 2 3 4 5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

14:43 09/07 2010 BMD BMDL Exponential0 1000 2000 3000 4000 5000 6000 7000 8000

-1 0 1 2 3 4 5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

14:43 09/07 2010 BMD BMDL Exponential

0 1000 2000 3000 4000 5000 6000 7000 8000

0 1 2 3 4 5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

14:43 09/07 2010 BMD BMDL Exponential0 1000 2000 3000 4000 5000 6000 7000 8000

0 1 2 3 4 5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

14:43 09/07 2010 BMD BMDL Exponential

### Chlorpyrifos 91107 acute oral (milk) Brain F (PND) 8

Chlorpyrifos 91107 acute oral (milk) Brain F (PND) 8 time (hrs)



## BMDS Exponential MODEL variance Bmr = 10% brain

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d) Gnuplot Plotting File: Thu Aug 19 12:45:48 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) The variance is to be modeled as Var(i) = exp(lalpha + log(mean(i)) * rho)

Total number of dose groups = 6 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 37.0859 37.0859 37.0859 37.0859rho -2.10567 -2.10567 -2.10567 - 2.10567 a 18140.5 18140.5 28317.9 28317.9 b 0.165586 0.165586 0.171736 0.171736 c -- -- 0.000406503 0.000406503 d -- 1 -- 1

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 46.7917 34.5211 46.7917 37.4376 rho -3.04757 -1.84029 -3.04757 - 2.13239 a 26995.9 26650.9 26995.9 26574.5 b 0.121787 0.177263 0.121787 0.468568 c -- -- 0 0.462647 d -- 1.63396 -- 13.4446

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 8 2.641e+004 3026 0.05 8 2.611e+004 2979 0.1 8 2.661e+004 2105 0.5 8 2.697e+004 1413 2 8 2.22e+004 4811 5 8 1.151e+004 5332

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 2.7e+004 2561 -0.6471 0.05 2.683e+004 2584 -0.7866 0.1 2.667e+004 2608 -0.0653 0.5 2.54e+004 2809 1.579 2 2.116e+004 3711 0.7934 5 1.468e+004 6476 -1.386 3 0 2.665e+004 2654 -0.2567 0.05 2.664e+004 2655 -0.5602 0.1 2.661e+004 2657 -0.005741 0.5 2.615e+004 2701 0.8608 2 2.218e+004 3143 0.02053 5 1.173e+004 5649 -0.1074 4 0 2.7e+004 2561 -0.6471 0.05 2.683e+004 2584 -0.78660.1 2.667e+004 2608 -0.0653 0.5 2.54e+004 2809 1.579 2 2.116e+004 3711 0.7934 5 1.468e+004 6476 -1.386 5 0 2.657e+004 2583 -0.18 0.05 2.657e+004 2583 -0.505 0.1 2.657e+004 2583 0.03767 0.5 2.657e+004 2583 0.4324 2 2.17e+004 3206 0.4433 5 1.229e+004 5876 -0.3771

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -413.3148 7 840.6295 A2 -404.6353 12 833.2706 A3 -409.0128 8 834.0255 R -444.9614 2 893.9228 2 -412.0599 4 832.1199 3 -409.9703 5 829.9406 4 -412.0599 4 832.1199 5 -409.4277 6 830.8553

Additive constant for all log-likelihoods = -44.11. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3)Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 80.65 10 < 0.0001 Test 2 17.36 5 0.003867 Test 3 8.755 4 0.06753 Test 4 6.094 4 0.1922 Test 5a 1.915 3 0.5902 Test 5b 4.179 1 0.04092 Test 6a 6.094 4 0.1922 Test 6b -1.137e-013 0 N/A Test 7a 0.8298 2 0.6604 Test 7b 1.085 1 0.2975 Test 7c 5.265 2 0.07192

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. A non-homogeneous variance model appears to be appropriate.

The p-value for Test 3 is less than .1. You may want to consider a different variance model.

The p-value for Test 4 is greater than .1. Model 2 seems to adequately describe the data.

The p-value for Test 5a is greater than .1. Model 3 seems to adequately describe the data.

The p-value for Test 5b is less than .05. Model 3 appears to fit the data better than Model 2.

The p-value for Test 6a is greater than .1. Model 4 seems to adequately describe the data.

Degrees of freedom for Test 6b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7a is greater than .1. Model 5 seems to adequately describe the data.

The p-value for Test 7b is greater than .05. Model 5 does not seem to fit the data better than Model 3.

The p-value for Test 7c is greater than .05. Model 5 does not seem to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.865123 0.648309 3 1.42315 0.911925 4 0.865123 0.639232 5 1.89749 1.02616

5000 10000 15000 20000 25000 30000

0 1 2 3 4 5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

12:45 08/19 2010 BMD BMDL Exponential5000 10000 15000 20000 25000 30000

0 1 2 3 4 5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

12:45 08/19 2010 BMD BMDL Exponential

5000 10000 15000 20000 25000 30000

0 1 2 3 4 5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

12:45 08/19 2010 BMD BMDL Exponential5000 10000 15000 20000 25000 30000

0 1 2 3 4 5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

12:45 08/19 2010 BMD BMDL ExponentialChlorpyrifos48139301acute oral (milk) BrainM(PND)8023910.8U/L1872.88RATBmdsExponentialModelvar====================================================================Exponential Model. (Version: 1.7; Date: 12/10/2009)Input Data File: C:/Usepa/BMDS212/Data/exp_Testrunscontinuous_Setting.(d)Gnuplot Plotting File:Mon Jan 31 09:39:23 2011====================================================================BMDS Model Run~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~The form of the response function by Model:Model 2: Y[dose] = a * exp{sign * b * dose}Model 3: Y[dose] = a * exp{sign * (b * dose)^d}Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}]Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]Note: Y[dose] is the median response for exposure = dose;sign = +1 for increasing trend in data;sign = -1 for decreasing trend.Model 2 is nested within Models 3 and 4.Model 3 is nested within Model 5.Model 4 is nested within Model 5.Dependent variable = MeanIndependent variable = DoseData are assumed to be distributed: normallyVariance Model: exp(lnalpha +rho *ln(Y[dose]))The variance is to be modeled as Var(i) = exp(lalpha + log(mean(i)) * rho)Total number of dose groups = 6Total number of records with missing values = 0Maximum number of iterations = 250Relative Function Convergence has been set to: 1e-008Parameter Convergence has been set to: 1e-008MLE solution provided: ExactInitial Parameter ValuesVariable Model 2 Model 3 Model 4 Model 5-------- ------- ------- ------- -------lnalpha 37.8267 37.8267 37.8267 37.8267rho -2.18811 -2.18811 -2.18811 -2.18811a 19562.5 19562.5 26590.8 26590.8b 0.109158 0.109158 0.116719 0.116719c -- -- 0.000523429 0.000523429d -- 1 -- 1Parameter Estimates by ModelVariable Model 2 Model 3 Model 4 Model 5-------- ------- ------- ------- -------lnalpha 77.061 36.1419 77.0609 36.1419rho -6.06079 -2.02234 -6.06079 -2.02234a 25091.2 24752.7 25091.2 24752.7b 0.0616144 0.191937 0.0616144 0.192434c -- -- 0 0.00042597d -- 13.4171 -- 14.3014Table of Stats From Input DataDose N Obs Mean Obs Std Dev----- --- ---------- -------------0 8 2.391e+004 18730.05 8 2.449e+004 34320.1 8 2.528e+004 21550.5 8 2.532e+004 28152 8 2.476e+004 27445 8 1.392e+004 4867Estimated Values of InterestModel Dose Est Mean Est Std Scaled Residual------- ------ ---------- --------- ----------------2 0 2.509e+004 2519 -1.3250.05 2.501e+004 2543 -0.58380.1 2.494e+004 2567 0.37660.5 2.433e+004 2766 1.0172 2.218e+004 3660 1.9925 1.844e+004 6408 -1.9953 0 2.475e+004 2543 -0.93620.05 2.475e+004 2543 -0.29310.1 2.475e+004 2543 0.58510.5 2.475e+004 2543 0.6362 2.475e+004 2543 0.008315 1.392e+004 4552 -1.248e-0074 0 2.509e+004 2519 -1.3250.05 2.501e+004 2543 -0.58380.1 2.494e+004 2567 0.37660.5 2.433e+004 2766 1.0172 2.218e+004 3660 1.9925 1.844e+004 6408 -1.9955 0 2.475e+004 2543 -0.93620.05 2.475e+004 2543 -0.29310.1 2.475e+004 2543 0.58510.5 2.475e+004 2543 0.6362 2.475e+004 2543 0.0082775 1.392e+004 4552 -6.062e-008Other models for which likelihoods are calculated:Model A1: Yij = Mu(i) + e(ij)Var{e(ij)} = Sigma^2Model A2: Yij = Mu(i) + e(ij)Var{e(ij)} = Sigma(i)^2Model A3: Yij = Mu(i) + e(ij)Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)Model R: Yij = Mu + e(i)Var{e(ij)} = Sigma^2Likelihoods of InterestModel Log(likelihood) DF AIC------- ----------------- ---- ------------A1 -407.2546 7 828.5092A2 -402.4099 12 828.8198A3 -404.19 8 824.3801R -432.9667 2 869.93342 -411.3514 4 830.70293 -405.0374 5 820.07484 -411.3514 4 830.70295 -405.0374 6 822.0748Additive constant for all log-likelihoods = -44.11. This constant added to theabove values gives the log-likelihood including the term that does notdepend on the model parameters.Explanation of TestsTest 1: Does response and/or variances differ among Dose levels? (A2 vs. R)Test 2: Are Variances Homogeneous? (A2 vs. A1)Test 3: Are variances adequately modeled? (A2 vs. A3)Test 4: Does Model 2 fit the data? (A3 vs. 2)Test 5a: Does Model 3 fit the data? (A3 vs 3)Test 5b: Is Model 3 better than Model 2? (3 vs. 2)Test 6a: Does Model 4 fit the data? (A3 vs 4)Test 6b: Is Model 4 better than Model 2? (4 vs. 2)Test 7a: Does Model 5 fit the data? (A3 vs 5)Test 7b: Is Model 5 better than Model 3? (5 vs. 3)Test 7c: Is Model 5 better than Model 4? (5 vs. 4)Tests of InterestTest -2*log(Likelihood Ratio) D. F. p-value-------- ------------------------ ------ --------------Test 1 61.11 10 < 0.0001Test 2 9.689 5 0.08453Test 3 3.56 4 0.4688Test 4 14.32 4 0.006333Test 5a 1.695 3 0.6381Test 5b 12.63 1 0.00038Test 6a 14.32 4 0.006333Test 6b 3.411e-013 0 N/ATest 7a 1.695 2 0.4285Test 7b 2.377e-007 1 0.9996Test 7c 12.63 2 0.001811The p-value for Test 1 is less than .05. There appears to be adifference between response and/or variances among the doselevels, it seems appropriate to model the data.The p-value for Test 2 is less than .1. A non-homogeneousvariance model appears to be appropriate.The p-value for Test 3 is greater than .1. The modeledvariance appears to be appropriate here.The p-value for Test 4 is less than .1. Model 2 may not adequatelydescribe the data; you may want to consider another model.The p-value for Test 5a is greater than .1. Model 3 seemsto adequately describe the data.The p-value for Test 5b is less than .05. Model 3 appearsto fit the data better than Model 2.The p-value for Test 6a is less than .1. Model 4 may not adequatelydescribe the data; you may want to consider another model.Degrees of freedom for Test 6b are less than or equal to 0.The Chi-Square test for fit is not valid.The p-value for Test 7a is greater than .1. Model 5 seemsto adequately describe the data.The p-value for Test 7b is greater than .05. Model 5 doesnot seem to fit the data better than Model 3.The p-value for Test 7c is less than .05. Model 5 appearsto fit the data better than Model 4.Benchmark Dose Computations:Specified Effect = 0.100000Risk Type = Relative deviationConfidence Level = 0.950000BMD and BMDL by ModelModel BMD BMDL------- ------------ ----------2 1.71 1.067613 4.40555 2.372474 1.71 1.067615 4.44013 2.3724510000150002000025000012345

Mean ResponsedoseExponential Model 2 with 0.95 Confidence Level09:39 01/31 2011BMDBMDLExponential10000150002000025000012345

Mean ResponsedoseExponential Model 3 with 0.95 Confidence Level09:39 01/31 2011BMDBMDLExponential10000150002000025000012345

Mean ResponsedoseExponential Model 4 with 0.95 Confidence Level09:39 01/31 2011BMDBMDLExponential10000150002000025000012345

Mean ResponsedoseExponential Model 5 with 0.95 Confidence Level09:39 01/31 2011BMDBMDLExponentialChlorpyrifos48139301acute oral (milk) RBCM(PND)806304.5U/L1459.88RATExponentialConstant var====================================================================Exponential Model. (Version: 1.7; Date: 12/10/2009)Input Data File: C:/Usepa/BMDS212/Data/exp_Testrunscontinuous_Setting.(d)Gnuplot Plotting File:Mon Jan 31 09:55:32 2011====================================================================BMDS Model Run~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~The form of the response function by Model:Model 2: Y[dose] = a * exp{sign * b * dose}Model 3: Y[dose] = a * exp{sign * (b * dose)^d}Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}]Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]Note: Y[dose] is the median response for exposure = dose;sign = +1 for increasing trend in data;sign = -1 for decreasing trend.Model 2 is nested within Models 3 and 4.Model 3 is nested within Model 5.Model 4 is nested within Model 5.Dependent variable = MeanIndependent variable = DoseData are assumed to be distributed: normallyVariance Model: exp(lnalpha +rho *ln(Y[dose]))rho is set to 0.A constant variance model is fit.Total number of dose groups = 6Total number of records with missing values = 0Maximum number of iterations = 250Relative Function Convergence has been set to: 1e-008Parameter Convergence has been set to: 1e-008MLE solution provided: ExactInitial Parameter ValuesVariable Model 2 Model 3 Model 4 Model 5-------- ------- ------- ------- -------lnalpha 14.4191 14.4191 14.4191 14.4191rho(S) 0 0 0 0a 3505.76 3505.76 6920.34 6920.34b 0.251523 0.251523 0.26177 0.26177c -- -- 0.000260825 0.000260825d -- 1 -- 1(S) = SpecifiedParameter Estimates by ModelVariable Model 2 Model 3 Model 4 Model 5-------- ------- ------- ------- -------lnalpha 14.4816 14.4412 14.4816 14.434rho 0 0 0 0a 6538.07 6365.61 6538.07 6308.03b 0.222939 0.237027 0.222939 0.4727c -- -- 0 0.286143d -- 1.47378 -- 11.3456Table of Stats From Input DataDose N Obs Mean Obs Std Dev----- --- ---------- -------------0 8 6305 14600.05 8 6017 13850.1 8 6591 923.90.5 8 6320 10282 8 4459 19165 8 1805 1706Estimated Values of InterestModel Dose Est Mean Est Std Scaled Residual------- ------ ---------- --------- ----------------2 0 6538 1395 -0.47350.05 6466 1395 -0.91040.1 6394 1395 0.39910.5 5848 1395 0.95662 4186 1395 0.55235 2145 1395 -0.68843 0 6366 1367 -0.12640.05 6356 1367 -0.70310.1 6340 1367 0.51870.5 6097 1367 0.46232 4563 1367 -0.2175 1762 1367 0.08894 0 6538 1395 -0.47350.05 6466 1395 -0.91040.1 6394 1395 0.39910.5 5848 1395 0.95662 4186 1395 0.55235 2145 1395 -0.68845 0 6308 1362 -0.0073180.05 6308 1362 -0.60520.1 6308 1362 0.58710.5 6308 1362 0.025482 4458 1362 2.577e-0095 1805 1362 3.2e-008Other models for which likelihoods are calculated:Model A1: Yij = Mu(i) + e(ij)Var{e(ij)} = Sigma^2Model A2: Yij = Mu(i) + e(ij)Var{e(ij)} = Sigma(i)^2Model A3: Yij = Mu(i) + e(ij)Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)Model R: Yij = Mu + e(i)Var{e(ij)} = Sigma^2Likelihoods of InterestModel Log(likelihood) DF AIC------- ----------------- ---- ------------A1 -370.0582 7 754.1163A2 -367.0576 12 758.1151A3 -370.0582 7 754.1163R -392.6363 2 789.27272 -371.5573 3 749.11463 -370.5881 4 749.17614 -371.5573 3 749.11465 -370.4166 5 750.8333Additive constant for all log-likelihoods = -44.11. This constant added to theabove values gives the log-likelihood including the term that does notdepend on the model parameters.Explanation of TestsTest 1: Does response and/or variances differ among Dose levels? (A2 vs. R)Test 2: Are Variances Homogeneous? (A2 vs. A1)Test 3: Are variances adequately modeled? (A2 vs. A3)Test 4: Does Model 2 fit the data? (A3 vs. 2)Test 5a: Does Model 3 fit the data? (A3 vs 3)Test 5b: Is Model 3 better than Model 2? (3 vs. 2)Test 6a: Does Model 4 fit the data? (A3 vs 4)Test 6b: Is Model 4 better than Model 2? (4 vs. 2)Test 7a: Does Model 5 fit the data? (A3 vs 5)Test 7b: Is Model 5 better than Model 3? (5 vs. 3)Test 7c: Is Model 5 better than Model 4? (5 vs. 4)Tests of InterestTest -2*log(Likelihood Ratio) D. F. p-value-------- ------------------------ ------ --------------Test 1 51.16 10 < 0.0001Test 2 6.001 5 0.3061Test 3 6.001 5 0.3061Test 4 2.998 4 0.5581Test 5a 1.06 3 0.7868Test 5b 1.938 1 0.1638Test 6a 2.998 4 0.5581Test 6b 0 0 N/ATest 7a 0.7169 2 0.6987Test 7b 0.3428 1 0.5582Test 7c 2.281 2 0.3196The p-value for Test 1 is less than .05. There appears to be adifference between response and/or variances among the doselevels, it seems appropriate to model the data.The p-value for Test 2 is greater than .1. A homogeneousvariance model appears to be appropriate here.The p-value for Test 3 is greater than .1. The modeledvariance appears to be appropriate here.The p-value for Test 4 is greater than .1. Model 2 seemsto adequately describe the data.The p-value for Test 5a is greater than .1. Model 3 seemsto adequately describe the data.The p-value for Test 5b is greater than .05. Model 3 doesnot seem to fit the data better than Model 2.The p-value for Test 6a is greater than .1. Model 4 seemsto adequately describe the data.Degrees of freedom for Test 6b are less than or equal to 0.The Chi-Square test for fit is not valid.The p-value for Test 7a is greater than .1. Model 5 seemsto adequately describe the data.The p-value for Test 7b is greater than .05. Model 5 doesnot seem to fit the data better than Model 3.The p-value for Test 7c is greater than .05. Model 5 doesnot seem to fit the data better than Model 4.Benchmark Dose Computations:Specified Effect = 0.100000Risk Type = Relative deviationConfidence Level = 0.950000BMD and BMDL by ModelModel BMD BMDL------- ------------ ----------2 0.472598 0.3599443 0.916354 0.4085194 0.472598 0.3376495 1.79072 0.423913010002000300040005000600070008000012345

Mean ResponsedoseExponential Model 2 with 0.95 Confidence Level09:55 01/31 2011BMDBMDLExponential010002000300040005000600070008000012345

Mean ResponsedoseExponential Model 3 with 0.95 Confidence Level09:55 01/31 2011BMDBMDLExponential010002000300040005000600070008000012345

Mean ResponsedoseExponential Model 4 with 0.95 Confidence Level09:55 01/31 2011BMDBMDLExponential010002000300040005000600070008000012345

Mean ResponsedoseExponential Model 5 with 0.95 Confidence Level09:55 01/31 2011BMDBMDLExponential

### Chlorpyrifos 48139301 acute oral (milk) RBC M (PND) 8

0 6304.5 U/L 1459.8 8 RAT



## R based model Acute milk Male Rbc

Mon Jan 31 10:06:04 2011

Redundancy Analysis

1 2 3 4 5 6 CondIndex 7.297 7.297 1.542 1.542 1.000 1.000 mu 0.198 0.198 0.936 0.936 1.444 1.444 lA.sexF 0.252 0.000 0.000 0.710 0.000 0.038 lA.sexM 0.000 0.252 0.710 0.000 0.038 0.000 lD.sexF 0.982 0.000 0.000 0.002 0.000 0.016 lD.sexM 0.000 0.982 0.002 0.000 0.016 0.000 lg.sexF 0.974 0.000 0.000 0.009 0.000 0.017 lg.sexM 0.000 0.974 0.009 0.000 0.017 0.000 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 0.002 0.000 -0.009 0.000 -0.005 0.000 lA.sexM 0.000 0.002 0.000 -0.009 0.000 -0.005 lD.sexF -0.009 0.000 0.179 0.000 0.112 0.000 lD.sexM 0.000 -0.009 0.000 0.179 0.000 0.112 lg.sexF -0.005 0.000 0.112 0.000 0.076 0.000 lg.sexM 0.000 -0.005 0.000 0.112 0.000 0.076 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 1.000 0.000 -0.509 0.000 -0.438 0.000 lA.sexM 0.000 1.000 0.000 -0.509 0.000 -0.438 lD.sexF -0.509 0.000 1.000 0.000 0.957 0.000 lD.sexM 0.000 -0.509 0.000 1.000 0.000 0.957 lg.sexF -0.438 0.000 0.957 0.000 1.000 0.000 lg.sexM 0.000 -0.438 0.000 0.957 0.000 1.000

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.575 0.041 lA.sexM 1.575 0.041

## lD.sexF -0.088 0.423 lD.sexM -0.088 0.423 lg.sexF 0.388 0.276 lg.sexM 0.388 0.276

---- Mean Absolute Value of Gradient Val lA.sexF 1.9904167 lA.sexM 1.9904167 lD.sexF 0.3800440 lD.sexM 0.3800440 lg.sexF 0.4907534 lg.sexM 0.4907534

## 01234560 2000 4000 6000 

# Male dose-respose curve for chlorpyrifos

doseResponse



## BMD BMDL



## Variance is heterogeneous ::: Power parameter set to 1

Intercept estimate (standard error): 4.935243 (0.1921248) BMD10 estimate (standard error) : 0.5006536 (0.1322799)

The confidence intervals for the variables are given by lower est. upper Intercept 4.5787490 4.9352427 5.3194924 BMD 0.3275609 0.5006536 0.7652135

BMDL 0.3510713

>

### Chlorpyrifos MRID 91107 acute oral (milk) RBC F (PND) 8



## BMDS Exponential MODEL variance Bmr = 10%

Warning: optimum may not have been found form model 2 (BMDL). Bad process completion in Optimization routine.

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d) Gnuplot Plotting File: Thu Aug 19 12:17:53 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) The variance is to be modeled as Var(i) = exp(lalpha + log(mean(i)) * rho)

Total number of dose groups = 6 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

-----lnalpha 21.2166 21.2166 21.2166 21.2166 rho -0.856979 -0.856979 -0.856979 - 0.856979 a 3134.79 108.759 6825.84 6825.84 b 0.300681 -0.156798 0.303809 0.303809 c -- -- 0.000201411 0.000201411 d -- 2 -- 1

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 27.6204 -21.9981 27.6204 22.6491 rho -1.58793 4.37075 -1.58793 -1.0078 a 6447.31 5145.45 6447.31 6362.67 b 0.210731 0.00531847 0.210731 0.239615 c -- -- 0 1.06378e-026 d -- 8.74725 -- 1.3054

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 8 6324 648.1 0.05 8 6197 613.7 0.1 8 6501 978.9 0.5 8 5877 1505 2 8 4600 1759 5 8 1375 1634

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 6447 940.1 -0.371 0.05 6380 948 -0.5467 0.1 6313 956 0.556 0.5 5803 1022 0.2055 2 4230 1314 0.7962 5 2248 2170 -1.138 3 0 5145 2158 1.545 0.05 5145 2158 1.378 0.1 5145 2158 1.776 0.5 5145 2158 0.9586 2 5145 2158 -0.7152 5 5145 2158 -4.9424 0 6447 940.1 -0.371 0.05 6380 948 -0.5467 0.1 6313 956 0.556 0.5 5803 1022 0.2055 2 4230 1314 0.7962 5 2248 2170 -1.138 5 0 6363 1004 -0.109 0.05 6343 1005 -0.4122 0.1 6314 1007 0.5243 0.5 5976 1036 -0.2713 2 4339 1217 0.6062 5 1794 1899 -0.6241

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -364.0961 7 742.1921 A2 -356.5333 12 737.0666 A3 -362.3269 8 740.6539 R -392.491 2 788.9821 2 -362.8476 4 733.6952 3 -392.491 5 794.9821 4 -362.8476 4 733.6952 5 -362.6837 6 737.3674

Additive constant for all log-likelihoods = -44.11. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 71.92 10 < 0.0001 Test 2 15.13 5 0.009839 Test 3 11.59 4 0.0207 Test 4 1.041 4 0.9035 Test 5a 60.33 3 < 0.0001 Test 5b -59.29 1 N/A Test 6a 1.041 4 0.9035 Test 6b 0 0 N/A Test 7a 0.7136 2 0.6999 Test 7b 59.61 1 < 0.0001 Test 7c 0.3278 2 0.8488

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. A non-homogeneous variance model appears to be appropriate.

The p-value for Test 3 is less than .1. You may want to consider a different variance model.

The p-value for Test 4 is greater than .1. Model 2 seems to adequately describe the data.

The p-value for Test 5a is less than .1. Model 3 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5b is less than .05. Model 3 appears to fit the data better than Model 2.

The p-value for Test 6a is greater than .1. Model 4 seems to adequately describe the data.

Degrees of freedom for Test 6b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7a is greater than .1. Model 5 seems to adequately describe the data.

The p-value for Test 7b is less than .05. Model 5 appears to fit the data better than Model 3.

The p-value for Test 7c is greater than .05. Model 5 does not seem to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviationConfidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.499976 0.364778 3 145.373 Bad completion 4 0.499976 0.283632 5 0.744403 0.297474

0 1000 2000 3000 4000 5000 6000 7000 8000

0 1 2 3 4 5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

12:18 08/19 2010 BMD BMDL Exponential0 1000 2000 3000 4000 5000 6000 7000 8000

0 20 40 60 80 100 120 140Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

12:18 08/19 2010 BMD BMDL Exponential

0 1000 2000 3000 4000 5000 6000 7000 8000

0 1 2 3 4 5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

12:18 08/19 2010 BMD BMDL Exponential0 1000 2000 3000 4000 5000 6000 7000 8000

0 1 2 3 4 5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

12:18 08/19 2010 BMD BMDL Exponential

### Chlorpyrifos 91107 Repeat oral (Corn Oil) Brain F (PND) 8 time (hrs)



## BMDS Exponential CONSTANT variance Bmr = 10%

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d) Gnuplot Plotting File: Fri Aug 20 09:45:34 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) rho is set to 0. A constant variance model is fit.

Total number of dose groups = 6 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 15.0166 15.0166 15.0166 15.0166rho(S) 0 0 0 0 a 28366.3 28366.3 46962.1 46962.1 b 0.27005 0.27005 0.294563 0.294563 c -- -- 0.0184661 0.0184661 d -- 1 -- 1

(S) = Specified

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 15.6891 15.5072 15.6891 15.1553 rho 0 0 0 0 a 44855.1 43998.3 44855.1 43504.8 b 0.258011 0.273817 0.258011 0.813136 c -- -- 0 0.398671 d -- 1.24794 -- 4.0167

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 8 4.229e+004 1616 0.05 8 4.473e+004 2598 0.1 8 4.35e+004 1409 0.5 8 4.281e+004 1258 1 8 3.427e+004 2009 3.5 8 1.734e+004 2414

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 4.486e+004 2552 -2.844 0.05 4.428e+004 2552 0.494 0.1 4.371e+004 2552 -0.2372 0.5 3.943e+004 2552 3.75 1 3.465e+004 2552 -0.4308 3.5 1.818e+004 2552 -0.9281 3 0 4.4e+004 2330 -2.075 0.05 4.379e+004 2330 1.135 0.1 4.351e+004 2330 -0.01066 0.5 4.047e+004 2330 2.842 1 3.607e+004 2330 -2.194 3.5 1.704e+004 2330 0.36334 0 4.486e+004 2552 -2.844 0.05 4.428e+004 2552 0.494 0.1 4.371e+004 2552 -0.2372 0.5 3.943e+004 2552 3.75 1 3.465e+004 2552 -0.4308 3.5 1.818e+004 2552 -0.9281 5 0 4.35e+004 1954 -1.759 0.05 4.35e+004 1954 1.767 0.1 4.35e+004 1954 -0.007437 0.5 4.281e+004 1954 -0.0007156 1 3.427e+004 1954 5.139e-005 3.5 1.734e+004 1954 -3.384e-006

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -384.3978 7 782.7956 A2 -381.053 12 786.106 A3 -384.3978 7 782.7956 R -465.1161 2 934.2321 2 -400.5375 3 807.075 3 -396.1729 4 800.3459 4 -400.5375 3 807.075 5 -387.7279 5 785.4558

Additive constant for all log-likelihoods = -44.11. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 168.1 10 < 0.0001 Test 2 6.69 5 0.2448 Test 3 6.69 5 0.2448 Test 4 32.28 4 < 0.0001 Test 5a 23.55 3 < 0.0001 Test 5b 8.729 1 0.003132 Test 6a 32.28 4 < 0.0001 Test 6b -1.137e-013 0 N/A Test 7a 6.66 2 0.03579 Test 7b 16.89 1 < 0.0001 Test 7c 25.62 2 < 0.0001

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is greater than .1. A homogeneous variance model appears to be appropriate here.

The p-value for Test 3 is greater than .1. The modeled variance appears to be appropriate here.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is less than .1. Model 3 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5b is less than .05. Model 3 appears to fit the data better than Model 2.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 6b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7a is less than .1. Model 5 may not adequately describe the data; you may want to consider another model.

The p-value for Test 7b is less than .05. Model 5 appears to fit the data better than Model 3.

The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviationConfidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.408357 0.374977 3 0.601713 0.48523 4 0.408357 0.372758 5 0.804549 0.692453

15000 20000 25000 30000 35000 40000 45000 50000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

09:45 08/20 2010 BMD BMDL Exponential15000 20000 25000 30000 35000 40000 45000 50000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

09:45 08/20 2010 BMD BMDL Exponential

15000 20000 25000 30000 35000 40000 45000 50000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

09:45 08/20 2010 BMD BMDL Exponential15000 20000 25000 30000 35000 40000 45000 50000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

09:45 08/20 2010 BMD BMDL Exponential

### Chlorpyrifos 91107 Repeat oral (Corn Oil) Brain M (PND) 8 time (hrs)



## BMDS Exponential MODEL variance Bmr = 10%

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d) Gnuplot Plotting File: Mon Aug 23 10:55:35 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) The variance is to be modeled as Var(i) = exp(lalpha + log(mean(i)) * rho)

Total number of dose groups = 6 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 6.42703 6.42703 6.42703 6.42703rho 0.823626 0.823626 0.823626 0.823626 a 25186.2 25186.2 45736.3 45736.3 b 0.332222 0.332222 0.352184 0.352184 c -- -- 0.0151936 0.0151936 d -- 1 -- 1

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 2.52917 3.6154 2.52917 10.209 rho 1.25748 1.148 1.25748 0.489746 a 44537.8 43863.5 44537.8 43166.6 b 0.330788 0.326994 0.330788 0.785613 c -- -- 0 0.321121 d -- 1.11249 -- 2.62367

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 8 4.326e+004 1500 0.05 8 4.356e+004 1981 0.1 8 4.266e+004 1576 0.5 8 4.078e+004 1865 1 8 3.101e+004 4154 3.5 8 1.39e+004 987.9

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 4.454e+004 2965 -1.215 0.05 4.381e+004 2934 -0.2399 0.1 4.309e+004 2904 -0.4206 0.5 3.775e+004 2672 3.209 1 3.199e+004 2408 -1.161 3.5 1.399e+004 1432 -0.1883 3 0 4.386e+004 2816 -0.6017 0.05 4.341e+004 2800 0.1455 0.1 4.29e+004 2780 -0.2455 0.5 3.839e+004 2609 2.594 1 3.288e+004 2387 -2.216 3.5 1.372e+004 1445 0.3418 4 0 4.454e+004 2965 -1.215 0.05 4.381e+004 2934 -0.23990.1 4.309e+004 2904 -0.4206 0.5 3.775e+004 2672 3.209 1 3.199e+004 2408 -1.161 3.5 1.399e+004 1432 -0.1883 5 0 4.317e+004 2248 0.1232 0.05 4.316e+004 2248 0.5004 0.1 4.313e+004 2248 -0.5949 0.5 4.075e+004 2217 0.04103 1 3.109e+004 2075 -0.1203 3.5 1.386e+004 1702 0.06025

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -391.2882 7 796.5765 A2 -381.0007 12 786.0015 A3 -391.1629 8 798.3258 R -470.3705 2 944.7409 2 -399.175 4 806.3501 3 -397.8473 5 805.6945 4 -399.175 4 806.3501 5 -391.4813 6 794.9625

Additive constant for all log-likelihoods = -44.11. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3)Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 178.7 10 < 0.0001 Test 2 20.58 5 0.0009743 Test 3 20.32 4 0.0004309 Test 4 16.02 4 0.002987 Test 5a 13.37 3 0.003903 Test 5b 2.656 1 0.1032 Test 6a 16.02 4 0.002987 Test 6b -1.137e-013 0 N/A Test 7a 0.6367 2 0.7273 Test 7b 12.73 1 0.0003594 Test 7c 15.39 2 0.0004556

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. A non-homogeneous variance model appears to be appropriate.

The p-value for Test 3 is less than .1. You may want to consider a different variance model.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is less than .1. Model 3 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5b is greater than .05. Model 3 does not seem to fit the data better than Model 2.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 6b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7a is greater than .1. Model 5 seems to adequately describe the data.

The p-value for Test 7b is less than .05. Model 5 appears to fit the data better than Model 3.

The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.318514 0.301226 3 0.404537 0.315934 4 0.318514 0.288611 5 0.632078 0.523231

10000 15000 20000 25000 30000 35000 40000 45000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

10:55 08/23 2010 BMD BMDL Exponential10000 15000 20000 25000 30000 35000 40000 45000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

10:55 08/23 2010 BMD BMDL Exponential

10000 15000 20000 25000 30000 35000 40000 45000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

10:55 08/23 2010 BMD BMDL Exponential10000 15000 20000 25000 30000 35000 40000 45000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

10:55 08/23 2010 BMD BMDL Exponential

### Chlorpyrifos 91107 Repeat oral (Corn Oil) RBC F (PND) 8 time (hrs) 

## BMDS Exponential MODEL variance Bmr = 10%

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d) Gnuplot Plotting File: Fri Aug 20 09:21:33 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) The variance is to be modeled as Var(i) = exp(lalpha + log(mean(i)) * rho)

Total number of dose groups = 6 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 1.24044 1.24044 1.24044 1.24044rho 1.5009 1.5009 1.5009 1.5009 a 2183.33 2183.33 6706.88 6706.88 b 0.618895 0.618895 0.695862 0.695862 c -- -- 0.02156 0.02156 d -- 1 -- 1

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 0.132479 0.132478 0.104344 0.175939 rho 1.63673 1.63673 1.64009 1.63095 a 6319.08 6319.08 6336.78 6231.54 b 0.618245 0.618245 0.635255 0.76121 c -- -- 0.00765865 0.0851656 d -- 1 -- 1.23628

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 8 5951 2082 0.05 8 6388 1021 0.1 8 5882 1282 0.5 8 4870 829 1 8 3333 821 3.5 8 723 264.2

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 6319 1377 -0.7558 0.05 6127 1343 0.5492 0.1 5940 1310 -0.1258 0.5 4639 1070 0.6114 1 3405 830.5 -0.2462 3.5 725.9 234.4 -0.0355 3 0 6319 1377 -0.7558 0.05 6127 1343 0.5492 0.1 5940 1310 -0.1258 0.5 4639 1070 0.6114 1 3405 830.5 -0.2462 3.5 725.9 234.4 -0.0355 4 0 6337 1382 -0.7898 0.05 6140 1346 0.51960.1 5950 1312 -0.146 0.5 4626 1067 0.6478 1 3380 825.2 -0.1613 3.5 729.2 234.6 -0.0745 5 0 6232 1357 -0.5847 0.05 6132 1339 0.5392 0.1 6000 1316 -0.2541 0.5 4742 1086 0.3344 1 3323 812.7 0.03417 3.5 729.1 235.9 -0.07263

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -360.5861 7 735.1722 A2 -346.6573 12 717.3147 A3 -349.8923 8 715.7846 R -394.7781 2 793.5561 2 -350.0992 4 708.1983 3 -350.0992 4 708.1983 4 -350.0925 5 710.185 5 -349.9949 6 711.9898

Additive constant for all log-likelihoods = -44.11. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3)Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 96.24 10 < 0.0001 Test 2 27.86 5 < 0.0001 Test 3 6.47 4 0.1667 Test 4 0.4137 4 0.9813 Test 5a 0.4137 4 0.9813 Test 5b 3.411e-013 0 N/A Test 6a 0.4004 3 0.9402 Test 6b 0.01335 1 0.908 Test 7a 0.2052 2 0.9025 Test 7b 0.2086 2 0.901 Test 7c 0.1952 1 0.6586

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. A non-homogeneous variance model appears to be appropriate.

The p-value for Test 3 is greater than .1. The modeled variance appears to be appropriate here.

The p-value for Test 4 is greater than .1. Model 2 seems to adequately describe the data.

The p-value for Test 5a is greater than .1. Model 3 seems to adequately describe the data.

Degrees of freedom for Test 5b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 6a is greater than .1. Model 4 seems to adequately describe the data.

The p-value for Test 6b is greater than .05. Model 4 does not seem to fit the data better than Model 2.

The p-value for Test 7a is greater than .1. Model 5 seems to adequately describe the data.

The p-value for Test 7b is greater than .05. Model 5 does not seem to fit the data better than Model 3.

The p-value for Test 7c is greater than .05. Model 5 does not seem to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.170419 0.154398 3 0.170419 0.154398 4 0.167206 0.129809 5 0.229627 0.130915

0 1000 2000 3000 4000 5000 6000 7000 8000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

09:21 08/20 2010 BMD BMDL Exponential0 1000 2000 3000 4000 5000 6000 7000 8000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

09:21 08/20 2010 BMD BMDL Exponential

0 1000 2000 3000 4000 5000 6000 7000 8000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

09:21 08/20 2010 BMD BMDL Exponential0 1000 2000 3000 4000 5000 6000 7000 8000

0 0.5 1 1.5 2 2.5 3 3.5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

09:21 08/20 2010 BMD BMDL Exponential

### 48139301 acute oral (milk) Brain F (PND) 8 0 26410.1 U/L 3026.3 8



## Wed Jan 26 11:55:42 2011

Redundancy Analysis

1 2 3 4 5 6 CondIndex 7.886 7.886 1.551 1.551 1.000 1.000 mu 0.184 0.184 0.933 0.933 1.448 1.448 lA.sexF 0.000 0.236 0.726 0.000 0.000 0.039 lA.sexM 0.236 0.000 0.000 0.726 0.039 0.000 lD.sexF 0.000 0.984 0.002 0.000 0.000 0.014 lD.sexM 0.984 0.000 0.000 0.002 0.014 0.000 lg.sexF 0.000 0.978 0.008 0.000 0.000 0.014 lg.sexM 0.978 0.000 0.000 0.008 0.014 0.000 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 0.001 0.000 -0.008 0.000 -0.006 0.000 lA.sexM 0.000 0.001 0.000 -0.008 0.000 -0.006 lD.sexF -0.008 0.000 0.174 0.000 0.144 0.000 lD.sexM 0.000 -0.008 0.000 0.174 0.000 0.144 lg.sexF -0.006 0.000 0.144 0.000 0.129 0.000 lg.sexM 0.000 -0.006 0.000 0.144 0.000 0.129 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 1.000 0.000 -0.496 0.000 -0.428 0.000 lA.sexM 0.000 1.000 0.000 -0.496 0.000 -0.428 lD.sexF -0.496 0.000 1.000 0.000 0.963 0.000 lD.sexM 0.000 -0.496 0.000 1.000 0.000 0.963 lg.sexF -0.428 0.000 0.963 0.000 1.000 0.000 lg.sexM 0.000 -0.428 0.000 0.963 0.000 1.000

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.595 0.038 lA.sexM 1.595 0.038 lD.sexF 0.412 0.417 lD.sexM 0.412 0.417 lg.sexF 0.552 0.359 lg.sexM 0.552 0.359

---- Mean Absolute Value of Gradient Val

## lA.sexF 2.1599569 lA.sexM 2.1599569 lD.sexF 0.3735085 lD.sexM 0.3735086 lg.sexF 0.3531427 lg.sexM 0.3531427

Variance is heterogeneous ::: Power parameter fit

Intercept estimate (standard error): 4.929073 (0.09827216) BMD10 estimate (standard error) : 1.529981 (0.4643814)

Power parameter estimate (s tandard error): 1.767668 (0.521671) The confidence intervals for the variables are given by lower est. upper Intercept 4.7400654 4.929073 5.125617

## BMD 0.9520477 1.529981 2.458744 Power Parameter 1.1115758 1.767668 2.811010

BMDL 1.028805

### 91107 acute oral (gavage) brain M (PND) 8 

## Exponential model Constant variance Bmr 10 Risk – relative

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d) Gnuplot Plotting File: Thu Sep 09 12:37:49 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) rho is set to 0. A constant variance model is fit.

Total number of dose groups = 6 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 15.5873 15.5873 15.5873 15.5873rho(S) 0 0 0 0 a 17425.5 17425.5 27185.6 27185.6 b 0.152906 0.152906 0.167351 0.167351 c -- -- 0.000410619 0.000410619 d -- 1 -- 1

(S) = Specified

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 16.0692 15.7341 16.0692 15.7311 rho 0 0 0 0 a 25353.3 24553.9 25353.3 24549.1 b 0.131784 0.180912 0.131784 0.438954 c -- -- 0 0.454717 d -- 2.35967 -- 13.1747

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 8 2.276e+004 3336 0.05 8 2.589e+004 2664 0.1 8 2.434e+004 1508 0.5 8 2.52e+004 1847 2 8 2.235e+004 2910 5 8 1.116e+004 2821

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 2.535e+004 3086 -2.373 0.05 2.519e+004 3086 0.6455 0.1 2.502e+004 3086 -0.6255 0.5 2.374e+004 3086 1.343 2 1.948e+004 3086 2.627 5 1.312e+004 3086 -1.792 3 0 2.455e+004 2610 -1.939 0.05 2.455e+004 2610 1.449 0.1 2.455e+004 2610 -0.231 0.5 2.447e+004 2610 0.7938 2 2.242e+004 2610 -0.08304 5 1.115e+004 2610 0.011594 0 2.535e+004 3086 -2.373 0.05 2.519e+004 3086 0.6455 0.1 2.502e+004 3086 -0.6255 0.5 2.374e+004 3086 1.343 2 1.948e+004 3086 2.627 5 1.312e+004 3086 -1.792 5 0 2.455e+004 2606 -1.937 0.05 2.455e+004 2606 1.456 0.1 2.455e+004 2606 -0.2282 0.5 2.455e+004 2606 0.7084 2 2.235e+004 2606 -7.294e-009 5 1.116e+004 2606 2.892e-009

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -398.0957 7 810.1915 A2 -394.8779 12 813.7557 A3 -398.0957 7 810.1915 R -437.7654 2 879.5308 2 -409.6608 3 825.3217 3 -401.6175 4 811.235 4 -409.6608 3 825.3217 5 -401.5452 5 813.0904

Additive constant for all log-likelihoods = -44.11. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 85.78 10 < 0.0001 Test 2 6.436 5 0.2661 Test 3 6.436 5 0.2661 Test 4 23.13 4 0.0001193 Test 5a 7.044 3 0.07052 Test 5b 16.09 1 < 0.0001 Test 6a 23.13 4 0.0001193 Test 6b 0 0 N/A Test 7a 6.899 2 0.03176 Test 7b 0.1446 1 0.7038 Test 7c 16.23 2 0.0002988

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is greater than .1. A homogeneous variance model appears to be appropriate here.

The p-value for Test 3 is greater than .1. The modeled variance appears to be appropriate here.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is less than .1. Model 3 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5b is less than .05. Model 3 appears to fit the data better than Model 2.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 6b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7a is less than .1. Model 5 may not adequately describe the data; you may want to consider another model.

The p-value for Test 7b is greater than .05. Model 5 does not seem to fit the data better than Model 3.

The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviationConfidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.799495 0.665064 3 2.12988 1.52673 4 0.799495 0.664911 5 2.01815 1.5418

10000 15000 20000 25000 30000

0 1 2 3 4 5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

12:37 09/09 2010 BMD BMDL Exponential10000 15000 20000 25000 30000

0 1 2 3 4 5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

12:37 09/09 2010 BMD BMDL Exponential

10000 15000 20000 25000 30000

0 1 2 3 4 5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

12:37 09/09 2010 BMD BMDL Exponential10000 15000 20000 25000 30000

0 1 2 3 4 5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

12:37 09/09 2010 BMD BMDL Exponential

### Chlorpyrifos 48139301 acute oral (Diet) RBC F (ADLT) 8

0 5337 U/L 304.8 8 RAT



## R based Acute diet Female Rbc

Mon Jan 31 10:28:21 2011

Redundancy Analysis

1 2 3 4 5 6 CondIndex 27.936 27.936 1.526 1.526 1.000 1.000 mu 0.052 0.052 0.949 0.949 1.448 1.448 lA.sexF 0.260 0.000 0.000 0.712 0.028 0.000 lA.sexM 0.000 0.260 0.712 0.000 0.000 0.028 lD.sexF 0.999 0.000 0.000 0.000 0.001 0.000 lD.sexM 0.000 0.999 0.000 0.000 0.000 0.001 lg.sexF 0.999 0.000 0.000 0.000 0.001 0.000 lg.sexM 0.000 0.999 0.000 0.000 0.000 0.001 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 0.002 0.000 -0.022 0.000 -0.030 0.000 lA.sexM 0.000 0.002 0.000 -0.022 0.000 -0.030 lD.sexF -0.022 0.000 1.004 0.000 1.427 0.000 lD.sexM 0.000 -0.022 0.000 1.004 0.000 1.427 lg.sexF -0.030 0.000 1.427 0.000 2.041 0.000 lg.sexM 0.000 -0.030 0.000 1.427 0.000 2.041 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 1.000 0.000 -0.516 0.000 -0.500 0.000 lA.sexM 0.000 1.000 0.000 -0.516 0.000 -0.500 lD.sexF -0.516 0.000 1.000 0.000 0.997 0.000 lD.sexM 0.000 -0.516 0.000 1.000 0.000 0.997 lg.sexF -0.500 0.000 0.997 0.000 1.000 0.000 lg.sexM 0.000 -0.500 0.000 0.997 0.000 1.000

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.594 0.042 lA.sexM 1.594 0.042 lD.sexF -0.001 1.002

## lD.sexM -0.001 1.002 lg.sexF 1.027 1.429 lg.sexM 1.027 1.429

---- Mean Absolute Value of Gradient Val lA.sexF 1.8318370 lA.sexM 1.8318370 lD.sexF 0.4212766 lD.sexM 0.4212766 lg.sexF 0.2934787 lg.sexM 0.2934787

Variance is heterogeneous ::: Power parameter fit

## Intercept estimate (standard error): 4.976937 (0.09421813) BMD10 estimate (standard error) : 0.6631076 (0.08758324)

Power parameter estimate (s tandard error): 1.744374 (0.1404777) The confidence intervals for the variables are given by lower est. upper Intercept 4.7953746 4.9769365 5.1653728 BMD 0.5245441 0.6631076 0.8382741 Power Parameter 1.5023927 1.7443738 2.0253292

BMDL 0.5450309

>

### Chlorpyrifos 48139301 Repeat oral (Corn Oil) RBC F (ADLT) 8 0 4903.3 U/L 275.8 8 RAT 0.05 4654.9 U/L 712.7 8 RAT

0.1 4119.3 U/L 488.7 8 RAT 0.5 3946 U/L 722.4 8 RAT

1 1334.8 U/L 395.1 8 RAT 3.5 134.5 U/L 184.9 8 RAT



## Mon Jan 31 13:39:43 2011

Redundancy Analysis

1 2 3 4 5 6 CondIndex 7.750 7.750 1.505 1.505 1.000 1.000 mu 0.185 0.185 0.953 0.953 1.434 1.434 lA.sexF 0.000 0.201 0.766 0.000 0.000 0.032 lA.sexM 0.201 0.000 0.000 0.766 0.032 0.000 lD.sexF 0.000 0.984 0.001 0.000 0.000 0.015 lD.sexM 0.984 0.000 0.000 0.001 0.015 0.000 lg.sexF 0.000 0.979 0.006 0.000 0.000 0.015 lg.sexM 0.979 0.000 0.000 0.006 0.015 0.000 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 0.002 0.000 -0.005 0.000 -0.006 0.000 lA.sexM 0.000 0.002 0.000 -0.005 0.000 -0.006 lD.sexF -0.005 0.000 0.065 0.000 0.080 0.000 lD.sexM 0.000 -0.005 0.000 0.065 0.000 0.080 lg.sexF -0.006 0.000 0.080 0.000 0.106 0.000 lg.sexM 0.000 -0.006 0.000 0.080 0.000 0.106 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 1.000 0.000 -0.456 0.000 -0.399 0.000 lA.sexM 0.000 1.000 0.000 -0.456 0.000 -0.399 lD.sexF -0.456 0.000 1.000 0.000 0.963 0.000 lD.sexM 0.000 -0.456 0.000 1.000 0.000 0.963 lg.sexF -0.399 0.000 0.963 0.000 1.000 0.000 lg.sexM 0.000 -0.399 0.000 0.963 0.000 1.000

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.538 0.044 lA.sexM 1.538 0.044

## lD.sexF -0.806 0.255 lD.sexM -0.806 0.255 lg.sexF 1.114 0.325 lg.sexM 1.114 0.325

---- Mean Absolute Value of Gradient Val lA.sexF 1.6112950 lA.sexM 1.6112950 lD.sexF 0.5774669 lD.sexM 0.5774669 lg.sexF 0.3618618 lg.sexM 0.3618618

## Variance is heterogeneous ::: Power parameter fit

Intercept estimate (standard error): 4.654547 (0.1300956) BMD10 estimate (standard error) : 0.4465991 (0.08046465)

Power parameter estimate (s tandard error): 3.047571 (0.722021) The confidence intervals for the variables are given by lower est. upper Intercept 4.4081308 4.6545473 4.9147385 BMD 0.3282987 0.4465991 0.6075284 Power Parameter 2.0677843 3.0475705 4.4916126

BMDL 0.3452339

>

### Chlorpyrifos 48139301 Repeat oral (Corn Oil) Brain F (PND) 8

0 42289.4 U/L 1615.9 8 0.05 44725.8 U/L 2598 8

0.1 43498.6 U/L 1409 8 0.5 42809.6 U/L 1257.7 8

1 34265.8 U/L 2009.1 8 3.5 17344.1 U/L 2413.8 8



## R based Repeat gavage Female pup Brain

Mon Jan 31 11:47:37 2011

Redundancy Analysis

1 2 3 4 5 6 CondIndex 7.416 7.416 1.565 1.565 1.000 1.000 mu 0.196 0.196 0.927 0.927 1.450 1.450 lA.sexF 0.000 0.432 0.000 0.533 0.035 0.000 lA.sexM 0.432 0.000 0.533 0.000 0.000 0.035 lD.sexF 0.000 0.983 0.000 0.001 0.016 0.000 lD.sexM 0.983 0.000 0.001 0.000 0.000 0.016 lg.sexF 0.000 0.970 0.000 0.014 0.016 0.000 lg.sexM 0.970 0.000 0.014 0.000 0.000 0.016 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 0.002 0.000 -0.016 0.000 -0.008 0.000 lA.sexM 0.000 0.002 0.000 -0.016 0.000 -0.008 lD.sexF -0.016 0.000 0.276 0.000 0.154 0.000 lD.sexM 0.000 -0.016 0.000 0.276 0.000 0.154 lg.sexF -0.008 0.000 0.154 0.000 0.094 0.000 lg.sexM 0.000 -0.008 0.000 0.154 0.000 0.094 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 1.000 0.000 -0.657 0.000 -0.584 0.000 lA.sexM 0.000 1.000 0.000 -0.657 0.000 -0.584 lD.sexF -0.657 0.000 1.000 0.000 0.956 0.000 lD.sexM 0.000 -0.657 0.000 1.000 0.000 0.956 lg.sexF -0.584 0.000 0.956 0.000 1.000 0.000 lg.sexM 0.000 -0.584 0.000 0.956 0.000 1.000

---- Parameter Values and Relative Standard Errors (sigma == 1)

## Parms Relative SE lA.sexF 1.593 0.045 lA.sexM 1.593 0.045 lD.sexF -0.508 0.525 lD.sexM -0.508 0.525 lg.sexF 0.222 0.306 lg.sexM 0.222 0.306

---- Mean Absolute Value of Gradient Val lA.sexF 2.0950189 lA.sexM 2.0950189 lD.sexF 0.3185851 lD.sexM 0.3185851 lg.sexF 0.3966182 lg.sexM 0.3966182

## Variance is homogeneous ::: Power parameter fit

Intercept estimate (standard error): 4.918671 (0.06002615) BMD10 estimate (standard error) : 0.601496 (0.09747683)

Power parameter estimate (s tandard error): 1.248135 (0.1109397) The confidence intervals for the variables are given by lower est. upper Intercept 4.8014791 4.918671 5.0387233 BMD 0.4542628 0.601496 0.7964496 Power Parameter 1.0597454 1.248135 1.4700145

BMDL 0.4755929

### Chlorpyrifos 48139301 Repeat oral (Corn Oil) RBC F (PND) 8

0 5951 U/L 2082 8 RAT



## R based Repeat gavage Female pup Rbc

Mon Jan 31 11:40:32 2011

Redundancy Analysis

1 2 3 4 5 6 CondIndex 6.409 6.409 1.649 1.649 1.000 1.000 mu 0.229 0.229 0.890 0.890 1.468 1.468 lA.sexF 0.406 0.000 0.549 0.000 0.000 0.044 lA.sexM 0.000 0.406 0.000 0.549 0.044 0.000 lD.sexF 0.978 0.000 0.003 0.000 0.000 0.020 lD.sexM 0.000 0.978 0.000 0.003 0.020 0.000 lg.sexF 0.951 0.000 0.028 0.000 0.000 0.021 lg.sexM 0.000 0.951 0.000 0.028 0.021 0.000 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 0.003 0.000 -0.015 0.000 -0.006 0.000 lA.sexM 0.000 0.003 0.000 -0.015 0.000 -0.006 lD.sexF -0.015 0.000 0.223 0.000 0.102 0.000 lD.sexM 0.000 -0.015 0.000 0.223 0.000 0.102 lg.sexF -0.006 0.000 0.102 0.000 0.054 0.000 lg.sexM 0.000 -0.006 0.000 0.102 0.000 0.054 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 1.000 0.000 -0.639 0.000 -0.528 0.000 lA.sexM 0.000 1.000 0.000 -0.639 0.000 -0.528 lD.sexF -0.639 0.000 1.000 0.000 0.935 0.000 lD.sexM 0.000 -0.639 0.000 1.000 0.000 0.935 lg.sexF -0.528 0.000 0.935 0.000 1.000 0.000 lg.sexM 0.000 -0.528 0.000 0.935 0.000 1.000

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.581 0.050 lA.sexM 1.581 0.050 lD.sexF -1.532 0.472 lD.sexM -1.532 0.472

## lg.sexF 0.103 0.232 lg.sexM 0.103 0.232

---- Mean Absolute Value of Gradient Val lA.sexF 1.7672291 lA.sexM 1.7672291 lD.sexF 0.3683798 lD.sexM 0.3683798 lg.sexF 0.6145329 lg.sexM 0.6145329

Variance is heterogeneous ::: Power parameter set to 1

Intercept estimate (standard error): 4.947321 (0.2076523)

## BMD10 estimate (standard error) : 0.1701286 (0.01035339)

The confidence intervals for the variables are given by lower est. upper Intercept 4.5641790 4.9473206 5.3626253 BMD 0.1516686 0.1701286 0.1908353

BMDL 0.1545417

### Chlorpyrifos 48139301 Repeat oral (Corn Oil) Brain M (PND) 8

0 43264.5 U/L 1500.4 8 RAT 0.05 43558.4 U/L 1981.1 8 RAT

0.1 42656.8 U/L 1575.7 8 RAT 0.5 40779.8 U/L 1864.7 8 RAT

1 31005.9 U/L 4154.3 8 RAT 3.5 13898 U/L 987.9 8 RAT



## R based Repeat gavage Male pup Brain

Mon Jan 31 11:58:46 2011

Redundancy Analysis

1 2 3 4 5 6 CondIndex 7.047 7.047 1.590 1.590 1.000 1.000 mu 0.207 0.207 0.915 0.915 1.456 1.456 lA.sexF 0.448 0.000 0.515 0.000 0.000 0.037 lA.sexM 0.000 0.448 0.000 0.515 0.037 0.000 lD.sexF 0.981 0.000 0.002 0.000 0.000 0.017 lD.sexM 0.000 0.981 0.000 0.002 0.017 0.000 lg.sexF 0.964 0.000 0.019 0.000 0.000 0.018 lg.sexM 0.000 0.964 0.000 0.019 0.018 0.000 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 0.002 0.000 -0.016 0.000 -0.007 0.000 lA.sexM 0.000 0.002 0.000 -0.016 0.000 -0.007 lD.sexF -0.016 0.000 0.256 0.000 0.122 0.000 lD.sexM 0.000 -0.016 0.000 0.256 0.000 0.122 lg.sexF -0.007 0.000 0.122 0.000 0.064 0.000 lg.sexM 0.000 -0.007 0.000 0.122 0.000 0.064 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 1.000 0.000 -0.668 0.000 -0.584 0.000 lA.sexM 0.000 1.000 0.000 -0.668 0.000 -0.584 lD.sexF -0.668 0.000 1.000 0.000 0.949 0.000 lD.sexM 0.000 -0.668 0.000 1.000 0.000 0.949 lg.sexF -0.584 0.000 0.949 0.000 1.000 0.000 lg.sexM 0.000 -0.584 0.000 0.949 0.000 1.000

## ---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.617 0.046 lA.sexM 1.617 0.046 lD.sexF -0.894 0.506 lD.sexM -0.894 0.506 lg.sexF 0.118 0.253 lg.sexM 0.118 0.253

---- Mean Absolute Value of Gradient Val lA.sexF 2.0573971 lA.sexM 2.0573971 lD.sexF 0.3422228 lD.sexM 0.3422228 lg.sexF 0.4932134 lg.sexM 0.4932134

Variance is heterogeneous ::: Power parameter set to 1

Intercept estimate (standard error): 5.112553 (0.06391137) BMD10 estimate (standard error) : 0.3182407 (0.01109938)

The confidence intervals for the variables are given by lower est. upper Intercept 4.9878482 5.1125531 5.2403758 BMD 0.2974895 0.3182407 0.3404395

BMDL 0.3007851

>

### Chlorpyrifos 48139301 Repeat oral (Corn Oil) RBC M (PND) 8

0 6410 U/L 1605 8 RAT 0.05 5516.5 U/L 1179.7 8 RAT

0.1 5449.8 U/L 1293.6 8 RAT 0.5 4054 U/L 1111.8 8 RAT

1 2478 U/L 532.5 8 RAT 3.5 539.8 U/L 332.2 8 RAT



## R based Repeat gavage Male pup Rbc

Mon Jan 31 12:06:40 2011

Redundancy Analysis

1 2 3 4 5 6 CondIndex 8.508 8.508 1.773 1.773 1.000 1.000 mu 0.176 0.176 0.846 0.846 1.501 1.501 lA.sexF 0.507 0.000 0.455 0.000 0.000 0.038 lA.sexM 0.000 0.507 0.000 0.455 0.038 0.000 lD.sexF 0.987 0.000 0.002 0.000 0.000 0.011 lD.sexM 0.000 0.987 0.000 0.002 0.011 0.000 lg.sexF 0.967 0.000 0.021 0.000 0.000 0.012 lg.sexM 0.000 0.967 0.000 0.021 0.012 0.000 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 0.004 0.000 -0.031 0.000 -0.009 0.000 lA.sexM 0.000 0.004 0.000 -0.031 0.000 -0.009 lD.sexF -0.031 0.000 0.488 0.000 0.161 0.000 lD.sexM 0.000 -0.031 0.000 0.488 0.000 0.161 lg.sexF -0.009 0.000 0.161 0.000 0.058 0.000 lg.sexM 0.000 -0.009 0.000 0.161 0.000 0.058 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 1.000 0.000 -0.720 0.000 -0.624 0.000 lA.sexM 0.000 1.000 0.000 -0.720 0.000 -0.624 lD.sexF -0.720 0.000 1.000 0.000 0.959 0.000 lD.sexM 0.000 -0.720 0.000 1.000 0.000 0.959 lg.sexF -0.624 0.000 0.959 0.000 1.000 0.000 lg.sexM 0.000 -0.624 0.000 0.959 0.000 1.000

---- Parameter Values and Relative Standard Errors (sigma == 1)

## Parms Relative SE lA.sexF 1.584 0.062 lA.sexM 1.584 0.062 lD.sexF -2.605 0.699 lD.sexM -2.605 0.699 lg.sexF -0.210 0.240 lg.sexM -0.210 0.240

---- Mean Absolute Value of Gradient Val lA.sexF 1.590903 lA.sexM 1.590903 lD.sexF 0.352936 lD.sexM 0.352936 lg.sexF 0.798925 lg.sexM 0.798925

## 012340 2000 4000 6000 

# Male dose-respose curve for chlorpyrifos

doseResponse



## BMD BMDL



## Variance is heterogeneous ::: Power parameter fit

Intercept estimate (standard error): 4.865244 (0.3004799) BMD10 estimate (standard error) : 0.07694372 (0.05494676)

Power parameter estimate (s tandard error): 0.825402 (0.1141079) The confidence intervals for the variables are given by lower est. upper Intercept 4.33035411 4.86524372 5.4662034 BMD 0.03168713 0.07694372 0.1868372 Power Parameter 0.64661820 0.82540200 1.0536178

BMDL 0.03663169

>

## Data: ChEDataDNTdams filename: dow1 che dam gd20 heart simple (SAP )

Parameter: Heart dams GD Day 20 (Dow study)

Model: R progr am – Single Sex

BMR: 10%

Power parameter:

Female Dose-Response Curve for Chlorpyrifos

Tue May 05 10:35:40 2009

Redundancy Analysis

1 2 3 4 5 6 CondIndex 6.477 6.477 1.790 1.790 1.000 1.000 mu 0.231 0.231 0.837 0.837 1.499 1.499 lA.sexF 0.388 0.000 0.563 0.000 0.000 0.049 lA.sexM 0.000 0.388 0.000 0.563 0.049 0.000 lD.sexF 0.978 0.000 0.004 0.000 0.000 0.018

## lD.sexM 0.000 0.978 0.000 0.004 0.018 0.000 lg.sexF 0.942 0.000 0.038 0.000 0.000 0.020 lg.sexM 0.000 0.942 0.000 0.038 0.020 0.000

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.654 0.082 lA.sexM 1.654 0.082 lD.sexF -1.792 0.732 lD.sexM -1.792 0.732 lg.sexF -0.102 0.278 lg.sexM -0.102 0.278

---- Mean Absolute Value of Gradient Val lA.sexF 1.6518257 lA.sexM 1.6518257 lD.sexF 0.4108376 lD.sexM 0.4108376 lg.sexF 0.8517210 lg.sexM 0.8517209

## Variance is heterogeneous ::: Power parameter set to 1

Intercept estimate (standard error): 4.802256 (0.3618797) BMD10 estimate (standard error) : 0.2943337 (0.02629857)

The confidence intervals for the variables are given by lower est. upper Intercept 4.1732895 4.8022559 5.5260154 BMD 0.2497107 0.2943337 0.3469309

BMDL 0.2566849

>

## Data: ChEDataDNTdams filename: dow1 che dams gd20 hindbrain 10bmr simple (SAP )

Parameter: Hindbrain dams GD Day 20

Model: R progr am – Single Sex

BMR: 10%

Power parameter:

Female Dose-Response Curve for Chlorpyrifos

## Mon May 04 08:50:13 2009

Redundancy Analysis

1 2 3 4 5 6 CondIndex 10.510 10.510 1.475 1.475 1.000 1.000 mu 0.136 0.136 0.969 0.969 1.429 1.429 lA.sexF 0.000 0.446 0.535 0.000 0.000 0.018 lA.sexM 0.446 0.000 0.000 0.535 0.018 0.000 lD.sexF 0.000 0.991 0.000 0.000 0.000 0.008 lD.sexM 0.991 0.000 0.000 0.000 0.008 0.000 lg.sexF 0.000 0.988 0.003 0.000 0.000 0.009 lg.sexM 0.988 0.000 0.000 0.003 0.009 0.000

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.608 0.071 lA.sexM 1.608 0.071 lD.sexF 0.095 0.779 lD.sexM 0.095 0.779 lg.sexF 0.546 0.512 lg.sexM 0.546 0.512

---- Mean Absolute Value of Gradient Val lA.sexF 1.9609364 lA.sexM 1.9609364 lD.sexF 0.4676926 lD.sexM 0.4676926 lg.sexF 0.5806155 lg.sexM 0.5806155

## Variance is homogeneous ::: Power parameter fit

Intercept estimate (standard error): 4.995013 (0.1009121) BMD10 estimate (standard error) : 1.099816 (0.2976588)

Power parameter estimate (s tandard error): 1.72704 (0.283509) The confidence intervals for the variables are given by lower est. upper Intercept 4.8010271 4.995013 5.196837 BMD 0.7135696 1.099816 1.695134 Power Parameter 1.3001708 1.727040 2.294058

BMDL 0.7673472

## Chemical: Chlorpyrifos (6/9/09)

Data: ChEDataDNTdams (original data) in C:\Documents and Settings\jliccion\My Documents\Jliccion\Benchmark dose fold er\Chlorpyrifos SAP BMD reruns 2009\Runs Dow1 study dam ChE DNT GD20 & LD 1 sap 4 & 5 & 11 & 12\original

Run file: chedamsGD20RBCsimple in C:\Doc uments and Settings\jliccion\My Documents\Jliccion\Benchmark dose fold er\Chlorpyrifos SAP BMD reruns 2009\Runs Dow1 study dam ChE DNT GD20 & LD1 sap 4 & 5 & 11 & 12\Runs

Parameter: Dams ChE RBC GD20

Model: R progr am – Single Sex

BMR: 10%

parameter



# Female Dose-Response Cu rve to Chlorpyrifos

## Tue Jun 09 13:23:30 2009

Redundancy Analysis

1 2 3 4 5 6 CondIndex 6.548 6.548 1.639 1.639 1.000 1.000 mu 0.224 0.224 0.895 0.895 1.466 1.466 lA.sexF 0.000 0.244 0.000 0.705 0.050 0.000 lA.sexM 0.244 0.000 0.705 0.000 0.000 0.050 lD.sexF 0.000 0.977 0.000 0.004 0.019 0.000 lD.sexM 0.977 0.000 0.004 0.000 0.000 0.019 lg.sexF 0.000 0.962 0.000 0.019 0.020 0.000 lg.sexM 0.962 0.000 0.019 0.000 0.000 0.020

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.609 0.089 lA.sexM 1.609 0.089 lD.sexF -1.938 0.598 lD.sexM -1.938 0.598 lg.sexF 0.370 0.346 lg.sexM 0.370 0.346

---- Mean Absolute Value of Gradient Val lA.sexF 1.2727391 lA.sexM 1.2727391 lD.sexF 0.4510209 lD.sexM 0.4510209 lg.sexF 0.6437339 lg.sexM 0.6437339

## Variance is heterogeneous ::: Power parameter fit

Intercept estimate (standard error): 5.000469 (0.2984647) BMD10 estimate (standard error) : 0.1439830 (0.077596)

Power parameter estimate (s tandard error): 1.447323 (0.3658707) The confidence intervals for the variables are given by lower est. upper Intercept 4.4671984 5.0004688 5.5973981 BMD 0.0692942 0.1439830 0.2991750 Power Parameter 0.9613044 1.4473232 2.1790646

BMDL 0.07844108

## >

### chlorpyrifos 44648101 DNT 870.6 forebrain F GD20 D



## Exponential Constant var 10 bmr

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Testrunscontinuous_Setting.(d) Gnuplot Plotting File: Thu Nov 04 09:45:13 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) rho is set to 0. A constant variance model is fit.

Total number of dose groups = 4 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 8.5137 8.5137 8.5137 8.5137rho(S) 0 0 0 0 a 704.396 704.396 2448.08 2448.08 b 0.423305 0.423305 0.402297 0.402297 c -- -- 0.000126426 0.000126426 d -- 1 -- 1

(S) = Specified

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 10.8009 8.53737 10.8009 8.53737 rho 0 0 0 0 a 2512.07 2317.51 2512.07 2317.51 b 0.312609 0.282322 0.312609 0.282322 c -- -- 0 0 d -- 2.02847 -- 2.02847

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 5 2332 85.8 0.3 5 2286 62.7 1 5 2149 85.9 5 5 309.5 79

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 2512 221.5 -1.823 0.3 2287 221.5 -0.0171 1 1838 221.5 3.14 5 526.3 221.5 -2.188 3 0 2318 71.43 0.438 0.3 2302 71.43 -0.5185 1 2146 71.43 0.08456 5 309.8 71.43 -0.009577 4 0 2512 221.5 -1.823 0.3 2287 221.5 -0.0171 1 1838 221.5 3.14 5 526.3 221.5 -2.188 5 0 2318 71.43 0.438 0.3 2302 71.43 -0.51851 2146 71.43 0.08456 5 309.8 71.43 -0.009577

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -95.13697 5 200.2739 A2 -94.83363 8 205.6673 A3 -95.13697 5 200.2739 R -144.8611 2 293.7222 2 -118.0094 3 242.0189 3 -95.37373 4 198.7475 4 -118.0094 3 242.0189 5 -95.37373 4 198.7475

Additive constant for all log-likelihoods = -18.38. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 100.1 6 < 0.0001Test 2 0.6067 3 0.8949 Test 3 0.6067 3 0.8949 Test 4 45.74 2 < 0.0001 Test 5a 0.4735 1 0.4914 Test 5b 45.27 1 < 0.0001 Test 6a 45.74 2 < 0.0001 Test 6b -2.842e-014 0 N/A Test 7a 0.4735 1 0.4914 Test 7b 0 0 N/A Test 7c 45.27 1 < 0.0001

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is greater than .1. A homogeneous variance model appears to be appropriate here.

The p-value for Test 3 is greater than .1. The modeled variance appears to be appropriate here.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is greater than .1. Model 3 seems to adequately describe the data.

The p-value for Test 5b is less than .05. Model 3 appears to fit the data better than Model 2.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 6b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7a is greater than .1. Model 5 seems to adequately describe the data.

Degrees of freedom for Test 7b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.337036 0.285382 3 1.16802 0.975837 4 0.337036 0.2853825 1.16802 0.975837

0 500 1000 1500 2000 2500

0 1 2 3 4 5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

08:45 11/04 2010 BMD BMDL Exponential0 500 1000 1500 2000 2500

0 1 2 3 4 5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

08:45 11/04 2010 BMD BMDL Exponential

0 500 1000 1500 2000 2500

0 1 2 3 4 5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

08:45 11/04 2010 BMD BMDL Exponential0 500 1000 1500 2000 2500

0 1 2 3 4 5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

08:45 11/04 2010 BMD BMDL Exponential

### chlorpyrifos 44648101 DNT 870.6 heart F GD20 D



## Dow1 dnt study

Exponential Constant var 10 bmr

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Testrunscontinuous_Setting.(d) Gnuplot Plotting File: Tue Nov 09 09:54:50 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) rho is set to 0. A constant variance model is fit.

Total number of dose groups = 4 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter ValuesVariable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 8.05686 8.05686 8.05686 8.05686 rho(S) 0 0 0 0 a 139.151 139.151 471.765 471.765 b 0.353526 0.353526 0.939432 0.939432 c -- -- 0.152416 0.152416 d -- 1 -- 1

(S) = Specified

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 8.49206 8.49206 8.38143 8.05695 rho 0 0 0 0 a 466.093 466.093 478.237 448.55 b 0.522439 0.522439 0.751646 0.990444 c -- -- 0.128482 0.16832 d -- 1 -- 12.2926

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 5 447.8 78.5 0.3 5 449.3 90.2 1 5 228.9 33.2 5 5 75.5 19.4

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 466.1 69.83 -0.5858 0.3 398.5 69.83 1.627 1 276.4 69.83 -1.522 5 34.2 69.83 1.323 3 0 466.1 69.83 -0.5858 0.3 398.5 69.83 1.627 1 276.4 69.83 -1.5225 34.2 69.83 1.323 4 0 478.2 66.07 -1.03 0.3 394.1 66.07 1.868 1 258 66.07 -0.9849 5 71.17 66.07 0.1467 5 0 448.6 56.18 -0.02986 0.3 448.5 56.18 0.02986 1 228.9 56.18 -2.243e-008 5 75.5 56.18 5.71e-009

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -90.56864 5 191.1373 A2 -84.43332 8 184.8666 A3 -90.56864 5 191.1373 R -112.4185 2 228.837 2 -94.92058 3 195.8412 3 -94.92058 3 195.8412 4 -93.8143 4 195.6286 5 -90.56953 5 191.1391

Additive constant for all log-likelihoods = -18.38. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 55.97 6 < 0.0001 Test 2 12.27 3 0.006511 Test 3 12.27 3 0.006511 Test 4 8.704 2 0.01288 Test 5a 8.704 2 0.01288 Test 5b -3.979e-013 0 N/A Test 6a 6.491 1 0.01084 Test 6b 2.213 1 0.1369 Test 7a 0.001783 0 N/A Test 7b 8.702 2 0.01289 Test 7c 6.49 1 0.01085

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. Consider running a non-homogeneous variance model.

The p-value for Test 3 is less than .1. You may want to consider a different variance model.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is less than .1. Model 3 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 5b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

The p-value for Test 6b is greater than .05. Model 4 does not seem to fit the data better than Model 2.

Degrees of freedom for Test 7a are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7b is less than .05. Model 5 appears to fit the data better than Model 3.

The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.20167 0.13615 3 0.20167 0.13615 4 0.162146 0.115278 5 0.854222 0.257375

0 100 200 300 400 500 600

0 1 2 3 4 5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

09:54 11/09 2010 BMD BMDL Exponential0 100 200 300 400 500 600

0 1 2 3 4 5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

09:54 11/09 2010 BMD BMDL Exponential

0 100 200 300 400 500 600

0 1 2 3 4 5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

09:54 11/09 2010 BMD BMDL Exponential0 100 200 300 400 500 600

0 1 2 3 4 5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

09:54 11/09 2010 BMD BMDL Exponential

### chlorpyrifos 44648101 DNT 870.6 heart F GD20 D



## Dow1 dnt study

Exponential Model var 10 bmr

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Testrunscontinuous_Setting.(d) Gnuplot Plotting File: Tue Nov 09 09:58:22 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) The variance is to be modeled as Var(i) = exp(lalpha + log(mean(i)) * rho)

Total number of dose groups = 4 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5-------- ------- ------- ------- --

----- lnalpha -1.4951 -1.4951 -1.4951 -1.4951 rho 1.66806 1.66806 1.66806 1.66806 a 139.151 139.151 471.765 471.765 b 0.353526 0.353526 0.939432 0.939432 c -- -- 0.152416 0.152416 d -- 1 -- 1

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha -2.28051 -2.28051 -2.60713 - 1.693 rho 1.89039 1.89039 1.89043 1.67257 a 431.455 431.455 497.51 449.824 b 0.357413 0.357413 0.936583 0.992581 c -- -- 0.144131 0.168922 d -- 1 -- 12.4899

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 5 447.8 78.5 0.3 5 449.3 90.2 1 5 228.9 33.2 5 5 75.5 19.4

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 431.5 98.93 0.3695 0.3 387.6 89.39 1.544 1 301.8 70.57 -2.31 5 72.25 18.27 0.3981 3 0 431.5 98.93 0.3695 0.3 387.6 89.39 1.544 1 301.8 70.57 -2.31 5 72.25 18.27 0.3981 4 0 497.5 96.14 -1.156 0.3 393.2 76.97 1.629 1 238.6 48 -0.4521 5 75.65 16.21 -0.020145 0 449.8 70.97 -0.06376 0.3 449.8 70.97 -0.01649 1 226.3 39.95 0.1463 5 75.98 16.04 -0.0676

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -90.56864 5 191.1373 A2 -84.43332 8 184.8666 A3 -84.93123 6 181.8625 R -112.4185 2 228.837 2 -91.24629 4 190.4926 3 -91.24629 4 190.4926 4 -87.8305 5 185.661 5 -84.93594 6 181.8719

Additive constant for all log-likelihoods = -18.38. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value-------- ------------------------ ------ -------------- Test 1 55.97 6 < 0.0001 Test 2 12.27 3 0.006511 Test 3 0.9958 2 0.6078 Test 4 12.63 2 0.001809 Test 5a 12.63 2 0.001809 Test 5b -5.684e-014 0 N/A Test 6a 5.799 1 0.01604 Test 6b 6.832 1 0.008956 Test 7a 0.009428 0 N/A Test 7b 12.62 2 0.001817 Test 7c 5.789 1 0.01613

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. A non-homogeneous variance model appears to be appropriate.

The p-value for Test 3 is greater than .1. The modeled variance appears to be appropriate here.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is less than .1. Model 3 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 5b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

The p-value for Test 6b is less than .05. Model 4 appears to fit the data better than Model 2.

Degrees of freedom for Test 7a are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7b is less than .05. Model 5 appears to fit the data better than Model 3.

The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.294787 0.2441833 0.294787 0.244183 4 0.132662 0.0982669 5 0.854689 0.219895

0 100 200 300 400 500 600

0 1 2 3 4 5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

09:58 11/09 2010 BMD BMDL Exponential0 100 200 300 400 500 600

0 1 2 3 4 5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

09:58 11/09 2010 BMD BMDL Exponential

0 100 200 300 400 500 600

0 1 2 3 4 5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

09:58 11/09 2010 BMD BMDL Exponential0 100 200 300 400 500 600

0 1 2 3 4 5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

09:58 11/09 2010 BMD BMDL Exponential

### Dow1 dnt study

chlorpyrifos 44648101 DNT 870.63 hindbrain F GD20 D

Exponential

Constant var

10 bmr

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Testrunscontinuous_Setting.(d) Gnuplot Plotting File: Thu Nov 04 09:07:16 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) rho is set to 0. A constant variance model is fit.

Total number of dose groups = 4 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- ------- lnalpha 8.33317 8.33317 8.33317 8.33317 rho(S) 0 0 0 0a 549.268 549.268 1361.22 1361.22 b 0.300404 0.300404 0.291204 0.291204 c -- -- 0.000225974 0.000225974 d -- 1 -- 1

(S) = Specified

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- ------- lnalpha 9.32203 8.35745 9.32203 8.33863 rho 0 0 0 0 a 1375.84 1295.11 1375.84 1289.65 b 0.258137 0.246907 0.258137 0.85294 c -- -- 0 0.238514 d -- 1.72667 -- 13.4568

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 5 1283 87.7 0.3 5 1296 92.2 1 5 1181 27.6 5 5 307.6 62

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 1376 105.7 -1.965 0.3 1273 105.7 0.4882 1 1063 105.7 2.493 5 378.5 105.7 -1.499 3 0 1295 65.28 -0.4182 0.3 1281 65.28 0.5372 1 1184 65.28 -0.127 5 307.2 65.28 0.01301 4 0 1376 105.7 -1.965 0.3 1273 105.7 0.4882 1 1063 105.7 2.493 5 378.5 105.7 -1.499 5 0 1290 64.67 -0.2334 0.3 1290 64.67 0.2334 1 1181 64.67 -2.186e-008 5 307.6 64.67 3.986e-009

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -93.33169 5 196.6634 A2 -89.98273 8 195.9655 A3 -93.33169 5 196.6634 R -130.6602 2 265.3204 2 -103.2203 3 212.4407 3 -93.5745 4 195.149 4 -103.2203 3 212.4407 5 -93.38631 5 196.7726

Additive constant for all log-likelihoods = -18.38. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 81.35 6 < 0.0001 Test 2 6.698 3 0.08218 Test 3 6.698 3 0.08218 Test 4 19.78 2 < 0.0001 Test 5a 0.4856 1 0.4859 Test 5b 19.29 1 < 0.0001 Test 6a 19.78 2 < 0.0001 Test 6b 0 0 N/A Test 7a 0.1092 0 N/A Test 7b 0.3764 1 0.5395 Test 7c 19.67 2 < 0.0001

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the doselevels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. Consider running a non-homogeneous variance model.

The p-value for Test 3 is less than .1. You may want to consider a different variance model.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is greater than .1. Model 3 seems to adequately describe the data.

The p-value for Test 5b is less than .05. Model 3 appears to fit the data better than Model 2.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 6b are less than or equal to 0. The Chi-Square test for fit is not valid.

Degrees of freedom for Test 7a are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7b is greater than .05. Model 5 does not seem to fit the data better than Model 3.

The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.408157 0.351427 3 1.10015 0.804926 4 0.408157 0.35139 5 1.01347 0.823914200 400 600 800 1000 1200 1400

0 1 2 3 4 5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

08:07 11/04 2010 BMD BMDL Exponential200 400 600 800 1000 1200 1400

0 1 2 3 4 5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

08:07 11/04 2010 BMD BMDL Exponential200 400 600 800 1000 1200 1400

0 1 2 3 4 5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

08:07 11/04 2010 BMD BMDL Exponential200 400 600 800 1000 1200 1400

0 1 2 3 4 5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

08:07 11/04 2010 BMD BMDL Exponential

### chlorpyrifos 44648101 DNT 870.63 RBC F GD20 D



## Dow1 dnt study

Exponential Constant var 10 bmr

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Testrunscontinuous_Setting.(d) Gnuplot Plotting File: Tue Nov 09 09:26:03 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) rho is set to 0. A constant variance model is fit.

Total number of dose groups = 4 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 7.00652 7.00652 7.00652 7.00652rho(S) 0 0 0 0 a 47.1399 47.1399 373.485 373.485 b 0.560906 0.560906 1.24423 1.24423 c -- -- 0.0446247 0.0446247 d -- 1 -- 1

(S) = Specified

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 7.30274 7.06976 7.28968 7.00652 rho 0 0 0 0 a 365.259 355.7 365.469 355.7 b 1.45943 1.46577 1.52086 1.58615 c -- -- 0.0235257 0.0491988 d -- 1.4469 -- 1.52088

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 5 355.7 46.8 0.3 4 262.3 51.6 1 4 62.5 24 5 4 17.5 9.8

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 365.3 38.53 -0.5548 0.3 235.8 38.53 1.378 1 84.87 38.53 -1.161 5 0.2474 38.53 0.8956 3 0 355.7 34.29 -1.017e-006 0.3 262.3 34.29 1.911e-006 1 62.5 34.29 -2.432e-006 5 6.3e-006 34.29 1.021 4 0 365.5 38.28 -0.5707 0.3 234.7 38.28 1.441 1 86.58 38.28 -1.258 5 8.776 38.28 0.4559 5 0 355.7 33.22 -2.305e-007 0.3 262.3 33.22 -3.354e-0081 62.5 33.22 7.06e-008 5 17.5 33.22 -1.161e-007

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -68.05544 5 146.1109 A2 -63.0613 8 142.1226 A3 -68.05544 5 146.1109 R -93.17983 2 190.3597 2 -70.57327 3 147.1465 3 -68.59298 4 145.186 4 -70.46232 4 148.9246 5 -68.05544 5 146.1109

Additive constant for all log-likelihoods = -15.62. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 60.24 6 < 0.0001Test 2 9.988 3 0.01867 Test 3 9.988 3 0.01867 Test 4 5.036 2 0.08063 Test 5a 1.075 1 0.2998 Test 5b 3.961 1 0.04658 Test 6a 4.814 1 0.02823 Test 6b 0.2219 1 0.6376 Test 7a 1.421e-013 0 N/A Test 7b 1.075 1 0.2998 Test 7c 4.814 1 0.02823

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. Consider running a non-homogeneous variance model.

The p-value for Test 3 is less than .1. You may want to consider a different variance model.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is greater than .1. Model 3 seems to adequately describe the data.

The p-value for Test 5b is less than .05. Model 3 appears to fit the data better than Model 2.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

The p-value for Test 6b is greater than .05. Model 4 does not seem to fit the data better than Model 2.

Degrees of freedom for Test 7a are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7b is greater than .05. Model 5 does not seem to fit the data better than Model 3.

The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.0721929 0.058417 3 0.144037 0.0822147 4 0.0710395 0.05665255 0.148684 0.0879464

0 50 100 150 200 250 300 350 400 450

0 1 2 3 4 5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

09:26 11/09 2010 BMD BMDL Exponential0 50 100 150 200 250 300 350 400 450

0 1 2 3 4 5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

09:26 11/09 2010 BMD BMDL Exponential

0 50 100 150 200 250 300 350 400 450

0 1 2 3 4 5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

09:26 11/09 2010 BMD BMDL Exponential0 50 100 150 200 250 300 350 400 450

0 1 2 3 4 5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

09:26 11/09 2010 BMD BMDL Exponential

### chlorpyrifos 44648101 DNT 871 forebrain F 1 D



## Dow1 dnt study

Exponential Constant var 10 bmr

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Testrunscontinuous_Setting.(d) Gnuplot Plotting File: Tue Nov 09 11:43:06 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) rho is set to 0. A constant variance model is fit.

Total number of dose groups = 4 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 9.20255 9.20255 9.20255 9.20255rho(S) 0 0 0 0 a 598.964 598.964 2348.33 2348.33 b 0.465332 0.465332 0.441469 0.441469 c -- -- 0.00010318 0.00010318 d -- 1 -- 1

(S) = Specified

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 10.9723 9.25474 10.9723 9.23247 rho 0 0 0 0 a 2405.58 2217 2405.58 2211.95 b 0.324417 0.295016 0.324417 0.876479 c -- -- 0 0.109541 d -- 2.04798 -- 18

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 5 2187 175.6 0.3 5 2237 86.2 1 5 2037 70.1 5 5 242.3 80.2

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 2406 241.3 -2.022 0.3 2182 241.3 0.5005 1 1739 241.3 2.757 5 475.1 241.3 -2.157 3 0 2217 102.2 -0.6473 0.3 2202 102.2 0.7634 1 2042 102.2 -0.1222 5 241.6 102.2 0.01637 4 0 2406 241.3 -2.022 0.3 2182 241.3 0.5005 1 1739 241.3 2.757 5 475.1 241.3 -2.157 5 0 2212 101.1 -0.5429 0.3 2212 101.1 0.54291 2037 101.1 -2.808e-009 5 242.3 101.1 -2.859e-009

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -102.0255 5 214.0509 A2 -99.06519 8 214.1304 A3 -102.0255 5 214.0509 R -144.5919 2 293.1837 2 -119.7229 3 245.4457 3 -102.5474 4 213.0947 4 -119.7229 3 245.4457 5 -102.3247 5 214.6493

Additive constant for all log-likelihoods = -18.38. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 91.05 6 < 0.0001Test 2 5.921 3 0.1155 Test 3 5.921 3 0.1155 Test 4 35.39 2 < 0.0001 Test 5a 1.044 1 0.3069 Test 5b 34.35 1 < 0.0001 Test 6a 35.39 2 < 0.0001 Test 6b -2.842e-014 0 N/A Test 7a 0.5984 0 N/A Test 7b 0.4454 1 0.5045 Test 7c 34.8 2 < 0.0001

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is greater than .1. A homogeneous variance model appears to be appropriate here.

The p-value for Test 3 is greater than .1. The modeled variance appears to be appropriate here.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is greater than .1. Model 3 seems to adequately describe the data.

The p-value for Test 5b is less than .05. Model 3 appears to fit the data better than Model 2.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 6b are less than or equal to 0. The Chi-Square test for fit is not valid.

Degrees of freedom for Test 7a are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7b is greater than .05. Model 5 does not seem to fit the data better than Model 3.

The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.324769 0.269365 3 1.12964 0.885242 4 0.324769 0.2693655 1.01374 0.904149

0 500 1000 1500 2000 2500

0 1 2 3 4 5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

11:43 11/09 2010 BMD BMDL Exponential0 500 1000 1500 2000 2500

0 1 2 3 4 5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

11:43 11/09 2010 BMD BMDL Exponential

0 500 1000 1500 2000 2500

0 1 2 3 4 5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

11:43 11/09 2010 BMD BMDL Exponential0 500 1000 1500 2000 2500

0 1 2 3 4 5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

11:43 11/09 2010 BMD BMDL Exponential

### chlorpyrifos 44648101 DNT 870.63 hindbrain F LD1 D



## Dow1 dnt study

Exponential model var 10 bmr

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Testrunscontinuous_Setting.(d) Gnuplot Plotting File: Thu Nov 04 14:19:02 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) The variance is to be modeled as Var(i) = exp(lalpha + log(mean(i)) * rho)

Total number of dose groups = 4 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5-------- ------- ------- ------- --

----- lnalpha 2.83544 2.83544 2.83544 2.83544 rho 0.938555 0.938555 0.938555 0.938555 a 440.493 440.493 1226.71 1226.71 b 0.338565 0.338565 0.328513 0.328513 c -- -- 0.000185862 0.000185862 d -- 1 -- 1

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha -1.44322 0.239702 -1.44322 0.239702 rho 1.69573 1.36946 1.69573 1.36946 a 1279.98 1141.14 1279.98 1141.14 b 0.334477 0.252108 0.334477 0.252108 c -- -- 0 0 d -- 2.0556 -- 2.0556

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 5 1168 100 0.3 5 1095 235.3 1 5 1089 58.6 5 5 228 52.6

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 1280 209.5 -1.192 0.3 1158 192.4 -0.7251 1 916.1 157.7 2.451 5 240.4 50.73 -0.5456 3 0 1141 139.8 0.4344 0.3 1136 139.3 -0.6436 1 1076 134.3 0.2181 5 228.2 46.44 -0.009721 4 0 1280 209.5 -1.192 0.3 1158 192.4 -0.7251 1 916.1 157.7 2.451 5 240.4 50.73 -0.54565 0 1141 139.8 0.4344 0.3 1136 139.3 -0.6436 1 1076 134.3 0.2181 5 228.2 46.44 -0.009721

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -105.6895 5 221.379 A2 -98.26598 8 212.532 A3 -102.9408 6 217.8816 R -130.0553 2 264.1107 2 -107.9575 4 223.915 3 -103.0751 5 216.1503 4 -107.9575 4 223.915 5 -103.0751 5 216.1503

Additive constant for all log-likelihoods = -18.38. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value-------- ------------------------ ------ -------------- Test 1 63.58 6 < 0.0001 Test 2 14.85 3 0.001952 Test 3 9.35 2 0.009327 Test 4 10.03 2 0.006626 Test 5a 0.2687 1 0.6042 Test 5b 9.765 1 0.001779 Test 6a 10.03 2 0.006626 Test 6b 2.842e-014 0 N/A Test 7a 0.2687 1 0.6042 Test 7b 5.684e-014 0 N/A Test 7c 9.765 1 0.001779

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. A non-homogeneous variance model appears to be appropriate.

The p-value for Test 3 is less than .1. You may want to consider a different variance model.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is greater than .1. Model 3 seems to adequately describe the data.

The p-value for Test 5b is less than .05. Model 3 appears to fit the data better than Model 2.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 6b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7a is greater than .1. Model 5 seems to adequately describe the data.

Degrees of freedom for Test 7b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.315 0.2822753 1.3273 0.645539 4 0.315 0.267246 5 1.3273 0.645539

200 400 600 800 1000 1200 1400

0 1 2 3 4 5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

13:19 11/04 2010 BMD BMDL Exponential200 400 600 800 1000 1200 1400

0 1 2 3 4 5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

13:19 11/04 2010 BMD BMDL Exponential

200 400 600 800 1000 1200 1400

0 1 2 3 4 5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

13:19 11/04 2010 BMD BMDL Exponential200 400 600 800 1000 1200 1400

0 1 2 3 4 5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

13:19 11/04 2010 BMD BMDL Exponential

### chlorpyrifos 44648101 DNT 870.6 RBC F LD1 D



## Dow1 dnt study

Exponential model var 10 bmr

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Testrunscontinuous_Setting.(d) Gnuplot Plotting File: Tue Nov 09 10:29:04 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) The variance is to be modeled as Var(i) = exp(lalpha + log(mean(i)) * rho)

Total number of dose groups = 4 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

-----lnalpha 1.66224 1.66224 1.66224 1.66224 rho 1.06602 1.06602 1.06602 1.06602 a 19.4368 19.4368 427.875 427.875 b 0.82397 0.82397 1.52843 1.52843 c -- -- 0.0109066 0.0109066 d -- 1 -- 1

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 5.32675 5.70122 1.3235 1.32448 rho 0.413499 0.339036 1.13547 1.11352 a 414.518 411.109 424.962 409.84 b 1.90988 1.87395 2.13123 1.92911 c -- -- 0.0128033 0.0137629 d -- 1.08687 -- 1.20447

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 5 407.5 77.2 0.3 5 247.9 37.6 1 5 51.8 13.3 5 5 4.9 6.4

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 414.5 49.87 -0.3146 0.3 233.7 44.3 0.7154 1 61.39 33.6 -0.6381 5 0.02953 6.925 1.573 3 0 411.1 47.99 -0.1682 0.3 240.8 43.83 0.3605 1 56.82 34.31 -0.3269 5 0.004695 6.971 1.57 4 0 425 60.2 -0.6486 0.3 226.8 42.15 1.12 1 55.23 18.9 -0.4063 5 5.451 5.076 -0.2426 5 0 409.8 55.23 -0.09474 0.3 246.5 41.62 0.072731 50.14 17.15 0.2168 5 5.641 5.081 -0.326

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -83.2534 5 176.5068 A2 -69.85589 8 155.7118 A3 -71.03719 6 154.0744 R -112.1756 2 228.3513 2 -75.75129 4 159.5026 3 -75.64199 5 161.284 4 -72.01299 5 154.026 5 -71.03719 6 154.0744

Additive constant for all log-likelihoods = -18.38. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 84.64 6 < 0.0001Test 2 26.8 3 < 0.0001 Test 3 2.363 2 0.3069 Test 4 9.428 2 0.008968 Test 5a 9.21 1 0.002407 Test 5b 0.2186 1 0.6401 Test 6a 1.952 1 0.1624 Test 6b 7.477 1 0.006251 Test 7a -5.684e-014 0 N/A Test 7b 9.21 1 0.002407 Test 7c 1.952 1 0.1624

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. A non-homogeneous variance model appears to be appropriate.

The p-value for Test 3 is greater than .1. The modeled variance appears to be appropriate here.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is less than .1. Model 3 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5b is greater than .05. Model 3 does not seem to fit the data better than Model 2.

The p-value for Test 6a is greater than .1. Model 4 seems to adequately describe the data.

The p-value for Test 6b is less than .05. Model 4 appears to fit the data better than Model 2.

Degrees of freedom for Test 7a are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7b is less than .05. Model 5 appears to fit the data better than Model 3.

The p-value for Test 7c is greater than .05. Model 5 does not seem to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.0551661 0.0458563 3 0.0673031 0.0462385 4 0.0501131 0.0435635 0.0810027 0.0465544

0 100 200 300 400 500

0 1 2 3 4 5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

10:29 11/09 2010 BMD BMDL Exponential0 100 200 300 400 500

0 1 2 3 4 5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

10:29 11/09 2010 BMD BMDL Exponential

0 100 200 300 400 500

0 1 2 3 4 5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

10:29 11/09 2010 BMD BMDL Exponential0 100 200 300 400 500

0 1 2 3 4 5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

10:29 11/09 2010 BMD BMDL Exponential

## Chemical: Chlorpyrifos (4/13/09)

Data: Dow2 study filename: dow2dntfemalebrainsimple (run file) file: ChEDataDNT2dams (original data file)

Parameter: Female Brain GD6-20

Sex: Female

Model: R progr am – Single Sex

BMR: 10%

Power parameter: 1.5070456

Female Dose-Response Curve for Chlorpyrifos

## Mon Apr 13 14:05:40 2009

Redundancy Analysis

1 2 3 4 5 6 CondIndex 5.404 5.404 1.479 1.479 1.000 1.000 mu 0.262 0.262 0.959 0.959 1.418 1.418 lA.sexF 0.474 0.000 0.000 0.493 0.033 0.000 lA.sexM 0.000 0.474 0.493 0.000 0.000 0.033 lD.sexF 0.969 0.000 0.000 0.001 0.031 0.000 lD.sexM 0.000 0.969 0.001 0.000 0.000 0.031 lg.sexF 0.940 0.000 0.000 0.029 0.031 0.000 lg.sexM 0.000 0.940 0.029 0.000 0.000 0.031

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.620 0.074 lA.sexM 1.620 0.074 lD.sexF -0.431 0.574 lD.sexM -0.431 0.574 lg.sexF 0.415 0.316 lg.sexM 0.415 0.316

---- Mean Absolute Value of Gradient Val lA.sexF 1.8215789 lA.sexM 1.8215789 lD.sexF 0.4067996 lD.sexM 0.4067996 lg.sexF 0.5373740 lg.sexM 0.5373740

## Variance is heterogeneous ::: Power parameter fit

Intercept estimate (standard error): 5.063403 (0.09803422) BMD10 estimate (standard error) : 0.6455593 (0.1037286)

Power parameter estimate (s tandard error): 1.507046 (0.1094634) The confidence intervals for the variables are given by lower est. upper Intercept 4.8746436 5.0634028 5.2594713 BMD 0.4885566 0.6455593 0.8530165 Power Parameter 1.3158881 1.5070456 1.7259724

BMDL 0.5119655

## Chemical: Chlorpyrifos (4/21/09)

Data: Dow2 study filename: dow2dntfemalerbcsimple (run file) file: ChEDataDNT2dams (original data file)

Parameter: Female RBC GD6-20

Sex: Female

Model: R progr am – Single Sex

BMR: 10%

Power parameter:

Female Dose-Response Curve for Chlorpyrifos

## Tue Apr 21 08:45:42 2009

Redundancy Analysis

1 2 3 4 5 6 CondIndex 8.365 8.365 1.822 1.822 1.000 1.000 mu 0.181 0.181 0.829 0.829 1.510 1.510 lA.sexF 0.000 0.151 0.012 0.778 0.059 0.001 lA.sexM 0.151 0.000 0.778 0.012 0.001 0.059 lD.sexF 0.000 0.983 0.000 0.006 0.011 0.000 lD.sexM 0.983 0.000 0.006 0.000 0.000 0.011 lg.sexF 0.000 0.974 0.000 0.014 0.011 0.000 lg.sexM 0.974 0.000 0.014 0.000 0.000 0.011

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.610 0.100 lA.sexM 1.610 0.100 lD.sexF -2.760 0.817 lD.sexM -2.760 0.817 lg.sexF 0.039 0.353 lg.sexM 0.039 0.353

---- Mean Absolute Value of Gradient Val lA.sexF 1.0170328 lA.sexM 1.0170328 lD.sexF 0.4122730 lD.sexM 0.4122730 lg.sexF 0.8811787 lg.sexM 0.8811787

## Variance is heterogeneous ::: Power parameter set to 1

Intercept estimate (standard error): 5.049589 (1.360545) BMD10 estimate (standard error) : 0.05605487 (0.01824432)

The confidence intervals for the variables are given by lower est. upper Intercept 3.28135405 5.04958947 7.77068046 BMD 0.03395319 0.05605487 0.09254351

BMDL 0.03693596

>

### chlorpyrifos 44556901 DNT 8706300 BRAIN F GD20



## Exponential Constant var 10 bmr

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Testrunscontinuous_Setting.(d) Gnuplot Plotting File: Wed Nov 03 12:57:52 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) rho is set to 0. A constant variance model is fit.

Total number of dose groups = 4 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 12.1573 12.1573 12.1573 12.1573rho(S) 0 0 0 0 a 2864.41 2864.41 11860.8 11860.8 b 0.476761 0.476761 0.457395 0.457395 c -- -- 9.6958e-005 9.6958e-005 d -- 1 -- 1

(S) = Specified

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 13.523 12.2365 13.523 12.1573 rho 0 0 0 0 a 12103 11434.8 12103 11296 b 0.360909 0.348145 0.360909 0.653324 c -- -- 0 0.101806 d -- 1.51419 -- 3.53297

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 5 1.13e+004 630 0.3 5 1.126e+004 334 1 5 9274 633 5 5 1150 208

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 1.21e+004 863.9 -2.089 0.3 1.086e+004 863.9 1.043 1 8436 863.9 2.168 5 1992 863.9 -2.178 3 0 1.143e+004 454.1 -0.6835 0.3 1.107e+004 454.1 0.9699 1 9340 454.1 -0.3245 5 1130 454.1 0.1004 4 0 1.21e+004 863.9 -2.089 0.3 1.086e+004 863.9 1.043 1 8436 863.9 2.168 5 1992 863.9 -2.178 5 0 1.13e+004 436.4 -1.273e-007 0.3 1.126e+004 436.4 -1.915e-0081 9274 436.4 9.442e-008 5 1150 436.4 8.077e-008

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -131.5731 5 273.1462 A2 -127.9929 8 271.9858 A3 -131.5731 5 273.1462 R -176.86 2 357.72 2 -145.2298 3 296.4596 3 -132.3653 4 272.7306 4 -145.2298 3 296.4596 5 -131.5731 5 273.1462

Additive constant for all log-likelihoods = -18.38. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 97.73 6 < 0.0001Test 2 7.16 3 0.06696 Test 3 7.16 3 0.06696 Test 4 27.31 2 < 0.0001 Test 5a 1.584 1 0.2081 Test 5b 25.73 1 < 0.0001 Test 6a 27.31 2 < 0.0001 Test 6b -5.684e-014 0 N/A Test 7a 0 0 N/A Test 7b 1.584 1 0.2081 Test 7c 27.31 2 < 0.0001

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. Consider running a non-homogeneous variance model.

The p-value for Test 3 is less than .1. You may want to consider a different variance model.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is greater than .1. Model 3 seems to adequately describe the data.

The p-value for Test 5b is less than .05. Model 3 appears to fit the data better than Model 2.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 6b are less than or equal to 0. The Chi-Square test for fit is not valid.

Degrees of freedom for Test 7a are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 7b is greater than .05. Model 5 does not seem to fit the data better than Model 3.

The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.291931 0.253693 3 0.649826 0.536917 4 0.291931 0.2536935 0.835998 0.576174

0 2000 4000 6000 8000 10000 12000

0 1 2 3 4 5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

11:57 11/03 2010 BMD BMDL Exponential0 2000 4000 6000 8000 10000 12000

0 1 2 3 4 5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

11:57 11/03 2010 BMD BMDL Exponential

0 2000 4000 6000 8000 10000 12000

0 1 2 3 4 5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

11:57 11/03 2010 BMD BMDL Exponential0 2000 4000 6000 8000 10000 12000

0 1 2 3 4 5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

11:57 11/03 2010 BMD BMDL Exponential

## Dow 2 study

Female rbc

chlorpyrifos 44556901 DNT 8706300 RBC F GD20 D



## Exponential Constant var 10 bmr

Warning: optimum may not have been found form model 1 (BMDL). Bad process completion in Optimization routine. Warning: optimum may not have been found form model 2 (BMDL). Bad process completion in Optimization routine.

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Testrunscontinuous_Setting.(d) Gnuplot Plotting File: Wed Nov 03 13:19:07 2010

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) rho is set to 0. A constant variance model is fit.

Total number of dose groups = 4 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: ExactInitial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 10.5247 10.5247 10.5247 10.5247 rho(S) 0 0 0 0 a 3.51007e-190 0 685.125 685.125 b 152.929 0.270806 136.147 2.35264 c -- -- 0 0 d -- 18 -- 2

(S) = Specified

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha 11.7348 11.5027 10.9678 10.5247 rho 0 0 0 0 a 652.498 359.643 652.5 652.5 b 152.929 0.270806 136.147 1.81593 c -- -- 0.247765 0 d -- 18 -- 1.0391

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 4 652.5 408 0.3 5 383.3 210 1 5 101.7 88.7 5 5 0 1.96

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 652.5 353.3 1.052e-005 0.3 7.758e-018 353.3 2.426 1 2.504e-064 353.3 0.6436 5 0 353.3 0 3 0 359.6 314.6 1.862 0.3 359.6 314.6 0.16811 359.6 314.6 -1.833 5 8.114e-100 314.6 -5.767e-102 4 0 652.5 240.8 2.079e-007 0.3 161.7 240.8 2.058 1 161.7 240.8 -0.5569 5 161.7 240.8 -1.501 5 0 652.5 192.9 2.692e-007 0.3 383.3 192.9 -7.861e-007 1 101.7 192.9 1.542e-006 5 0.03282 192.9 -0.0003804

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 -109.485 5 228.9701 A2 -83.82269 8 183.6454 A3 -109.485 5 228.9701 R -118.6826 2 241.3652 2 -117.4627 3 240.9254 3 -116.276 4 240.552 4 -113.694 4 235.3879 5 -109.485 4 226.9701

Additive constant for all log-likelihoods = -17.46. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3)Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 69.72 6 < 0.0001 Test 2 51.32 3 < 0.0001 Test 3 51.32 3 < 0.0001 Test 4 15.96 2 0.000343 Test 5a 13.58 1 0.0002284 Test 5b 2.373 1 0.1234 Test 6a 8.418 1 0.003716 Test 6b 7.537 1 0.006043 Test 7a 1.447e-007 1 0.9997 Test 7b 13.58 0 N/A Test 7c 8.418 0 N/A

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. Consider running a non-homogeneous variance model.

The p-value for Test 3 is less than .1. You may want to consider a different variance model.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is less than .1. Model 3 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5b is greater than .05. Model 3 does not seem to fit the data better than Model 2.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

The p-value for Test 6b is less than .05. Model 4 appears to fit the data better than Model 2.

The p-value for Test 7a is greater than .1. Model 5 seems to adequately describe the data.

Degrees of freedom for Test 7b are less than or equal to 0. The Chi-Square test for fit is not valid.

Degrees of freedom for Test 7c are less than or equal to 0. The Chi-Square test for fit is not valid.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.000688953 Bad completion 3 3.25871 Bad completion 4 0.00104772 0.000850368 5 0.0631473 0.0310436

0 200 400 600 800 1000 1200 1400

-1 0 1 2 3 4 5Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

12:19 11/03 2010 BMD BMDL Exponential0 200 400 600 800 1000 1200 1400

-1 0 1 2 3 4 5Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

12:19 11/03 2010 BMD BMDL Exponential

0 200 400 600 800 1000 1200 1400

0 1 2 3 4 5Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

12:19 11/03 2010 BMD BMDL Exponential0 200 400 600 800 1000 1200 1400

0 1 2 3 4 5Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

12:19 11/03 2010 BMD BMDL Exponential

### Ginger Moser

Male PND 17

Brain ChE

EPA BMDS

Exponential model ‐ Continuous

– rel dev

10 bmr

Constant variance

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Testrunscontinuous_Setting.(d) Gnuplot Plotting File: Mon Jan 24 09:20:04 2011

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) rho is set to 0. A constant variance model is fit.

Total number of dose groups = 6 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: ExactInitial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- ------- lnalpha -2.14586 -2.14586 -2.14586 -2.14586 rho(S) 0 0 0 0 a 1.43098 1.45527 4.87714 4.87714 b 0.0864062 0.00184562 0.22167 0.22167 c -- -- 0.174028 0.174028 d -- 2 -- 1

(S) = Specified

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- ------- lnalpha -1.35768 -1.35768 -1.57278 -1.954 rho 0 0 0 0 a 4.7949 4.7949 4.91829 4.65164 b 0.125487 0.125487 0.188557 0.2373 c -- -- 0.151415 0.24986 d -- 1 -- 2.3993

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 10 4.645 0.3025 0.5 10 4.635 0.2166 2 10 4.118 0.5246 5 10 1.931 0.4058 10 10 1.437 0.4232 20 10 0.8912 0.1495

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 4.795 0.5072 -0.9352 0.5 4.503 0.5072 0.8224 2 3.731 0.5072 2.414 5 2.56 0.5072 -3.921 10 1.367 0.5072 0.4365 20 0.3898 0.5072 3.126 3 0 4.795 0.5072 -0.9352 0.5 4.503 0.5072 0.8224 2 3.731 0.5072 2.414 5 2.56 0.5072 -3.921 10 1.367 0.5072 0.4365 20 0.3898 0.5072 3.126 4 0 4.918 0.4555 -1.898 0.5 4.543 0.4555 0.6416 2 3.607 0.4555 3.546 5 2.37 0.4555 -3.049 10 1.378 0.4555 0.410220 0.8408 0.4555 0.3499 5 0 4.652 0.3764 -0.0566 0.5 4.631 0.3764 0.03756 2 4.114 0.3764 0.03042 5 1.935 0.3764 -0.03284 10 1.163 0.3764 2.298 20 1.162 0.3764 -2.277

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 34.37589 7 -54.75177 A2 43.48535 12 -62.9707 A3 34.37589 7 -54.75177 R -58.16722 2 120.3344 2 10.73031 3 -15.46062 3 10.73031 3 -15.46062 4 17.18333 4 -26.36665 5 28.62013 5 -47.24027

Additive constant for all log-likelihoods = -55.14. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)

Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of InterestTest -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 203.3 10 < 0.0001 Test 2 18.22 5 0.002684 Test 3 18.22 5 0.002684 Test 4 47.29 4 < 0.0001 Test 5a 47.29 4 < 0.0001 Test 5b 0 0 N/A Test 6a 34.39 3 < 0.0001 Test 6b 12.91 1 0.0003275 Test 7a 11.51 2 0.003165 Test 7b 35.78 2 < 0.0001 Test 7c 22.87 1 < 0.0001

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. Consider running a non-homogeneous variance model.

The p-value for Test 3 is less than .1. You may want to consider a different variance model.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is less than .1. Model 3 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 5b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

The p-value for Test 6b is less than .05. Model 4 appears to fit the data better than Model 2.

The p-value for Test 7a is less than .1. Model 5 may not adequately describe the data; you may want to consider another model.

The p-value for Test 7b is less than .05. Model 5 appears to fit the data better than Model 3.

The p-value for Test 7c is less than .05. Model 5 appears to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviation

Confidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL------- ------------ ---------- 2 0.839612 0.741626 3 0.839612 0.741626 4 0.664975 0.576235 5 1.87391 1.47351

0.5

1 1.5

2 2.5

3 3.5

4 4.5

5

0 5 10 15 20Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

09:20 01/24 2011 BMD BMDL Exponential0.5 1 1.5

2 2.5

3 3.5

4 4.5

5

0 5 10 15 20Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

09:20 01/24 2011 BMD BMDL Exponential0.5 1 1.5

2 2.5

3 3.5

4 4.5

5

0 5 10 15 20Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

09:20 01/24 2011 BMD BMDL Exponential0.5 1 1.5

2 2.5

3 3.5

4 4.5

5

0 5 10 15 20Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

09:20 01/24 2011 BMD BMDL Exponential

## Ginger Moser Male PND 17 RBC ChE EPA BMDS Exponential model - Continuous – rel dev 10 bmr nonhomogeneous variance

==================================================================== Exponential Model. (Version: 1.7; Date: 12/10/2009) Input Data File: C:/Usepa/BMDS212/Data/exp_Testrunscontinuous_Setting.(d) Gnuplot Plotting File: Mon Jan 24 09:52:58 2011

====================================================================

BMDS Model Run

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The form of the response function by Model: Model 2: Y[dose] = a * exp{sign * b * dose} Model 3: Y[dose] = a * exp{sign * (b * dose)^d} Model 4: Y[dose] = a * [c-(c-1) * exp{-b * dose}] Model 5: Y[dose] = a * [c-(c-1) * exp{-(b * dose)^d}]

Note: Y[dose] is the median response for exposure = dose; sign = +1 for increasing trend in data; sign = -1 for decreasing trend.

Model 2 is nested within Models 3 and 4. Model 3 is nested within Model 5. Model 4 is nested within Model 5.

Dependent variable = Mean Independent variable = Dose Data are assumed to be distributed: normally Variance Model: exp(lnalpha +rho *ln(Y[dose])) The variance is to be modeled as Var(i) = exp(lalpha + log(mean(i)) * rho)

Total number of dose groups = 6 Total number of records with missing values = 0 Maximum number of iterations = 250 Relative Function Convergence has been set to: 1e-008 Parameter Convergence has been set to: 1e-008

MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

-----lnalpha -4.97525 -4.97525 -4.97525 - 4.97525 rho 0.755844 0.755844 0.755844 0.755844 a 0.0705752 0.281819 0.76965 0.76965 b 0.167635 0.0102239 0.30979 0.30979 c -- -- 0.0318017 0.0318017 d -- 2 -- 1

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- --

----- lnalpha -5.16362 -5.16362 -5.06982 - 5.32654 rho 0.280456 0.280456 0.443427 0.299807 a 0.740524 0.740524 0.746358 0.73489 b 0.273522 0.273522 0.301401 0.313235 c -- -- 0.0333737 0.0533944 d -- 1 -- 1.15886

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

----- --- ---------- ------------- 0 10 0.733 0.04383 0.5 10 0.6571 0.06872 2 10 0.4398 0.08411 5 10 0.1469 0.04648 10 10 0.083 0.05824 20 10 0.0257 0.01267

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

------- ------ ---------- --------- ---------------- 2 0 0.7405 0.07252 -0.3281 0.5 0.6459 0.07114 0.4992 2 0.4285 0.06716 0.5316 5 0.1886 0.05986 -2.204 10 0.04805 0.04942 2.237 20 0.003117 0.03367 2.121 3 0 0.7405 0.07252 -0.3281 0.5 0.6459 0.07114 0.4992 2 0.4285 0.06716 0.5316 5 0.1886 0.05986 -2.204 10 0.04805 0.04942 2.237 20 0.003117 0.03367 2.1214 0 0.7464 0.07429 -0.5686 0.5 0.6454 0.07194 0.513 2 0.4197 0.06539 0.9701 5 0.1848 0.05451 -2.196 10 0.06033 0.04253 1.686 20 0.02665 0.03549 -0.08445 5 0 0.7349 0.06657 -0.08978 0.5 0.6583 0.06548 -0.05741 2 0.4281 0.06139 0.6024 5 0.1686 0.05339 -1.288 10 0.05551 0.0452 1.923 20 0.0394 0.04294 -1.009

Other models for which likelihoods are calculated:

Model A1: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma^2

Model A2: Yij = Mu(i) + e(ij) Var{e(ij)} = Sigma(i)^2

Model A3: Yij = Mu(i) + e(ij) Var{e(ij)} = exp(lalpha + log(mean(i)) * rho)

Model R: Yij = Mu + e(i) Var{e(ij)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

------- ----------------- ---- ------------ A1 145.1405 7 -276.2809 A2 158.7751 12 -293.5502 A3 150.3958 8 -284.7916 R 45.50894 2 -87.01788 2 141.8192 4 -275.6385 3 141.8192 4 -275.6385 4 143.6454 5 -277.2909 5 144.0066 6 -276.0133

Additive constant for all log-likelihoods = -55.14. This constant added to the above values gives the log-likelihood including the term that does not depend on the model parameters.

Explanation of Tests

Test 1: Does response and/or variances differ among Dose levels? (A2 vs. R) Test 2: Are Variances Homogeneous? (A2 vs. A1) Test 3: Are variances adequately modeled? (A2 vs. A3) Test 4: Does Model 2 fit the data? (A3 vs. 2)

Test 5a: Does Model 3 fit the data? (A3 vs 3) Test 5b: Is Model 3 better than Model 2? (3 vs. 2)

Test 6a: Does Model 4 fit the data? (A3 vs 4) Test 6b: Is Model 4 better than Model 2? (4 vs. 2)Test 7a: Does Model 5 fit the data? (A3 vs 5) Test 7b: Is Model 5 better than Model 3? (5 vs. 3) Test 7c: Is Model 5 better than Model 4? (5 vs. 4)

Tests of Interest

Test -2*log(Likelihood Ratio) D. F. p-value

-------- ------------------------ ------ -------------- Test 1 226.5 10 < 0.0001 Test 2 27.27 5 < 0.0001 Test 3 16.76 4 0.002153 Test 4 17.15 4 0.001805 Test 5a 17.15 4 0.001805 Test 5b -4.945e-012 0 N/A Test 6a 13.5 3 0.00367 Test 6b 3.652 1 0.05599 Test 7a 12.78 2 0.00168 Test 7b 4.375 2 0.1122 Test 7c 0.7224 1 0.3954

The p-value for Test 1 is less than .05. There appears to be a difference between response and/or variances among the dose levels, it seems appropriate to model the data.

The p-value for Test 2 is less than .1. A non-homogeneous variance model appears to be appropriate.

The p-value for Test 3 is less than .1. You may want to consider a different variance model.

The p-value for Test 4 is less than .1. Model 2 may not adequately describe the data; you may want to consider another model.

The p-value for Test 5a is less than .1. Model 3 may not adequately describe the data; you may want to consider another model.

Degrees of freedom for Test 5b are less than or equal to 0. The Chi-Square test for fit is not valid.

The p-value for Test 6a is less than .1. Model 4 may not adequately describe the data; you may want to consider another model.

The p-value for Test 6b is greater than .05. Model 4 does not seem to fit the data better than Model 2.

The p-value for Test 7a is less than .1. Model 5 may not adequately describe the data; you may want to consider another model.

The p-value for Test 7b is greater than .05. Model 5 does not seem to fit the data better than Model 3.

The p-value for Test 7c is greater than .05. Model 5 does not seem to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 0.100000

Risk Type = Relative deviationConfidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ---------- 2 0.3852 0.346963 3 0.3852 0.346963 4 0.362322 0.322049 5 0.481397 0.327596

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0 5 10 15 20Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

09:52 01/24 2011 BMD BMDL Exponential0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0 5 10 15 20Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

09:52 01/24 2011 BMD BMDL Exponential

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0 5 10 15 20Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

09:52 01/24 2011 BMD BMDL Exponential0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0 5 10 15 20Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

09:52 01/24 2011 BMD BMDL Exponential

## Ginger Moser Male PND 17 Brain ChE

R-based single sex BMD Program

Mon Jan 24 10:06:19 2011

Redundancy Analysis

1 2 3 4 5 6 CondIndex 9.122 9.122 1.816 1.816 1.000 1.000 mu 0.166 0.166 0.832 0.832 1.510 1.510 lA.sexF 0.000 0.450 0.000 0.509 0.041 0.000 lA.sexM 0.450 0.000 0.509 0.000 0.000 0.041 lD.sexF 0.000 0.988 0.000 0.003 0.009 0.000 lD.sexM 0.988 0.000 0.003 0.000 0.000 0.009 lg.sexF 0.000 0.973 0.000 0.017 0.010 0.000 lg.sexM 0.973 0.000 0.017 0.000 0.000 0.010 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 0.003 0.000 -0.019 0.000 -0.006 0.000 lA.sexM 0.000 0.003 0.000 -0.019 0.000 -0.006 lD.sexF -0.019 0.000 0.258 0.000 0.090 0.000 lD.sexM 0.000 -0.019 0.000 0.258 0.000 0.090 lg.sexF -0.006 0.000 0.090 0.000 0.034 0.000 lg.sexM 0.000 -0.006 0.000 0.090 0.000 0.034 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 1.000 0.000 -0.685 0.000 -0.588 0.000 lA.sexM 0.000 1.000 0.000 -0.685 0.000 -0.588 lD.sexF -0.685 0.000 1.000 0.000 0.964 0.000 lD.sexM 0.000 -0.685 0.000 1.000 0.000 0.964 lg.sexF -0.588 0.000 0.964 0.000 1.000 0.000 lg.sexM 0.000 -0.588 0.000 0.964 0.000 1.000

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.657 0.053 lA.sexM 1.657 0.053 lD.sexF -0.448 0.508 lD.sexM -0.448 0.508 lg.sexF -0.117 0.184 lg.sexM -0.117 0.184

## ---- Mean Absolute Value of Gradient Val lA.sexF 1.5677228 lA.sexM 1.5677228 lD.sexF 0.4792042 lD.sexM 0.4792042 lg.sexF 1.0780830 lg.sexM 1.0780830

Variance is homogeneous ::: Power parameter set to 1

Intercept estimate (standard error): 5.162048 (0.1236352) BMD10 estimate (standard error) : 0.8385296 (0.06262208)

The confidence intervals for the variables are given by

## lower est. upper Intercept 4.9260808 5.1620479 5.4093182 BMD 0.7295612 0.8385296 0.9637738

BMDL 0.7462883

## Ginger Moser Male PND 17 RBC ChE

R-based single sex BMD Program

Mon Jan 24 10:37:30 2011

Redundancy Analysis

1 2 3 4 5 6 CondIndex 8.372 8.372 1.745 1.745 1.000 1.000 mu 0.179 0.179 0.857 0.857 1.495 1.495 lA.sexF 0.385 0.000 0.000 0.571 0.045 0.000 lA.sexM 0.000 0.385 0.571 0.000 0.000 0.045 lD.sexF 0.986 0.000 0.000 0.003 0.011 0.000 lD.sexM 0.000 0.986 0.003 0.000 0.000 0.011 lg.sexF 0.971 0.000 0.000 0.017 0.012 0.000 lg.sexM 0.000 0.971 0.017 0.000 0.000 0.012 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 0.003 0.000 -0.018 0.000 -0.006 0.000 lA.sexM 0.000 0.003 0.000 -0.018 0.000 -0.006 lD.sexF -0.018 0.000 0.242 0.000 0.096 0.000 lD.sexM 0.000 -0.018 0.000 0.242 0.000 0.096 lg.sexF -0.006 0.000 0.096 0.000 0.041 0.000 lg.sexM 0.000 -0.006 0.000 0.096 0.000 0.041 lA.sexF lA.sexM lD.sexF lD.sexM lg.sexF lg.sexM lA.sexF 1.000 0.000 -0.635 0.000 -0.537 0.000 lA.sexM 0.000 1.000 0.000 -0.635 0.000 -0.537 lD.sexF -0.635 0.000 1.000 0.000 0.960 0.000 lD.sexM 0.000 -0.635 0.000 1.000 0.000 0.960 lg.sexF -0.537 0.000 0.960 0.000 1.000 0.000 lg.sexM 0.000 -0.537 0.000 0.960 0.000 1.000

---- Parameter Values and Relative Standard Errors (sigma == 1) Parms Relative SE lA.sexF 1.619 0.058 lA.sexM 1.619 0.058 lD.sexF -0.895 0.492 lD.sexM -0.895 0.492 lg.sexF 0.036 0.204 lg.sexM 0.036 0.204

## Variance is heterogeneous ::: Power parameter set to 1

Intercept estimate (standard error): 5.052217 (0.1244639) BMD10 estimate (standard error) : 0.3846067 (0.02402530)

The confidence intervals for the variables are given by lower est. upper Intercept 4.8149786 5.0522175 5.3011454 BMD 0.3418926 0.3846067 0.4326573

BMDL 0.3485101