Page 1 of 61 

# APPENDIX C: Summaries OF BMD Analyses of

Studies on LH SurgePage 2 of 61 TABLE OF CONTENTS

C.1 NHEERL STUDY

................................ ................................ ................................ ................................ ................ 3

C.2 MORSETH ONE MON TH LH STUDY

................................ ................................ ................................ ................ 11

C.3. MORSETH SIX MO NTH LH STUDY

................................ ................................ ................................ ................. 27

C.4. MINNEMA ONE MO NTH LH STUDY

................................ ................................ ................................ .............. 43

C.5 BMD OUTPUT SUM MARIES

– AUC

................................ ................................ ................................ ................ 51

C.6. PREDICTED DAI LY STEADY STATE SERU M AUC

................................ ................................ ............................ 56Page 3 of 61 C.1 NHEERL STUDY

Model

(Constant

Variance) BMD BMDL p

- value AIC

Exponential 2 58.5 29.6 0.004 515.8

Exponential 3 58.5 29.6 0.004 515.8

Exponential 4 4.23 1.96 0.94 501.3

Exponential 5 4.31 1.96 0.86 503.3

Hill 4.16 1.68 0.87 503.3

Linear 78.4 53.1 0.002 518.1

Polynomial 78.4 53.1 0.002 518.1

Power 78.4 53.1 0.002 518.1

NHEERL Study: BMD OUTPUT for Exponential Model Runs (Constant Variance;

Best Model Fit)

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

== Exponential Model. (Version: 1.7; Date: 12/10/2009)

Input Data File: C:/Usepa/BMDS212/Data/exp_Actualr uns_Setting.(d)

Gnuplot Plotting File:

Tue Jul 20 14:07:40 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.Page 4 of 61 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 = 7

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 4.07816 4.07816 4.07816 4.07816

rho(S) 0 0 0 0

a 11.0009 11.0009 2 2.9404 22.9404

b 0.00794702 0.00794702 0.0449039 0.0449039

c

-- -- 0.352881 0.352881

d

-- 1

-- 1

(S) = Specified

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- ------- lnalpha 4.25603 4.25603 4.08595 4.08581

rho 0 0 0 0

a 16.8823 16.8823 21.9754 21.9037

b 0.0117718 0.0117718 0.219622 0.22213

c

-- -- 0.422997 0.427096

d

-- 1

-- 1.07687Page 5 of 61 Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

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

0 16 21.85 7

1.56 11 18. 57 10.8

3.12 12 16.05 9.3

6.25 17 11.81 7

12.5 7 11.88 9

25 14 8.5 7.6

75 20 9.6 6.7

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

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

2 0 16.88 8.398 2.365

1.56 16.58 8.398 0.7878

3.12 16.27 8.398

- 0.09425

6.25 15.68 8.398

- 1.902

12.5 14.57 8.398

- 0.8482

25 12.58 8.398

- 1.817

75 6.982 8.398 1.394

3 0 16.88 8.398 2.365

1.56 16.58 8.398 0.7878

3.12 16.27 8.398

- 0.09425

6.25 15.68 8.398

- 1.902

12.5 14.57 8.398

- 0.8482

25 12.58 8.398

- 1.817

75 6.982 8.398 1.394

4 0 21.98 7.714

- 0.06604

1.56 18.3 7.714 0.1173

3.12 15.6 9 7.714 0.1613

6.25 12.51 7.714

- 0.3737

12.5 10.11 7.714 0.6071

25 9.348 7.714

- 0.4113

75 9.296 7.714 0.1765

5 0 21.9 7.713

- 0.02886

1.56 18.47 7.713 0.04191

3.12 15.75 7.713 0.1316

6 .25 12.38 7.713

- 0.303

12.5 9.978 7.713 0.6526

25 9.377 7.713

- 0.4255

75 9.355 7.713 0.1421Page 6 of 61 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

- 246.2909 8 508.5818

A2

- 243.8179 14 515.6358

A3

- 2 46.2909 8 508.5818

R

- 261.9739 2 527.9478

2

- 254.9177 3 515.8354

3

- 254.9177 3 515.8354

4

- 246.6687 4 501.3374

5

- 246.6617 5 503.3234

Additive constant for all log

- likelihoods =

- 89.14. This constant added to the

above values gives the l og

- 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)Page 7 of 61 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 36.31 12 0.0002886

Test 2 4.946 6 0.5508

Test 3 4.946 6 0.5508

Test 4 17.25 5 0.004043

Test 5a 17.25 5 0.004043

Test 5b

- 7.39e

- 013 0 N/A

Test 6a 0.7556 4 0.9443

Test 6b 16.5 1 < 0.0001

Test 7a 0.7416 3 0.8634

Test 7b 16.51 2 0.0002597

Test 7c 0.01397 1 0.9059

The p

- val ue 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.

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 seemsPage 8 of 61 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.

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 = 1.000000

Risk Type = Estimated standard deviations from control

Confidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

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

2 58.451 29.553

3 58.451 29.553

4 4.26795 1.9583

5 4.30751 1.962Page 9 of 61 5 10 15 20 25

0 10 20 30 40 50 60 70Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

14:07 07/20 2010 BMD BMDL Exponential

5 10 15 20 25

0 10 20 30 40 50 60 70Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

14:07 07/20 2010 BMD BMDL ExponentialPage 10 of 6 1 5 10 15 20 25

0 10 20 30 40 50 60 70Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

14:07 07/20 2010 BMD BMDL Exponential

5 10 15 20 25

0 10 20 30 40 50 60 70Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

14:07 07/20 2010 BMD BMDL ExponentialPage 11 of 61 ________________________________________________________________

C.2 MORSETH ONE MON TH LH STUDY

Model

(Constant

Variance) a BMD BMDL p

- value AIC

Exponential 2 152.4 25.6 0.78 819.9

Exponential 3 152.4 25.6 0.78 819.9

Exponential 4 129.6 11.8 0.85 821.2

Exponential 5 Not Computed Not Computed

-- --

Hill Not Computed Not Computed

-- --

Linear 226.5 126.7 0.54 821.0

Polynomial 215.9 128.8 0.59 820.8

Power 215.9 128.8 0.59 820.8

a Nonhomogeneous Variance recommended by BMDS output

Model

(Nonhomogeneous

Variance) BMD BMDL p

- value AIC

Exponential 2 523.7 174.7 0.017 795.3

Exponential 3 523.7 174.7 0.017 795.3

Exponential 4 Not

Computed Not

Computed -- --

Exponential 5 Not

Computed Not

Computed -- --

Hill Not

Computed Not

Computed -- --

Linear 291.3 197.3 0.006 797.5

Polynomial 291.3 197.3 0.006 797.5

Power 291.3 197.3 0.006 797.5

Morseth One Month LH Study: BMD OUTPUT for Exponential Model Runs

(Constant Variance)Page 12 of 61 ==================================================================

== Exponential Model. (Version: 1.7; Date: 12/10/2009)

Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d)

Gnuplot Plotting File:

Thu Jul 22 13:45:53 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]))

rho is set to 0.

A constant variance model is fit.

Total number of dose groups = 5

To tal 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: ExactPage 13 of 61 Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- ------- lnalpha 15.2574 15.2574 15.2574 15.2574

rho(S) 0 0 0 0

a 1413.18 1413.18 3165.75 3165.75

b 0.00610828 0.00610828 0.0210595 0.0210595

c

-- -- 0.244281 0.244281

d

-- 1

-- 1

(S) = Specified

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- ------- lnalpha 15.27 94 15.2794 15.2638 15.2609

rho 0 0 0 0

a 2764.3 2764.3 2913.32 2803.53

b 0.0 0915046 0.00915045 0.0267425 0.0265255

c

-- -- 0.269019 0.28969

d

-- 1

-- 2.20144

NC = No Converg ence

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

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

0 10 2650 2389

2.5 10 3015 3220

5 1 0 2717 2542

40 10 1450 857

200 10 812 470

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

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

2 0 2764 2079

- 0.1738

2.5 2702 2079 0.4764Page 14 of 61 5 2641 2079 0.1161

40 1917 2079

- 0.7103

200 443.4 2079 0.5606

3 0 2764 2079

- 0.1738

2.5 2702 20 79 0.4764

5 2641 2079 0.1161

40 1917 2079

- 0.7103

200 443.4 2079 0.5606

4 0 2913 2063

- 0.4036

2.5 2776 2063 0.367

5 2647 2063 0.1076

40 1514 2063

- 0.09875

200 793.9 2063 0.0278

5 0 2804 2060

- 0.2357

2.5 2798 2060 0.3324

5 2780 2060

- 0.09725

40 1449 2060 0.0007715

200 812.2 2060

- 0.0002382

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)} = Sigm a^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

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

A1

- 406.4352 6 824.8704

A2

- 388.3924 10 796.7849

A3

- 406.4352 6 824.8704

R

- 410.3636 2 824.7272

2

- 406.986 3 819.9719

3

- 406.986 3 819.9719Page 15 of 61 4

- 406.5955 4 821.1911

5

- 406.5231 5 823.0462

Additive constant for all log

- likelihoods =

- 45.95. 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 Mo del 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 Mod el 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 43.94 8 < 0.0001

Test 2 36.09 4 < 0.0001

Test 3 36.09 4 < 0.0001

Test 4 1.102 3 0.7767

Test 5a 1.102 3 0.7767

Test 5b

- 1.819e

- 011 0 N/A

Test 6a 0.3207 2 0.8518

Test 6b 0.7808 1 0.3769

Test 7a 0.1758 1 0.675

Test 7b 0.9257 2 0.6295

Test 7c 0.1449 1 0.7034

The p

- value for Test 1 is less than .05. There appears to be a

difference between response and/or variances among the dosePage 16 of 61 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 m odel.

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 = 1.000000

Risk Type = Estimated standard deviations from control

Confidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

------- ------------ ----------Page 17 of 61 2 152.443 25.6454

3 152.443 25.6454

4 129.57 11.8371

5 0 0 Not computed

0 1000 2000 3000 4000 5000

0 50 100 150 200Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

13:45 07/22 2010 BMD BMDL ExponentialPage 18 of 61 0 1000 2000 3000 4000 5000

0 50 100 150 200Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

13:45 07/22 2010 BMD BMDL Exponential

0 1000 2000 3000 4000 5000

0 50 100 150 200Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

13:45 07/22 2010 BMD BMDL ExponentialPage 19 of 61 ________________________________________________________________

Morseth One Month LH Study: BMD OUTPUT for Exponential Model Runs

(Nonhomogeneous Variance)

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

== Exponential Model. (Version: 1.7; Date: 12/10/2009)

Input Data File: C:/Usepa/BMDS212/D ata/exp_Actualruns_Setting.(d)

Gnuplot Plotting File:

Fri Aug 06 09:35:01 2010

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

==

BMDS Model Run

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

~~~ 0 1000 2000 3000 4000 5000

0 50 100 150 200Mean Response

dose Exponential Model 5

13:45 07/22 2010 BMD ExponentialPage 20 of 61 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 = 5

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

- 7.57613

- 7.57613

- 7.57613

- 7.57613

rho 2.94035 2.94035 2.94035 2.94035

a 14 13.18 1413.18 3165.75 3165.75

b 0.00610828 0.00610828 0.0210595 0.0210595

c

-- -- 0.244281 0.244281

d

-- 1

-- 1

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5Page 21 of 61 -------- ------- ------- ------- ------- lnalpha

- 7.94834

- 7.94834

- 7.47498

- 7.51913

rho 2.99971 2.99971 2.91611 2.92104

a 2583.98 2583.98 3033.27 2862.72

b 0.00589281 0.00589281 0.035215 0.0290775

c

-- -- 0.273458 0.29 1831

d

-- 1

-- 2.06427

NC = No Convergence

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

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

0 10 2650 2389

2.5 10 3015 3220

5 10 2717 2542

40 10 1450 857

200 10 812 470

Estima ted Values of Interest

Model Dose Est Mean Est Std Scaled Residual

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

2 0 2584 2466 0.08467

2.5 2546 2412 0.6146

5 2509 2359 0.2788

40 2041 1732

- 1.08

200 795.1 421 0.1266

3 0 2584 2466 0.08467

2.5 2546 2412 0.6146

5 2509 2359 0.2788

40 2041 1732

- 1.08

200 795.1 421 0.1266

4 0 3033 2842

- 0.4264

2.5 2848 2592 0.2043

5 2677 2369 0.05275

40 1368 890.3 0.2903

200 831.4 430.6

- 0.1425

5 0 2863 2606

- 0.2581

2.5 2854 2594 0.1967

5 2825 2556

- 0.1339

40 1353 871.8 0.3532

200 835.4 431.3

- 0.1718Page 22 of 61 Other models for which likel ihoods 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)} = e xp(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

- 406.4352 6 824.8704

A2

- 388.3924 10 796.7849

A3

- 388.5412 7 791.0824

R

- 410.3636 2 824.7272

2

- 393.6414 4 795.2827

3

- 393.6414 4 795.2827

4

- 389.3963 5 788.7925

5

- 389.0987 6 790.1974

Additive constant for all log

- likelihoods =

- 45.95. This constant added to the

above values gives the log

- likelihood includ ing 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)Page 23 of 61 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 43.94 8 < 0.0001

Test 2 36.09 4 < 0.0001

Test 3 0.2976 3 0.9605

Test 4 10.2 3 0.01694

Test 5a 10.2 3 0.01694

Test 5b 0 0 N/A

Test 6a 1.71 2 0.4253

Test 6b 8.49 1 0.003571

Test 7a 1.115 1 0.291

Test 7b 9.085 2 0.01065

Test 7c 0.5951 1 0.4405

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

- Squar e test for fit is not valid.

The p

- value for Test 6a is greater than .1. Model 4 seemsPage 24 of 61 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.

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 = 1.000000

Risk Type = Estimated standard deviations from control

Confidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

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

2 523.696 174.68

3 523.696 174.68

4 0 0 Not computed

5 0 0 Not computedPage 25 of 61 0 1000 2000 3000 4000 5000

0 100 200 300 400 500Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

09:35 08/06 2010 BMD BMDL Exponential

0 1000 2000 3000 4000 5000

0 100 200 300 400 500Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

09:35 08/06 2010 BMD BMDL ExponentialPage 26 of 61 0 1000 2000 3000 4000 5000

0 50 100 150 200Mean Response

dose Exponential Model 4

09:35 08/06 2010 BMD Exponential

0 1000 2000 3000 4000 5000

0 50 100 150 200Mean Response

dose Exponential Model 5

09:35 08/06 2010 BMD ExponentialPage 27 of 61 ________________________________________________________________

C. 3. MORSETH SIX MONT H LH STUDY

Model

(Constant

Variance) BMD BMDL p

- value AIC

Exponential 2 20.3 4.6 0.67 661.7

Exponential 3 20.3 Not Computed 0.67 661.7

Exponential 4 16.7 3.4 0.38 663.7

Exponential 5 3.9 3.7 N/A 664.5

Hill 4.2 3.3 0.58 663.2

Linear 26.2 16.3 0.61 661.9

Polynomial 20.6 12.9 0.05 666.9

Power 25.9 16.3 0.61 661.9

Model

(Nonhomogeneous

Variance) BMD BMDL p

- value AIC

Exponential 2 36.2 17.1 0.52 636.1

Exponential 3 36.2 17.1 0.52 636.1

Exponential 4 Not

Computed Not

Computed -- --

Exponential 5 Not

Computed Not

Computed -- --

Hill Not

Computed Not

Computed -- --

Linear 32.6 22.9 0.36 636.9

Polynomial 32.6 22.9 0.36 636.9

Power 32.6 22.9 0.36 636.9

Morseth 6 Month LH Study

- Exponential (Constant Variance) Model Results

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

completion in Optimization routine.Page 28 of 61 ==================================================================

== Exponential Model. (Version: 1.7; Date: 12/10/2009)

Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d)

G nuplot Plotting File:

Thu Jul 29 13:57:08 2010

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

==

BMDS Model Run

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

~~~

The form of the response function b y 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: ExactPage 29 of 61 Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- ------- lnalpha 15.3723 15.3723 15.3723 15.3723

rho(S) 0 0 0 0

a 1497.19 1497.19 3812.55 3812.55

b 0.0471466 0.0471466 0.14585 0.14585

c

-- -- 0.21433 0.21433

d

-- 1

-- 1

(S) = Specified

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- ------- lnalpha 15.3924 15.3924 15.3913 15.3746

rho 0 0 0 0

a 3458.46 3458.46 3544.45 3483. 5

b 0.0498021 0.0498021 0.0805078 0.262698

c

-- -- 0.16111 0.246304

d

-- 1

-- 18

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

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

0 10 3336 3138

1.8 10 3631 2732

3.65 10 2500 1897

29.4 10 858 416

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

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

2 0 3458 2200

- 0.176

1.8 3162 2200 0.6743

3.65 2884 2200

- 0.5514

29.4 799.8 2200 0.08362

3 0 3458 2200

- 0.176Page 30 of 61 1.8 3162 2200 0.6743

3.65 2884 22 00

- 0.5514

29.4 799.8 2200 0.08362

4 0 3544 2199

- 0.2998

1.8 3143 2199 0.7014

3.65 2787 2199

- 0.4133

29.4 849.9 2199 0.01171

5 0 3484 2180

- 0.2139

1.8 3483 2180 0.2139

3.65 2500 2180

- 9.656e

- 007

29.4 858 2180 2.459e

- 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

- 327.446 5 664.892

A2

- 313.3212 8 642.6425

A3

- 327.446 5 664.892

R

- 331.828 2 667.6559

2

- 3 27.8484 3 661.6967

3

- 327.8484 3 661.6967

4

- 327.826 4 663.652

5

- 327.4918 5 664.9836

Additi ve constant for all log

- likelihoods =

- 36.76. This constant added to the

above values gives the log

- likelihood including the term that does not

depend on the model parameters.Page 31 of 61 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 37.01 6 < 0.0001

Test 2 28.25 3 < 0.0001

Test 3 28.25 3 < 0.0001

Test 4 0.8047 2 0.6687

Test 5a 0.8047 2 0.6687

Test 5b

- 1.137e

- 013 0 N/A

Test 6a 0.7599 1 0.3833

Test 6b 0.04476 1 0.8325

Test 7a 0.09163 0 N/A

Test 7b 0.7131 2 0.7001

Test 7c 0.6683 1 0.4136

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

cons ider a different variance model.Page 32 of 61 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 f reedom 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.

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 greater than .05. Model 5 does

not seem to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 1.000000

Risk Typ e = Estimated standard deviations from control

Confidence Level = 0.950000

BMD and BMDL by Model

Model BMD BMDL

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

2 20.2989 4.57536

3 20.2989 Bad completion

4 16.7073 3.43496

5 3.92994 3.66097Page 33 of 61 1000 2000 3000 4000 5000 6000

0 5 10 15 20 25 30Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

13:57 07/29 2010 BMD BMDL Exponential

1000 2000 3000 4000 5000 6000

0 5 10 15 20 25 30Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

13:57 07/29 2010 BMD BMDL ExponentialPage 34 of 61 1000 2000 3000 4000 5000 6000

0 5 10 15 20 25 30Mean Response

dose Exponential Model 4 with 0.95 Confidence Level

13:57 07/29 2010 BMD BMDL Exponential

1000 2000 3000 4000 5000 6000

0 5 10 15 20 25 30Mean Response

dose Exponential Model 5 with 0.95 Confidence Level

13:57 07/29 2010 BMD BMDL ExponentialPage 35 of 61 Morseth 6 Month LH Study

- Exponential (Nonhomogeneous Variance) Model

Results

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

== Exponential Model. (Version: 1.7; Date: 12/10/2009)

Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d)

Gnuplot Plotting File:

Thu Jul 29 14: 00:34 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 = 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

- 008Page 36 of 61 MLE solution provided: Exact

Initial Parameter Values

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- ------- lnalpha

- 6.65625

- 6.65625

- 6.65625

- 6.65625

rho 2.77485 2.77485 2.77485 2.77485

a 1497.19 1497.19 3812.55 3812.55

b 0.0471466 0.0471466 0.14585 0.14585

c

-- -- 0.21433 0.21433

d

-- 1

-- 1

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- ------- lnalpha

- 7.30123

- 7.30123

- 7.07367

- 6.96831

rho 2.84795 2.84795 2.81591 2.80183

a 3418.7 3418.7 3725.66 3553.6 3

b 0.0471113 0.0471113 0.124728 0.213467

c

-- -- 0.210352 0.241407

d

-- 1

-- 2.9 234

NC = No Convergence

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

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

0 10 3336 3138

1.8 10 3631 2732

3.65 10 2500 1897

29.4 10 858 416

Estimated Values of Interest

Model Dose Est Mean Est Std Scaled Residual

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

2 0 3419 2797

- 0.0935Page 37 of 61 1.8 3141 2479 0.6254

3.65 2879 2190

- 0.5468

29.4 855.7 389.2 0.01852

3 0 3419 2797

- 0.0935

1.8 3141 2479 0.6254

3.65 2879 21 90

- 0.5468

29.4 855.7 389.2 0.01852

4 0 3726 3105

- 0.3968

1.8 3134 2434 0.6456

3.65 2650 1922

- 0.2464

29.4 858.9 393.4

- 0.007039

5 0 3554 2891

- 0.2381

1.8 3394 2711 0.2765

3.65 2522 1789

- 0.03959

29.4 857.9 394.8 0.001045

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)} = Sigm a^2

Likelihoods of Interest

Model Log(likelihood) DF AIC

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

A1

- 327.446 5 664.892

A2

- 313.3212 8 642.6425

A3

- 313.4172 6 638.8345

R

- 331.828 2 667.6559

2

- 314.0727 4 636.1453

3

- 314.0727 4 636.1453

4

- 313.7385 5 637.477

5

- 313.4172 6 638.8345Page 38 of 61 Additive constant for all log

- likelihoods =

- 36.76. 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 37.01 6 < 0.0001

Test 2 28.25 3 < 0.0001

Test 3 0.192 2 0.9084

Test 4 1.311 2 0.5192

Test 5a 1.311 2 0.5192

Test 5b

- 1.137e

- 013 0 N/A

Test 6a 0.6425 1 0.4228

Test 6b 0.6683 1 0.4136

Test 7a

- 1.137e

- 013 0 N/A

Test 7b 1.311 2 0 .5192

Test 7c 0.6425 1 0.4228

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 mod el the data.

The p

- value for Test 2 is less than .1. A non

- homogeneousPage 39 of 61 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 T est 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.

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 greater than .05. Model 5 does

not seem to fit the data better than Model 4.

Benchmark Dose Computations:

Specified Effect = 1.000000

Risk Type = Estimated standard deviations from control

Confidence Leve l = 0.950000

BMD and BMDL by Model

Model BMD BMDL

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

2 36.1811 17.0918

3 36.1811 17.0918

4 0 0 Not computedPage 40 of 61 5 0 0 Not computed

1000 2000 3000 4000 5000 6000

0 5 10 15 20 25 30 35Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

14:00 07/29 2010 BMD BMDL ExponentialPage 41 of 61 1000 2000 3000 4000 5000 6000

0 5 10 15 20 25 30 35Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

14:00 07/29 2010 BMD BMDL ExponentialPage 42 of 61 1000 2000 3000 4000 5000 6000

0 5 10 15 20 25 30Mean Response

dose Exponential Model 4

14:00 07/29 2010 BMD ExponentialPage 43 of 61 ________________________________________________________________

C. 4. MINNEMA ONE MONT H LH STUDY

Model

(C onstant

Variance) BMD BMDL p

- value AIC

Exponential 2 653.3 245.1 0.40 274.6

Exponential 3 653.3 245.1 0.40 274.6

Exponential 4 Not Computed Not Computed

-- -- Exponential 5 Not Computed Not Computed

-- --

Hill Not Computed Not Computed

-- --

Linear 500.8 240.7 0.38 274.7

Polynomial 500.8 239.9 0.38 274.7

Power 500.8 240.7 0.38 274.7 1000 2000 3000 4000 5000 6000

0 5 10 15 20 25 30Mean Response

dose Exponential Model 5

14:00 07/29 2010 BMD ExponentialPage 44 of 61 Minnema One Month LH Study

- Exponential (Constant Variance) Model Results

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

== Exponential Model. (Version: 1.7; Date: 12/10/2009)

Input Data File: C:/Usepa/BMDS212/Data/exp_Actualruns_Setting.(d)

Gnuplot Plotting File:

Fri Jul 23 08:35:00 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: nor mally

Variance Model: exp(lnalpha +rho *ln(Y[dose]))

rho is set to 0.

A constant variance model is fit.

Total number of dose groups = 5

Total number of records with missing values = 0

Maximum number of iterations = 250

Relative Function Convergence has been set to: 1e

- 008Page 45 of 61 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 1.65681 1.65681 1.65681 1.65681

rho(S) 0 0 0 0

a 3.0427 3.0427 4.305 4.305

b 0.00143069 0.00143069 0.0127345 0.0127345

c

-- -- 0.588463 0.588463

d

-- 1

-- 1

(S) = Specified

Parameter Estimates by Model

Variable Model 2 Model 3 Model 4 Model 5

-------- ------- ------- ------- ------- lnalpha 1.68618 1.68618 1.65872 1.65872

rho 0 0 0 0

a 3.55899 3.55899 4.04195 4.04195

b 0.00161956 0.00161956 0.128035 0.128035

c

-- -- 0.669638 0.669638

d

-- 1

-- 1

NC = No Convergence

Table of Stats From Input Data

Dose N Obs Mean Obs Std Dev

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

0 20 4.1 2.52

2.5 20 3.51 2.41

5 20 3.53 2.57

40 20 2.66 1.71

200 20 2.75 2.43

Estimated Values of InterestPage 46 of 61 Model Dose Est Mean Est Std Scaled Residual

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

2 0 3.559 2.324 1.041

2.5 3.545 2.324

- 0.06662

5 3.53 2.324

- 0.0005584

40 3.336 2.324

- 1.301

200 2.574 2.324 0.3382

3 0 3.559 2.324 1.041

2.5 3.545 2.324

- 0.06662

5 3.53 2.324

- 0.0005584

40 3.336 2.324

- 1.301

200 2.574 2.324 0.3382

4 0 4.042 2.292 0.1133

2.5 3.676 2.292

- 0.3243

5 3.411 2.292 0.233

40 2.715 2.292

- 0.1066

200 2.707 2.292 0.08461

5 0 4.04 2 2.292 0.1133

2.5 3.676 2.292

- 0.3243

5 3.411 2.292 0.233

40 2.715 2.292

- 0.1066

200 2.707 2.292 0.08461

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

- 132.8405 6 277.6809Page 47 of 61 A2

- 130.8789 10 281.7577

A3

- 132.8405 6 277.6809

R

- 135.53 2 275.0599

2

- 134.3092 3 274.6183

3

- 134.3092 3 274.6183

4

- 132.9359 4 273.8719

5

- 132.9359 4 273.8719

Additive constant for all log

- likelihoods =

- 91.89. 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 9.302 8 0.3174

Test 2 3.923 4 0.4165

Test 3 3.923 4 0.4165

Test 4 2.937 3 0.4014

Test 5a 2.937 3 0.4014

Test 5b 0 0 N/A

Test 6a 0.191 2 0.9089

Test 6b 2.746 1 0.09747

Test 7a 0.191 2 0.9089

Test 7b 2.746 1 0.09747Page 48 of 61 Test 7c 2.274e

- 013 0 N/A

The p

- value for Test 1 is greater than .05. There may not be a

diffence between responses and/or variances among the dose levels

Modelling the data with a dose/response curve may not be appropriate.

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 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.

De grees 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 great er 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 = 1.0000 00

Risk Type = Estimated standard deviations from control

Confidence Level = 0.950000Page 49 of 61 BMD and BMDL by Model

Model BMD BMDL

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

2 653.287 245.122

3 653.287 245.122

4 0 0 Not computed

5 0 0 Not computed

1 1.5

2 2.5

3 3.5

4 4.5

5 5.5

0 100 200 300 400 500 600Mean Response

dose Exponential Model 2 with 0.95 Confidence Level

08:35 07/23 2010 BMD BMDL ExponentialPage 50 of 61 1 1.5

2 2.5

3 3.5

4 4.5

5 5.5

0 100 200 300 400 500 600Mean Response

dose Exponential Model 3 with 0.95 Confidence Level

08:35 07/23 2010 BMD BMDL Exponential

1.5

2 2.5

3 3.5

4 4.5

5 5.5

0 50 100 150 200Mean Response

dose Exponential Model 4

08:35 07/23 2010 BMD ExponentialPage 51 of 61 C.5 BMD OUTPUT SUMMARIES

– AUC

1. AUC for Steady State Serum Levels of Total Triazine Equivalents

Model

(Constant

Variance) BMD BMDL p

- value AIC

Exponential 2 6.73 3.66 0.25 407.8

Exponential 3 6.73 3.66 0.25 407.8

Exponential 4 2.19 0.82 0.89 405.1

Exponential 5 2.19 0.82 0.89 405.1

Hill 2.08 0.65 0.92 404.9

Linear 9.91 7.00 0.10 410.1

Polynomial 9.91 7.00 0.10 410.1

Power 9.91 7.00 0.10 410.1 1.5

2 2.5

3 3.5

4 4.5

5 5.5

0 50 100 150 200Mean Response

dose Exponential Model 5

08:35 07/23 2010 BMD ExponentialPage 52 of 61 ==================================================================

== Hill Model. (Version: 2.15; Date: 10/28/2009)

Input Data File: C:/Usepa/BMDS212/Data/hil_Actualruns_Opt.(d)

Gnuplot Plotting File: C:/Usepa/BMDS212/Data/hil_Actualruns_Opt.plt

Fri Jul 30 12:15:38 2010

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

==

BMDS Model Run

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

~~~

The form of the response function is:

Y[dose] = intercept + v*dose^n/(k^n + dose^n)

Dependent variable = Mean

Independent variable = Dose

rho is set to 0

Power parameter restricted to be greater than 1

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

Default Initial Parameter Values

alpha = 68.6432

rho = 0 Specified

intercept = 21.848

v =

- 13.348

n = 1.1704

k = 1.67918

Asymptotic Correlation Matrix of Parameter EstimatesPage 53 of 61 ( *** The model parameter(s)

- rho

- n

have been estimated at a boundary point, or have been specified by the user,

and do not appear in the correla tion matrix )

alpha intercept v k

alpha 1 2.4e

- 009 1.1e

- 008

- 2.1e

- 008

intercept 2.4e

- 009 1

- 0.37

- 0.52

v 1.1e

- 008

- 0.37 1

- 0.48

k

- 2.1e

- 008

- 0.52

- 0.48 1

Parameter Estimates

95.0% Wald Confidence Interval

Variable Estimate Std. Err. Lower Conf. Limit Upper Conf. Limit

alpha 63.7164 10.2688 43.5899 83.843

intercept 21.819 1.88733 18.1199 25.5181

v

- 14.7403 3.24361

- 21.0977

- 8.38296

n 1 NA

k 1.76071 1.38154

- 0.947057 4.46847

NA

- Indi cates that this parameter has hit a bound

implied by some inequality constraint and thus

has no standard error.

Table of Data and Estimated Values of Interest

Dose N Obs Mean Est Mean Obs Std Dev Est Std Dev Scaled Res

.

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

0 16 21.8 21.8 7 7.98 0.0145

0.4545 11 18.6 18.8 10.8 7.98

- 0.0933

1.331 12 16 15.5 9.3 7.98 0.248

3.024 17 11.8 12.5 7 7.98

- 0.358

6.406 7 11.9 10.3 9 7.98 0.538

13.17 14 8.5 8.82 7.6 7.98

- 0.149Page 54 of 61 Model Descriptions for likelihoods 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)} = Sigma^2

Model A3 uses any fixed variance parameters that

were specified by the user

Model R: Yi = Mu + e(i)

Var{e(i)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) # Param's AIC

A1

- 198.190188 7 410.380377

A2

- 196.288688 12 416.577376

A3

- 198.190188 7 410.380377

fitted

- 198.446043 4 404.892085

R

- 209.919649 2 423.839298

Explanation of Tests

Test 1: Do responses and/or variances differ among Dose levels?

(A2 vs. R)

Test 2: Are Variances Homogeneous? (A1 vs A2)

Test 3: Are variances adequately modeled? (A2 vs. A3)

Test 4: Does the Model for the Mean Fit? (A3 vs. fitted)

(Note: When rho=0 the results of Test 3 and Test 2 will be the same.)

Tests of Interest

Test

- 2*log(Likelihood Ratio) Test df p

- value

Test 1 27.2619 10 0.002367

Test 2 3.803 5 0.5781

Test 3 3.803 5 0.5781

Test 4 0.511709 3 0.9163Page 55 of 61 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 Te st 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 greater than .1. The model chosen seems

to adequately describe the data

Benchmark Dose Computation

Specified effect = 1

Risk Type = Estimated standard deviations from the control mean

Confidence level = 0.95

BMD = 2.07966

BMDL = 0.64721Page 56 of 61 C.6 . PREDICTED DAILY STEADY STATE SERUM A UC

Model

(Constant

Variance) BMD BMDL p

- value AIC

Exponential 2 161.7 88.0 0.25 407.8

Exponential 3 161.7 88.0 0.25 407.8

Exponential 4 52.5 19.8 0.87 405.1

Exponential 5 52.5 19.8 0.87 405.1

Hill 49.9 15.6 0.92 404.9

Linear 238.2 168.2 0.10 410.1

Polynomial 238.2 168.2 0.10 410.1

Power 238.2 168.2 0.10 410.1 5 10 15 20 25

0 2 4 6 8 10 12Mean Response

dose Hill Model with 0.95 Confidence Level

12:15 07/30 2010 BMD BMDL HillPage 57 of 61 ==================================================================

== Hill Model. (Version: 2.15; Date: 10/28/2009)

Input Data File: C:/Usepa/BMDS212/Data/hil_Actualruns_Opt.(d)

Gnuplot Plotting File: C:/Usepa/BMDS212/Data/hil_Actualruns_Opt.plt

Fri Jul 30 12:45:13 2010

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

==

BMDS Model Run

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

~~~

The form of the response function is:

Y[dose] = intercept + v*dose^n/(k^n + dose^n)

Dependent variable = Mean

Independent variable = Dose

rho is set to 0

Power parameter restricted to be greater than 1

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

Default Initial Parameter Values

alpha = 68.6432

rho = 0 Specified

intercept = 21.848

v =

- 13.348

n = 1.17066

k = 40.3586

Asymptotic Correlation Matrix of Para meter Estimates

( *** The model parameter(s)

- rho

- nPage 58 of 61 have been estimated at a boundary point, or have been specified by the user,

and do not appear in the correlation matrix )

alpha intercept v k

alpha 1 8.8e

- 010 9.7e

- 009

- 1.9e

- 008

intercept 8.8e

- 010 1

- 0.37

- 0.52

v 9.7e

- 009

- 0.37 1

- 0.48

k

- 1.9e

- 008

- 0.52

- 0.48 1

Parameter Estimates

95.0% Wald Confidence Interval

Variable Estimate Std. Err. Lower Conf. Limit Upper Conf. Limit

alpha 63.7164 10.2688 43.5899 83.8429

intercept 21.8193 1.88737 18.1201 25.5185

v

- 14.7416 3.24394

- 21.0996

- 8.38356

n 1 NA

k 42.3243 33.205

- 22.7563 107.405

NA

- Indicates that this parameter has hit a bound

implied by some inequality constraint and thus

has no standard error.

Table of Data and Estimated Values of Interest

Dose N Obs Mean Est Mean Obs Std Dev Est Std Dev Scaled Res.

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

0 16 21.8 21.8 7 7.98 0.0144

10.93 11 18.6 18.8 10.8 7.98

- 0.093

31.99 12 16 15.5 9.3 7.98 0.248

72.68 17 11.8 12.5 7 7.98

- 0.358

153.9 7 11.9 10.3 9 7.98 0.538

316.4 14 8.5 8.82 7.6 7.98

- 0.149

Model Descriptions for likelihoods calculatedPage 59 of 61 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)} = Sigma^2

Model A3 uses any fixed variance parameters that

were specified by the user

Model R: Yi = Mu + e(i)

Var{e(i)} = Sigma^2

Likelihoods of Interest

Model Log(likelihood) # Param's AIC

A1

- 198.190188 7 410.380377

A2

- 196.288688 12 416.577376

A3

- 198.190188 7 410.380377

fitted

- 198.446021 4 404. 892043

R

- 209.919649 2 423.839298

Explanation of Tests

Test 1: Do responses and/or variances differ among Dose levels?

(A2 vs. R)

Test 2: Are Variances Homogeneous? (A1 vs A2)

Tes t 3: Are variances adequately modeled? (A2 vs. A3)

Test 4: Does the Model for the Mean Fit? (A3 vs. fitted)

(Note: When rho=0 the results of Test 3 and Test 2 will be the same.)

Tests of Interest

Test

- 2*log(Likelihood Ratio) Test df p

- value

Test 1 27.2619 10 0.002367

Test 2 3.803 5 0.5781

Test 3 3.803 5 0.5781

Test 4 0.511666 3 0.9163

The p

- value for Test 1 is less than .05. There appears to be aPage 60 of 61 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 greater than .1. The model chosen seems

to adequately describe th e data

Benchmark Dose Computation

Specified effect = 1

Risk Type = Estimated standard deviations from the control mean

Confidence level = 0.95

BMD = 49.9819

BMDL = 15.5588Page 61 of 61 5 10 15 20 25

0 50 100 150 200 250 300Mean Response

dose Hill Model with 0.95 Confidence Level

12:45 07/30 2010 BMD BMDL Hill