# AP Statistics Curriculum 2007 StudentsT

### From Socr

##  General Advance-Placement (AP) Statistics Curriculum \- Student's T Distribution

Very frequently in practice, we do not know the population variance. Therefore we need to estimate it using the sample-variance. This requires us to introduce the T-distribution, which is a one-parameter distribution connecting . 

###  Student's T Distribution

The Student's t-distribution arises in the problem of estimating the mean of a normally distributed population when the sample size is small and the population variance is unknown. It is the basis of the popular Student's t-tests for the statistical significance of the difference between two sample means, and for confidence intervals for the difference between two population means. 

Suppose _X_ 1, ..., _X_ _n_ are independent random variables that are Normally distributed with expected value μ and variance σ2. Let 

     be the sample mean, and 

     be the sample variance. We already  discussed the following statistic: 
    

is normally distributed with mean 0 and variance 1, since the sample mean  is normally distributed with mean μ and standard deviation . 

Gosset studied a related quantity under the pseudonym _Student_ , 

    

which differs from _Z_ in that the (unknown) population standard deviation  is replaced by the sample standard deviation _S_ _n_. Technically,  has a Chi-square distribution  distribution. Gosset's work showed that _T_ has a specific probability density function, which approaches Normal(0,1) as the degree of freedom (df=sample-size -1) increases. 

###  Computing with T-distribution

  * You can see the discretized T-table or 
  * Use the interactive SOCR T-distribution or 
  * Use the high precision T-distribution calculator. 



###  Example

Suppose a researcher wants to examine CD4 counts for HIV(+) patients seen at his clinic. She randomly selects a sample of n = 25 HIV(+) patients and measures their CD4 levels (cells/uL). Suppose she obtains the following results and we are interested in calculating a 95% (α = 0.05) confidence interval for μ: 

Variable | N |  N* |  Mean | SE of Mean| StDev | Minimum |  Q1|  Median |  Q3 | Maximum   
---|---|---|---|---|---|---|---|---|---|---  
CD4 |  25|  0 | 321.4|  14.8 |  73.8 | 208.0 | 261.5 |  325.0 | 394.0 |  449.0   
  
What do we know from the background information? 

    
     _s_ = 73.8
     _S_ _E_ = 14.8
     _n_ = 25

    

    

    

     [290.85,351.95]

####  CI Interpretation

Still, does this CI (290.85, 351.95) mean anything to us? Consider the following information: The U.S. Government classification of AIDS has three official categories of CD4 counts – asymptomatic = greater than or equal to 500 cells/uL 

  * AIDS related complex (ARC) = 200-499 cells/uL 
  * AIDS = less than 200 cells/uL 
  * Now how can we interpret our CI? 



###  SOCR CI Experiments

The SOCR Confidence Interval Experiment provides empirical evidence that the definition and the construction protocol for Confidence intervals are consistent. 

###  Activities

  * A biologist obtained body weights of male reindeer from a herd during the seasonal round-up. He measured the weight of a random sample of 102 reindeer in the herd, and found the sample mean and standard deviation to be 54.78 kg and 8.83 kg, respectively. Suppose these data come from a normal distribution. Calculate a 99% confidence interval. 


  * Suppose the proportion of blood type O in the population is 0.44. If we take a random sample of 12 subjects and make a note of their blood types. What is the probability that exactly 6 subjects have type O blood type in the sample? 



####  Approach I (exact)

     _P_(_X_ = 6) = ? Where 
     , using SOCR Binomial interactive GUI or calculator. 

####  Approach II (Approximate)

     . 
     . , where  and _X_ 1 = 5.5, _X_ 2 = 6.5. So, 

####  Approach III (Approximate)

     . The sample proportion is . Thus, , where _p_ 1 = 0.5 − 1 / 24 and _p_ 2 = 0.5 + 1 / 24. Note that approach II is very similar to approach III, however, the former uses the total sum of successes (X), whereas the latter employs the proportion (X/n). This is why the left-right additive term of 0.5 in approach II becomes a 0.5*(1/12) = 1/24 in the III approximation. Finally, standardize each of the 2 limits (_p_ 1 and _p_ 2), using the _Z_ = (_p_ − 0.44) / 0.1433 transformation, to get