Two Sample T-Tests

1. Two sample t-tests are used to compare population means of two populations. For instance, when comparing a treatment group to a control group.
    
2. Hypotheses in two sample tests are also left-, right- or two-tailed. You must be more careful in two sample tests when determining which kind of hypothesis you are testing.
    
   For example, let be the population mean change in cholesterol levels of adolescent boys who took Lovastatin. Also, let be the population mean change in cholesterol levels of adolescent boys in a control group who took a placebo.
    
   1. The two-tailed hypothesis is
      
      H_o:  =
      H_a:  
       
      We could re-write these hypotheses as
       
      H_o:  - = 0
      H_a:  - 0
        
   2. Now consider the hypotheses
       
      H_o:  
      H_a:  <
       
      which can be re-written as
       
      H_o:  -   0
      H_a:  -  < 0
        
      Written this way this hypothesis is left-tailed. We must write the hypothesis in the second form to decide if they are left- or right-tailed.
       
3. The protocol for conducting two sample hypothesis tests are identical to those for one sample tests. The only real difference is the test statistic that should be used, which is:
    
   
   
   where
    
    
    
   This statistic has a t-distribution with n₂ + n₂ - 2 degrees of freedom
    

4. As an example we'll look at the Lovastatin data again.
    
   Let be the population mean change in cholesterol levels of adolescent boys who took Lovastatin. Let be the population mean change in cholesterol levels of adolescent boys in a control group who took a placebo.
    
   For the treatment group the sample mean was = -20, the sample sample variance was = 244 and the sample size was n₁ = 61. For the control group = -3, = 65 and n₂ = 49.
    
   We'll test if the reduction in cholesterol levels in the treatment group is greater than the control group. In other words:
    
   H_o:  
   H_a:  <
    
   or
    
   H_o:  -   0
   H_a:  -  < 0
    
   We'll use a significance level of = 0.05
    
   Before we can calculate the test statistic we must calculate = [(61 - 1)*244 + (49 - 1)*65]/(61 + 49 - 2) = 164.44
    
   So, t = [(-20 - -3) - 0]/sqrt[164.44*(1/61 + 1/49)] = -6.91
    
   This t-statistic has 61 + 49 - 2 degrees of freedom (108 df). Since this is greater than 30 we'll use the z-tables to find p-values.
    
   Since this is a left-tailed test the p-value is P(T < -6.91). Since 6.91 is off the bottom of the z-table the p-value is zero.
    
   Since the p-value is less than we reject the null hypothesis. We conclude that Lovastatin lowers cholesterol levels more than a placebo.
    

5. Assumptions: The assumptions of this test are:
    
   1. Both parent populations are normally distributed
       
   2. Random samples were taken from each population
       
   3. The population variances of the two populations are equal

E-mail Mr. Callahan at stat110@edcallahan.com with questions or comments about this web site or about the class itself.