Example 1 Two school systems that are in similar communities

Example 1:

Two school systems that are in similar communities differ in that system A has comprehensive sex education programs starting in first grade and continuing through high school. System B has no formal program but is considering adopting the same program as system A. Those who are opposed say that the program will lead to earlier sexual experience. To provide data to support or refute this argument, random samples of 15 tenth-graders are chosen from each school population. They are asked to provide answers to questions about their sexual experience in a setting that ensures anonymity. Based on their reports, a mean level of ‘sexual experience’ is derived for each group (higher scores indicate more sexual experience). Assuming a = .05, are there differences between the two school districts?

School District A

School District B

= 45

= 55

s_A = 10

s_B = 12

n_A = 15

n_B = 15

Solution

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   =   0  
Ha:   u1 - u2   =/   0  
At level of significance =    0.05          
As we can see, this is a    two   tailed test.      
Calculating the means of each group,              
              
X1 =    45          
X2 =    55          
              
Calculating the standard deviations of each group,              
              
s1 =    10          
s2 =    12          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    15          
n2 = sample size of group 2 =    15          
Thus, df = n1 + n2 - 2 =    28          
Also, sD =    4.03319559          
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    -2.479423518          
              
where uD = hypothesized difference =    0          
              
Now, the critical value for t is              
              
tcrit =    +/-   2.048407142      
              
              
Also, using p values,              
              
p =    0.019446542          
              
As |t| > 2.0484, and P < 0.05,    WE REJECT THE NULL HYPOTHESIS.          

Thus, there is significant difference between the mean level of sexual experience between the two school districts at 0.05 level. [CONCLUSION]

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Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!