a statistics teacher wants to see if there is any differnce

a statistics teacher wants to see if there is any differnce in the abilities of students enrolled in statistics today and those enrolled five years ago. A sample of final examination scores from students enrolled today and from students enrolled five years ago was taken. You are given the following information.

X Bar=82 today and 88 five years ago

S2= 112.5 today, 54 five years ago

n= 45 today and 36 five years ago

what is the 95% confidence interval for the difference of the two population means?

Solution

Let the difference bet u1 - u2, where

u1 = today

u2 = 5 years ago

As  
              
X1 =    82          
X2 =    88          
              
Calculating the standard deviations of each group,              
              
s1 =    10.60660172          
s2 =    7.348469228          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    45          
n2 = sample size of group 2 =    36          
Thus, df = n1 + n2 - 2 =    79          

Thus,sD =    2          
              
              
For the   0.95   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.025          
t(alpha/2) =    1.99045021          
              
lower bound = [X1 - X2] - t(alpha/2) * sD =    -9.98090042          
upper bound = [X1 - X2] + t(alpha/2) * sD =    -2.01909958          
              
Thus, the confidence interval is              
              
(   -9.98090042   ,   -2.01909958   ) [ANSWER]