A random sample of students at a college reported what they
Solution
Consider the table:
Calculating the standard deviation of the differences (third column):              
               
 s =    3.237304581          
               
 Thus, the standard error of the difference is sD = s/sqrt(n):              
               
 sD =    0.835868449          
For the   0.95   confidence level,      
               
 alpha/2 = (1 - confidence level)/2 =    0.025       As df = n - 1 =    14          
t(alpha/2) =    2.144786688          
               
 lower bound = [X1 - X2] - t(alpha/2) * sD =    -2.387426188          
 upper bound = [X1 - X2] + t(alpha/2) * sD =    1.198092855          
               
 Thus, the confidence interval is              
               
 (   -2.387426188   ,   1.198092855   ) [ANSWER]
If the order has to be the other way around, then please answer (-1.198, 2.387).
****************************
YES, IT INCLUDES 0, AS YOU CAN SEE.
| 165.1 | 168 | 2.9 | 
| 170.18 | 173 | 2.82 | 
| 185.42 | 183 | -2.42 | 
| 172.72 | 164 | -8.72 | 
| 190.5 | 189 | -1.5 | 
| 170.18 | 171 | 0.82 | 
| 177.8 | 174 | -3.8 | 
| 190.5 | 193 | 2.5 | 
| 177.8 | 177 | -0.8 | 
| 177.8 | 178 | 0.2 | 
| 182.88 | 181 | -1.88 | 
| 177.8 | 176 | -1.8 | 
| 167.64 | 170 | 2.36 | 
| 172.72 | 170 | -2.72 | 
| 182.88 | 186 | 3.12 | 

