Construct a 90 confidence interval for the mean difference D
Construct a 90% confidence interval for the mean difference D.. (Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)
Using the confidence interval, test whether the mean difference differs from zero.
| A sample of 31 paired observations generates the following data:   | 
Solution
Note that              
               
 Lower Bound = dbar - z(alpha/2) * s / sqrt(n)              
 Upper Bound = dbar + z(alpha/2) * s / sqrt(n)              
               
 where              
               
 X = sample mean =    2          
 z(alpha/2) = critical z for the confidence interval =    1.644853627          
 s = sample standard deviation =    2          
 n = sample size =    31          
               
 Thus,              
               
 Lower bound =    1.409151135          
 Upper bound =    2.590848865          
               
 Thus, the confidence interval is              
               
 (1.41,2.59) [ANSWER, PART A]
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As 0 is not a part of this interval, then:
OPTION 1: The mean difference differs from zero.

