A company institutes an exercise break for its workers to se

A company institutes an exercise break for its workers to see if it will improve job satisfaction, as measured by a quedtionnaire that assesses workers satisfaction. Scores for 10 randomly selected workers before and after the implementation of exercise program arc shown in the table Test an appropriate hypothesis and state your conclusion. How big might that difference be? Check a 90% confidence interval.

Solution

1.

Let ud = u2 - u1.              
Formulating the null and alternative hypotheses,              
              
Ho:   ud   <=   0  
Ha:   ud   >   0  
At level of significance =    0.1          
As we can see, this is a    right   tailed test.      
              
Calculating the standard deviation of the differences (third column):              
              
s =    7.525210155          
              
Thus, the standard error of the difference is sD = s/sqrt(n):              
              
sD =    2.379680396          
              
Calculating the mean of the differences (third column):              
              
XD =    8.5          
              
As t = [XD - uD]/sD, where uD = the hypothesized difference =    0   , then      
              
t =    3.571908233          
              
As df = n - 1 =    9          
              
Then the critical value of t is              
              
tcrit =    +   1.833112933      
              
As t > 1.833,    WE REJECT THE NULL HYPOTHESIS.          
              
Also, using p values,              
              
p =        0.003003142      
              
As P < 0.1, WE REJECT THE NULL HYPOTHESIS.          
Thus, there is significant evidence that the mean job satisfaction after the institution of exercise break is greater. [CONCLUSION]
              
********************************************       2.      
              
For the   0.9   confidence level,      
              
alpha/2 = (1 - confidence level)/2 =    0.05          
t(alpha/2) =    1.833112933          
              
lower bound = [X1 - X2] - t(alpha/2) * sD =    4.13777709          
upper bound = [X1 - X2] + t(alpha/2) * sD =    12.86222291          
              
Thus, the confidence interval is              
              
(   4.13777709   ,   12.86222291   ) [ANSWER]