Question

According to a national​ study, 38​% of taxpayers used computer software to do their taxes for a certain tax year. A sample of 145 taxpayers was selected. Complete parts a through d below. a. Calculate the standard error of the proportion. sigma Subscript pequals nothing ​(Round to four decimal places as​ needed.) b. What is the probability that less than 34​% of the taxpayers from the sample used computer software to do their​ taxes? ​P(Less than 34​% of the taxpayers sampled used computer ​software)equals nothing ​(Round to four decimal places as​ needed.) c. What is the probability that between 32​% and 43​% of the taxpayers from the sample used computer software to do their​ taxes? ​P(Between 32​% and 43​% of the taxpayers sampled used computer ​software)equals nothing ​(Round to four decimal places as​ needed.) d. What impact would changing the sample size to 245 taxpayers have on the results of parts​ a, b, and​ c? Choose the correct answer below. A. The standard error would be​ reduced, which​ would, in​ turn, reduce the probabilities that the sample proportions will be closer to the population proportion. B. The standard error would be​ increased, which​ would, in​ turn, reduce the probabilities that the sample proportions will be closer to the population proportion. C. The standard error would be​ reduced, which​ would, in​ turn, increase the probabilities that the sample proportions will be closer to the population proportion. D. The standard error would be​ increased, which​ would, in​ turn, increase the probabilities that the sample proportions will be closer to the population proportion. E. Changing the sample size would have no effect on the standard error or the probabilities that the sample proportions will be closer to the population proportion.

Answer 1

a)

p=0.38

n-145

Standard Error ,    SE = √( p(1-p)/n ) =    0.0403

b)

population proportion ,p=   0.38                          
n=   145                          
                              
std error , SE = √( p(1-p)/n ) =    0.040                          
                              
sample proportion , p̂ =   0.34                          
Z=( p̂ - p )/SE= (   0.340   -   0.38   ) /    0.040   =   -0.992  
P ( p̂ <    0.340   ) =P(Z<( p̂ - p )/SE) =                      
                              
=P(Z <    -0.992   ) =    0.1605                   (answer)

c)

                       population proportion ,p=   0.38                      
                       n=   145                      
                                                  
                       std error , SE = √( p(1-p)/n ) =    0.0403                      
                                                  
                       we need to compute probability for                           
                       0.32   < p̂ <   0.43                  
                                                  
                       Z1 =( p̂1 - p )/SE= (   0.32   -   0.38   ) /    0.0403   =   -1.488
                       Z2 =( p̂2 - p )/SE= (   0.43   -   0.38   ) /    0.0403   =   1.240
P(   0.32   < p̂ <   0.43   ) =    P[( p̂1-p )/SE< Z <(p̂2-p)/SE ]    =P(    -1.488   < Z <   1.240   )          
                                                  
= P ( Z <   1.240   ) - P (    -1.488   ) =    0.8926   -   0.068   =   0.8243 (answer)

d)

              The standard error would be​ reduced, which​ would, in​ turn, increase the probabilities that the sample proportions will be closer to the population proportion