Question

Q19. A major dish network chain is considering opening a new office in an area that currently does

         not have any office to serve residents of that area. The chain will open store only if more than

         7,200 of the 24,000 households in the area shows interest to get dish network installation in their

         houses. A telephone poll of 625 randomly selected households in the area shows that 425

         households are not interested in the dish network installations. Using 95% confidence level, the

         chain should

          a. use sample proportion 0.68 for making decision to open the office  

          b. not open its office in the area        

        c. use Z_α=1.96 to make decision to open the office     

          d. consider all of a, b, c suggestions for making a decision to open office

Answer 1

p=7200/24000=0.3

Ho :   p =    0.3                  
H1 :   p >   0.3       (Right tail test)          
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   200                  
Sample Size,   n =    625                  
                          
Sample Proportion ,    p̂ = x/n =    0.3200                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.01833                  
Z Test Statistic = ( p̂-p)/SE = (   0.3200   -   0.3   ) /   0.0183   =   1.0911
                          
critical z value =        1.645 [Excel function =NORMSINV(α)              
                          

Decision: test stat < critical value ,do not reject null hypothesis                       

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answer:   b. not open its office in the area