Question

Q18. Health Department, NYC plans to release a new drug commercially that is expected to lower

         blood pressure effectively. The drug was administered to a random sample of 20 patients.

         Among these patients the decrease in the blood pressure of 7 patients was recorded and no

         change in blood pressure of other patients was recorded . The decrease in the blood pressure of

         the 7 patients was 21, 22, 16, 19, 20, 23, 19. At 1% level of significance, to support the

         release of drug, Health Department used

         a. mean decrease in blood pressure =20 for test of hypothesis

         b. Critical value t_α/2 = 3.707 was valid value to compare t_STAT

         c. Degree of freedom = 6 were used to get t_α/2 for the conclusion                     

        d none of the above a, b, c are correct for conclusion       

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

18)

         d none of the above a, b, c are correct for conclusion        

19)

required sample proportion = (625-425)/625= 0.32

critical z value =        1.645
Ho :   p =    0.3                  
H1 :   p >   0.3       (Right tail test)          
                                
Standard Error ,    SE = √( p(1-p)/n ) =    0.01833                  
Z Test Statistic = ( p̂-p)/SE = (   0.3200   -   0.3   ) /   0.0183   =   1.0911

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answer: b. not open its office in the area