Question

A television network maintains that switching its highest rated situation comedy (sitcom) from Monday to Friday evening will have no effect on the show’s popularity, which is currently 92% of the viewing audience. The sponsor believes the change will damage the program’s rating. A trial Friday evening is agreed to and 85 homes in which someone is watching television are randomly surveyed. Of these viewers, 74 of them are watching the sitcom. At the 0.05 significance level, can we conclude that proportion of viewers is less than 92%?

1. What is the decision rule? (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

2. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round the intermediate values and final answer to 2 decimal places.)

3. What is your decision regarding the null hypothesis?

Answer 1

SOLUTION :

Given,

n= 85

x= number of homes watching a sitcom = 74

= 0.05 Level of significance.

The hypothesis regarding this problem will be

Ho : p= 0.92. Against Ha : p < 0.92

Hence The test statistic is,

  

Z_c_a_l= -1.70

One tailed test so

= 1.64.............( by using critical value of Z)

Decision : | Z | >

i.e. 1.70 > 1.64

Reject Ho at 0.05 level of significance.

Conclusion : There is sufficient evidence to conclude that propertion of viewers is less than 92%.