Question

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 Zα=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

The sample proportion here is computed as:
p = 425 / 625 = 0.68

Now for 95% confidence level, we have from the standard normal tables:
P( -1.96 < Z < 1.96) = 0.95

Therefore the lower bound of the confidence interval for proportion here is computed as:

Now the required proportion here is computed as:
P = 7200/24000 = 0.3

As the lower bound of confidence interval is greater than 0.3, therefore the store should be opened here and we use the z = 1.96 to make the decision here.

Therefore c is the correct answer here.