It is believed that 25% of U. S. homes have a direct satellite television receiver. How large a sample is necessary to estimate the true population of homes which do with 95% confidence and within 3 percentage points?

A sample of 801 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the z-score that has a p-value of .

The margin of error is of:

95% confidence level

So , z is the value of Z that has a p-value of , so .

25% of U.S. homes have a direct satellite television receiver.

This means that

How large a sample is necessary to estimate the true population of homes which do with 95% confidence and within 3 percentage points?

This is n for which M = 0.03. So

Rounding up:

A sample of 801 is needed.

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is symmetric about the line y = 120,000.

step-by-step explanation: