Skipping Lunch A nutritionist wishes to determine, within 3%, the true proportion of adults who do not eat any lunch. If he wishes to be 95% confident that his estimate contains the population proportion, how large a sample will be necessary? A previous study found that 15% of the 125 people surveyed said they did not eat lunch.

A sample of at least 545 adults 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 zscore that has a pvalue of .

For this problem, we have that:

The margin of error is:

95% confidence level

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

If he wishes to be 95% confident that his estimate contains the population proportion, how large a sample will be necessary?

We need a sample of at least n.

n is found when M = 0.03. So

Rounding up

A sample of at least 545 adults is needed.