Question

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.

Answer 1

A sample of at least 545 adults is needed.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

`\pi = 0.15`

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

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

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.03 = 1.96\sqrt{(0.15*0.85)/(n)}`

`0.03√(n) = 1.96√(0.15*0.85)`

`√(n) = (1.96√(0.15*0.85))/(0.03)`

`(√(n))^(2) = ((1.96√(0.15*0.85))/(0.03))^(2)`

`n = 544.23`

Rounding up

A sample of at least 545 adults is needed.