Question

According to a Harris Poll in 2009, 72% of those who drive and own cell phones say they use them to talk while they are driving. If you wish to conduct a survey in your city to determine what percent of the drivers with cell phones use them to talk while driving, how large a sample should be if you want your estimate to be within 0.02 with 95% confidence.

Answer 1

We need a sample of at least 1937.

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)`.

The margin of error is:

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

For this problem, we have that:

`\pi = 0.72`

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`.

How large a sample should be if you want your estimate to be within 0.02 with 95% confidence.

We need a sample of at least n.

n is found when M = 0.02. So

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

`0.02 = 1.96\sqrt{(0.72*0.28)/(n)}`

`0.02√(n) = 1.96√(0.72*0.28)`

`√(n) = (1.96√(0.72*0.28))/(0.02)`

`(√(n))^(2) = ((1.96√(0.72*0.28))/(0.02))^(2)`

`n = 1936.16`

Rounding up to the nearest number.

We need a sample of at least 1937.