Question

An automobile manufacturer would like to know what proportion of its customers are not satisfied by the service provided by the local dealer. The customer relations department will survey a random sample of customers and compute a​ 95% confidence interval for the proportion who are not satisfied. Past studies suggest this proportion will be about 0.15. Find the sample size needed if the margin of error of the confidence interval is to be about 0.04

Answer 1

Answer:

The sample size needed if the margin of error of the confidence interval is to be about 0.04 is 18.

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

Past studies suggest this proportion will be about 0.15

This means that $p = 0.15$

Find the sample size needed if the margin of error of the confidence interval is to be about 0.04

This is n when M = 0.04. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.15*0.85}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.15*0.85}$

$\sqrt{n} = \frac{1.96\sqrt{0.15*0.85}}{0.04}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.15*0.85}}{0.04})^{2}$

$n = 17.5$

Rounding up

The sample size needed if the margin of error of the confidence interval is to be about 0.04 is 18.