Question

In a sample of 70 stores of a certain​ company, 62 violated a scanner accuracy standard. It has been demonstrated that the conditions for a valid​ large-sample confidence interval for the true proportion of the stores that violate the standard were not met. Determine the number of stores that must be sampled in order to estimate the true proportion to within 0.04 with 95​% confidence using the​ large-sample method.

Answer 1

Answer:

We need at least 243 stores.

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 of the interval is:

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

For this problem, we have that:

$n = 70, p = \frac{62}{70} = 0.886$

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 $Z = 1.96$.

Determine the number of stores that must be sampled in order to estimate the true proportion to within 0.04 with 95​% confidence using the​ large-sample method.

We need at least n stores.

n is found when M = 0.04. So

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

$0.04 = 1.96\sqrt{\frac{0.886*0.114}}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.886*0.114}$

$\sqrt{n} = \frac{1.96\sqrt{0.886*0.114}}{0.04}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.886*0.114}}{0.04})^{2}$

$n = 242.5$

Rounding up

We need at least 243 stores.