Question

A local outdoor equipment store is being sold. The buyers are trying to estimate the percentage of items that are outdated. They will randomly sample among its 100,000 items in order to determine the proportion of merchandise that is outdated. The current owners have never determined their outdated percentage and can not help the buyers. Approximately how large a sample do the buyers need in order to insure that they are 98 % confident that the margin of error is within 3%

Answer 1

Using the z-distribution, as we are working with a proportion, it is found that the buyers need a sample of 1505.

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

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

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

The margin of error is given by:

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

In this problem, we have a 98% confidence level, hence$\alpha = 0.98$, z is the value of Z that has a p-value of $\frac{1+0.98}{2} = 0.99$, so the critical value is z = 2.327.

There is no estimate of the percentage, hence $\pi = 0.5$, and the margin of error is of M = 0.03, hence we solve for n to find the desired sample size.

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

$0.03 = 2.327\sqrt{\frac{0.5(0.5)}{n}}$

$0.03\sqrt{n} = 2.327(0.5)$

$\sqrt{n} = \frac{2.327(0.5)}{0.03}$

$(\sqrt{n})^2 = \left(\frac{2.327(0.5)}{0.03}\right)^2$

$n = 1504.2$

Rounding up, the buyers need a sample of 1505.

More can be learned about the z-distribution at brainly.com/question/25890103