Question

A random sample of 200 shoppers at a local grocery store found that 135 of the 200 sampled

Answer 1

Using the z-distribution, as we are working with a proportion, it is found that samples of 937 should be taken.

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 that:

• The estimate is of $\pi = 0.675$.

• The margin of error is of M = 0.03.

• 95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so $z = 1.96$.

Then, we solve for n to find the minimum sample size.

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

$0.03 = 1.96\sqrt{\frac{0.675(0.325)}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.675(0.325)}$

$\sqrt{n} = \frac{1.96\sqrt{0.675(0.325)}}{0.03}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.675(0.325)}}{0.03}\right)^2$

$n = 936.4$

Rounding up, it is found that samples of 937 should be taken.

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