Question

Suppose that a fast food restaurant decides to survey its customers to gauge interest in a breakfast menu. After surveying multiple people, the restaurant created a 95% confidence interval for the proportion of customers interested in a breakfast menu. The confidence interval is (0.688,0.766) . Use the confidence interval to find the point estimate and margin of error for the proportion. Give your answer precise to three decimal places.

Answer 1

Answer:

$\hat p = 0.727$ point of estimate for the proportion of customers interested in a breakfast menu

$ME= 0.039$ represent the margin of error.

Step-by-step explanation:

For this case we want to find a confidence interval for the proportion of customers interested in a breakfast menu.

The confidence interval for this case is given by this formula:

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And the margin of error is given by:

$ME = z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The 95% confidence for this case is given by:

$0.688 \leq p \leq 0.766$

We can find the point of estimate for the proportion using the fact that the interval is symmetrical

$\hat p = \frac{Lower+Upper}{2} =\frac{0.688+0.766}{2}=0.727$

And we can find the margin of error with this difference:

$ME = Upper- \hat p = 0.766-0.727 = 0.039$

Or equivalently with:

$ME = \hat p -Lower = 0.727- 0.688 =0.039$

So then the final answer for this case would be:

$\hat p = 0.727$ point of estimate for the proportion of customers interested in a breakfast menu

$ME= 0.039$ represent the margin of error.