Question

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 295 accurate orders and 55 that were not accurate. a. Construct a 95​% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part​ (a) to this 95​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: 0.138less thanpless than0.211. What do you​ conclude? a. Construct a 95​% confidence interval. Express the percentages in decimal form.

Answer 1

Answer: (a)  $14.5\%<p<23.5\%$

(b) Restaurant B has a significantly lower percentage of orders that are not accurate.

Step-by-step explanation:

Confidence interval for population proportion is given by :-

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

Given: Significance level : $\alpha: 1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

For Restaurant A , The proportion of orders not accurate : $p=\dfrac{55}{295}\approx0.19$

Then , the Confidence interval for population proportion will be :-

$0.19\pm (1.96)\sqrt{\dfrac{0.19(1-0.19)}{295}}\\\\=0.19\pm0.045=(0.145,0.235)=(14.5\%,23.5\%)\\\\\text{i.e.}14.5\%<p<23.5\%$

Also, 95​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B:$0.138<p<0.211$

By comparing both the lower confidence limit and upper confidence limit of the interval for Restaurant B is lower than the lower confidence limit of t and the upper confidence limit of the interval for Restaurant A.

Therefore, Restaurant B has a significantly lower percentage of orders that are not accurate.