Question

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 315 accurate orders and 62 that were not accurate. a. Construct a 90​% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part​ (a) to this 90​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: 0.148 less than p less than 0.215. What do you​ conclude?

a. Construct a 90​% confidence interval. Express the percentages in decimal form.

Answer 1

Total number of orders ​​​​​​ 315+62=377

Hence the proportion of not accurate orders is

The confidence interval for the population proportion is given by

For 90% confidence interval also n=292​​​​​​

Hence the interval is

Hence 0.1331<p<0.1959

(13.31%,19.59%) is  90​% confidence interval.

b) given 0.148 <p <215.

  for restaurant B

0.1331<p<0.1959 for restaurant A

by looking the confidence interval we conclude that the  confidence interval for restaurant A is wider than the confidence interval for restaurant B