Question

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 253 accurate orders and

74 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: 0.2080 <p < 0.288

What do you​ conclude?

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

Answer 1

Ans:

a)n=74+253=327

sample proportion=74/(74+253)=0.2263

90% confidence interval for p

=0.2263+/-1.645*sqrt(0.2263*(1-0.2263)/327)

=0.2263+/-0.0381

=(0.1882, 0.2644)

or(18.82% , 26,44%)

b)Confidence intervals for both restaurants are overlapping,so percentage of orders that are not accurate is not different.