Question

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 245 accurate orders and 71 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: 0.204< p < 0.288. What do you​ conclude?

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

Answer 1

a)
sample proportion, = 0.2247
sample size, n = 316
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.2247 * (1 - 0.2247)/316) = 0.0235

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.2247 - 1.64 * 0.0235 , 0.2247 + 1.64 * 0.0235)
CI = (0.186 , 0.263)

b)
As the calculated CI overlaps with the CI for restaurant B, we can conclude that there is not significant difference between two proportions