Question

In a study of the accuracy of fast food drive-through orders, Restaurant A had 296 accurate orders and66 that were not accurate.

a. Construct a 90% confidence interval estimate of the percentage of orders that are not accurate. Express the percentages in decimal form.

b. Compare the results from part (a) to this 90 % confidence interval for the percentage of orders that are not accurate at Restaurant B:0.165 <p <0.237 What do you conclude?

Answer 1

sample proportion, pcap = 0.223
sample size, n = 296
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.223 * (1 - 0.223)/296) = 0.0242

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 = (0.223 - 1.64 * 0.0242 , 0.223 + 1.64 * 0.0242)
CI = (0.183 , 0.263)

b)
As the CI for restaurant B overlaps with calculated CI for restaurant A, we can conclude that there is no difference in the proportion of orders that are not accurate for Restaurant A and B