Question

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 270 accurate orders and 64 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: 0.165<p<0.259 What do you​ conclude?

a. Construct a 95% confidence interval. Express the percentages in decimal form.

Answer 1

a)
sample proportion, = 0.1916
sample size, n = 334
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.1916 * (1 - 0.1916)/334) = 0.0215

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.1916 - 1.96 * 0.0215 , 0.1916 + 1.96 * 0.0215)
CI = (0.149 , 0.234)

14.9% < p < 23.4%

b)
for B, CI is 0.165<p<0.259

As the CI of A and B overlaps there is no difference in the proportion of inaccurate orders between two restaurants A and B