Question

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

343343

accurate orders and

5252

that were not accurate.

a. Construct a

9090​%

confidence interval estimate of the percentage of orders that are not accurate.

b. Compare the results from part​ (a) to this

9090​%

confidence interval for the percentage of orders that are not accurate at Restaurant​ B:

0.1190.119less than<pless than<0.1740.174.

What do you​ conclude?

Answer 1

a)
sample proportion, = 0.1516
sample size, n = 343
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.1516 * (1 - 0.1516)/343) = 0.0194

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

Margin of Error, ME = zc * SE
ME = 1.64 * 0.0194
ME = 0.0318

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.1516 - 1.64 * 0.0194 , 0.1516 + 1.64 * 0.0194)
CI = (0.12 , 0.183)

b)

For B, CI is 0.119 < p < 0.174

As the CI of A and B overlaps there is nno differenc ein the proportion of inaccurate orders between two restaurants A and B