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

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