Question

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

241241

accurate orders and

6565

that were not accurate.

a. Construct a

9595%

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

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

9595%

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

0.1860.186less than<pless than<0.2780.278.

What do you conclude?

Answer #1

a)

sample proportion, pcap = 0.2124

sample size, n = 306

Standard error, SE = sqrt(pcap * (1 - pcap)/n)

SE = sqrt(0.2124 * (1 - 0.2124)/306) = 0.0234

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 = (0.2124 - 1.96 * 0.0234 , 0.2124 + 1.96 * 0.0234)

CI = (0.1665 , 0.2583)

b)

The calculated CI operlaps with (0..186 < 0.278)

There is not sufficient evidence to conclude that the proportion is
different for two restaurants

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confidence interval for the percentage of orders that are not
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