In a study of the accuracy of fast food drive-through orders, Restaurant A had
accurate orders and
that were not accurate.
a. Construct a
confidence interval estimate of the percentage of orders that are not accurate.
b. Compare the results from part (a) to this
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?
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)
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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