Question

In: Statistics and Probability

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 211 accurate...

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 211 accurate orders and 73 that were not accurate.
a. Construct a 95​% confidence interval estimate of the percentage of orders that are not accurate. (Round to three decimal places as needed.)
b. Compare the results from part​ (a) to this 95​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: 0.235<p<0.325. What do you conclude?

Solutions

Expert Solution

(a)

n = 211

p = 73/211 = 0.3460

q = 1 - p = 0.6540

SE =

= 0.05

From Table, critical values of Z = 1.96

Confidence Interval:

0.3460 (1.96 X 0.0327)

= 0.3460 0.0642

= ( 0.282 ,0.410)

Confidence Interval:

0.282 < P < 0.410

(b)

Point estimate of the percentage of orders that are not accurate for Restaurant A = 0.3460

Confidence interval of the percentage of orders that are not accurate for Restaurant A = 0.282 < P < 0.410

Point estimate of the percentage of orders that are not accurate for Restaurant B = (0.235 +0.325)/2 = 0.28

Confidence interval of the percentage of orders that are not accurate for Restaurant A = 0.235 < P < 0.325

Even though the Point estimate of Restaurant B is less than Restaurant A , we cannot conclude that the percentage of orders that are not accurate for Restaurant B are less than the percentage of orders that are not accurate for Restaurant A since the Confidence Intervals of both overlap.


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