Question

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 277 accurate orders and 67 that were not accurate. a. Construct a 90​% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part​ (a) to this 90​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: 0.174less thanpless than0.244. What do you​ conclude?

Answer 1

Answer:

Step-by-step explanation:

Given that in a study of the accuracy of fast food​ drive-through orders, Restaurant A had 277 accurate orders and 67 that were not accurate.

Sample proportion for non accurate =$\frac{67}{277} =0.2419$

Std error = $\sqrt{\frac{pq}{n} }$

Margin of error = 1.645 * std error

Confidence Interval:

±4.23

Range for the true population proportion:

19.96% to 28.42%

b) 17.40 to 24.40 is given. Mid value is 20.9% which must have been the sample proportion.

24.19% observed proportion lies in both the intervals.