Question

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 238 accurate orders and 58 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: 0.174 less than p less than 0.259. What do you​ conclude?

Answer 1

Restaurant A had 238 accurate orders and 58 that were not accurate.

Hence total number of orders = n= 238+58 = 296

x = number of not accurate orders = 58

The percentage of orders that are not accurate is :

Hence the 90% confidence interval is given by,

c = 0.90 ,

Zc =

Hence margin of error is,

= 0.04

Hence the 90% confidence interval is given by,

( 0.20 - 0.038 , 0.20 + 0.038 )
( 0.162 , 0.238 )

0.162 < P < 0.238

b)

The 90​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B is :

0.174 < P < 0.259

Here we can see that 90% confidence interval for restaurant A is wider that restaurant B.

We can say that the percentage of orders that are not accurate for restaurant A can be greater than restaurant B.