Question

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

245245

accurate orders and

6666

that were not accurate.

a. Construct a

9090​%

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

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

9090​%

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

0.1920.192less than<pless than<0.2750.275.

What do you​ conclude?

Answer 1

Total orders = 245 + 66 = 311

a) The sample proportion of orders that are not accurate is computed here as:
p = x/n = 66/311 = 0.2122

From standard normal tables, we have:
P( -1.645 < Z < 1.645) = 0.9

Therefore the 90% confidence interval here is computed as:

Therefore 17.41%, 25.03% is the required 90% confidence interval here for the percentage of orders that are not accurate.

b) For restaurant B, we have here:

0.192 < p < 0.275

The confidence interval for B does not wholly lie above the confidence interval for A nor does the whole confidence interval lies below the lower value for confidence interval for A.

​​​​​​​Therefore we cannot conclude anything from the given interval.