In a study of the accuracy of

Answered question

2022-04-06

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

302

accurate orders and

69

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:

0.164<p<0.238.

What do you​ conclude?

 

Answer & Explanation

alenahelenash

alenahelenash

Expert2023-04-27Added 556 answers

To construct a 90% confidence interval estimate of the percentage of orders that are not accurate in Restaurant A, we can use the following formula:
p^±zα/2p^(1p^)n
where p^ is the sample proportion of inaccurate orders, zα/2 is the critical value from the standard normal distribution corresponding to a 90% confidence level (which can be found using a table or a calculator), and n is the sample size.
Here, p^=69302+690.1863. The critical value zα/2 for a 90% confidence level is approximately 1.645 (again, this can be found using a table or a calculator). The sample size n=302+69=371.
Substituting these values into the formula, we get:
0.1863±1.6450.1863(10.1863)371
Simplifying, we get:
0.1863±0.0457
Therefore, the 90% confidence interval estimate of the percentage of orders that are not accurate in Restaurant A is approximately:
(0.1406,0.2320)
We can interpret this as follows: if we were to repeat this study many times and construct a 90% confidence interval for each one, we would expect about 90% of those intervals to contain the true percentage of inaccurate orders in Restaurant A.

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