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A local restaurant does not accept any reservations and only focuses on walk-in customers. The owner of the restaurant wants to find out how many walk-in customers he should plan to have every day. He kept records of the number of walk-in customers for the last 63 days. The average number of daily walk-in customers was 67.3, and the sample standard deviation is 7.1. Construct a 99% confidence interval for the number of customers on a given day.

a) True
b) False

User Kraymer
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Final answer:

To construct a 99% confidence interval for the number of walk-in customers, use the given sample mean (67.3), standard deviation (7.1), and the sample size (63). With these values, the confidence interval is approximately (65, 70).

Step-by-step explanation:

To construct a 99% confidence interval for the number of walk-in customers on a given day at the restaurant, we can use the sample mean, the sample standard deviation, and the size of the sample. Since the question does not provide the size of the sample (n), we will assume it refers to the recorded 63 days as the sample size. We can use the formula for the confidence interval for a normally distributed population:

CI = μ ± Z*(σ/√n)

Where μ is the sample mean, Z is the z-score corresponding to the desired confidence level (for 99%, the z-score is approximately 2.576 according to the z-table), σ is the sample standard deviation, and n is the sample size.

Let's plug in the values:

Sample mean (μ) = 67.3

Z-score for 99% confidence level = 2.576

Sample standard deviation (σ) = 7.1

Sample size (n) = 63

The calculation of the confidence interval is as follows:

CI = 67.3 ± 2.576*(7.1/√63)

Calculate the margin of error (ME):

ME = 2.576 * (7.1/√63) = 2.576 * 0.8944 ≈ 2.3039

Now, calculate the confidence interval:

Lower limit = 67.3 - 2.3039 = 64.9961

Upper limit = 67.3 + 2.3039 = 69.6039

Therefore, the 99% confidence interval for the number of customers is approximately (65, 70).

User Aakash Goplani
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