A retailer wants to estimate the average amount spent by customers on holiday shopping. In a random sample of 50 customers, the average amount spent was \$250, and the population standard deviation $\sigma$ is known to be \$40. Construct and interpret an 80% confidence interval for the average amount spent by all customers.

• A. (210.00, 290.00)
• B. (248.976, 251.024)
• C. (242.76, 257.24)
• D. (248.72, 251.28)

Answer: (C)

Verified step by step guidance

1. Identify the sample mean (\( \bar{x} \)) which is given as \$250, the sample size (n) which is 50, and the population standard deviation (\( \sigma \)) which is \$40.
2. Determine the z-score corresponding to the desired confidence level. For an 80% confidence interval, the z-score is approximately 1.28. This value can be found using a standard normal distribution table or calculator.
3. Calculate the standard error of the mean using the formula \( \text{SE} = \frac{\sigma}{\sqrt{n}} \). Substitute \( \sigma = 40 \) and \( n = 50 \) into the formula to find the standard error.
4. Construct the confidence interval using the formula \( \bar{x} \pm z \times \text{SE} \). Substitute the sample mean, z-score, and standard error into the formula to find the lower and upper bounds of the confidence interval.
5. Interpret the confidence interval: The interval provides a range within which we can be 80% confident that the true average amount spent by all customers falls. This means that if we were to take many samples and construct confidence intervals in the same way, approximately 80% of those intervals would contain the true average amount spent.