Question

Some fast food chains offer a lower priced combination meal in an effort to attract budget conscious customers. One chain test-marketed a burger, fries, and drinks combination for $1.71. The weekly sales volumes for these meals was impressive. Suppose the chain wants to estimate the average amount its customers spent on a meal at their restaurant while this combination offer was in effect. An analyst gathers data from 28 randomly selected customers. The following data presents the sample meal totals.

$ 3.21, 5.40, 3.50, 4.39, 5.60, 8.65, 5.02, 4.20, 1.25, 7.64, 3.28, 5.75, 3.26, 3.80, 5.46, 9.87, 4.67, 5.86, 3.73, 4.08 5.47, 4.49, 5.19, 5.82, 7.62, 4.83, 8.42, 9.10

Use the data to construct a 90% confidence interval to estimate the population mean value. Assume the amounts spent are normally distributes.

Answer 1

To construct a 90% confidence interval to estimate the population mean value, we can use the provided sample data. The 90% confidence interval is ($4.638, $5.542).

Step-by-step explanation:

To construct a 90% confidence interval to estimate the population mean value, we can use the sample data provided. Firstly, we calculate the sample mean and the standard error. The sample mean is $5.09 and the standard error is approximately $0.238. Using the t-distribution with degrees of freedom equal to n - 1 (28 - 1 = 27), we find the critical value for a 90% confidence interval, which is approximately 1.703. Finally, we can calculate the confidence interval by subtracting and adding the margin of error from the sample mean. The margin of error is approximately $0.452. Therefore, the 90% confidence interval to estimate the population mean value is ($4.638, $5.542).