Some fast-food chains offer a lower-priced combination meal in an effort to attract budget-conscious customers. One chain tested a burger, chips and drink combination for $6.95. The weekly sales volume for these meals was impressive. Suppose the chain wants to estimate the average amount its customers spent (in dollars) 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 represent the sample meal totals in dollars. The average spending from the sample is $7.87 and the sample standard deviation is $0.87.

Use the above information to construct a 90% confidence interval( Upper and lower Limit ) to estimate the population mean value in dollars. Assume that the amounts spent are normally distributed.

Question

Some fast-food chains offer a lower-priced combination meal in an effort to attract budget-conscious customers. One chain tested a burger, chips and drink combination for $6.95. The weekly sales volume for these meals was impressive. Suppose the chain wants to estimate the average amount its customers spent (in dollars) 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 represent the sample meal totals in dollars. The average spending from the sample is $7.87 and the sample standard deviation is $0.87.

Use the above information to construct a 90% confidence interval( Upper and lower Limit )  to estimate the population mean value in dollars. Assume that the amounts spent are normally distributed.

BlackTom AI · Accepted Answer

The task is to construct a 90% confidence interval for the population mean using the given sample data.
First, note the sample size n = 28, sample mean x̄ = 7.87, and sample standard deviation s = 0.87. Since the population distribution is assumed normal and the sample size is moderate, we use the t-distribution with df = n − 1 = 27 to form the confidence interval.
Next, determine the t critical value for a 90% two-tailed interval. The two-tailed alpha is 0.10, so each tail has 0.05. The critical value t_{0.95,27} is approximately 1.703.
Compute the standard error of the mean: SE = s / sqrt(n) = 0.87 / sqrt(28). sqrt(28) ≈ 5.2915, so SE ≈ 0.87 / 5.2915 ≈ 0.1645.
Now calculate the margin of error: ME = t*SE ≈ 1.703 × 0.1645 ≈ 0.28 (to two decimals, about 0.28).
Construct the confidence interval around the sample mean: lower limit = x̄ − ME ≈ 7.87 − 0.28 = 7.59, upper limit = x̄ + ME ≈ 7.87 + 0.28 = 8.15.
Thus, the 90% confidence interval for the population mean spending is approximately [7.59, 8.15].

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