In an examination of holiday spending (known to be normally distributed) of a sample of 16 holiday shoppers at a local mall, an average of $54 was spent per hour of shopping. Based on the current sample, the standard deviation is equal to $21. Find a 90% confidence interval for the population mean level of spending per hour.
A) [$44.83, $63.17]
B) [$44.79, $63.20]
C) [$45.36, $62.63]
D) [$46.96, $61.04]

Question

Quizplus · Accepted Answer

The 90% confidence interval is calculated using the formula: mean ± (t-value * (standard deviation/√n)). With a sample size of 16, the t-value for 90% confidence is approximately 1.753. Thus, the interval is $54 ± (1.753 * (21/√16)) = [$44.79, $63.20].

In an Examination of Holiday Spending (Known to Be Normally