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

(45.6965,63.2035)

Step-by-step explanation:

Given that 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.

i.e.

Margin of error = t critical * std error=9.2035

df = 15

(Here since sample std dev is known we use t critical value)

Confidence interval lower bound =

Upper bound =

98%

step-by-step explanation: