In an examination of purchasing patterns of shoppers, a sample of 20 shoppers revealed that they spent, on average, $54 per hour of shopping. based on previous years, the population standard deviation is thought to be $21 per hour of shopping. assuming that the amount spent per hour of shopping is normally distributed, find a 90% confidence interval for the mean amount.

Confidence interval for the mean amount = 54+1.645*21/sqrt(16) =(62.64 , 45.36)

Explanation:

confidence interval = mean + z*, where z* is the upper (1-C)/2 critical value for the standard normal distribution.

z score for 90% confidence interval = 1.645

confidence interval for the mean amount = 54+1.645*21/sqrt(16) =(62.64 , 45.36)

u = 30

continuing your method

= +

subtract from both sides

= - thus

= -

= ⇒ u = 30