Question

1. A sociologist conducted a study at a local grocery store to determine how much money shoppers spent on average. She sampled 27 shoppers and calculated the mean amount of money spent to be $95.5 with a standard deviation of $15.39.
   1. (5 points) Construct a 95% confidence interval for the mean amount of money spent at the grocery store.

2. (2 points) Three months prior to this study, the manager of the grocery store started an advertising campaign with the goal of increasing the mean amount of money spent to be at least $100. Based on the confidence interval from part (a), did the manager meet his goal? Explain your answer.

Answer 1

a)

95% Confidence Interval

X̅ ± t(α/2, n-1) S/√(n)

t(α/2, n-1) = t(0.05 /2, 27- 1 ) = 2.056

95.5 ± 2.056 * 15.39/√(27)

Lower Limit = 95.5 - 2.056 * 15.39/√(27)

Lower Limit = 89.41

Upper Limit = 95.5 + 2.056 * 15.39/√(27)

Upper Limit = 101.59

95% Confidence interval is ( 89.41 , 101.59  )

b)

The manager claims that mean amount be at least $100

Since some values in confidence interval are less than 100, we conclude that

the manager did not meet his goal.