Question

Exhibit: Lunch Customers

A random sample of 68 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed in the restaurant was 42 minutes with a standard deviation of 11 minutes.

(For this exhibit, avoid rounding intermediate steps and round your final solutions to 4 decimal places)

34. Construct 88% confidence interval for the population mean and provide the lower bound of the confidence interval below.

35. Refer to the Exhibit Lunch Customers.

Provide the upper bound of the confidence interval that you calculated above (i.e. 88% confidence interval for the population mean).

36. Refer to the Exhibit Lunch Customers.

Construct 96% confidence interval for the population mean and provide the lower bound of the confidence interval below.

37. Refer to the Exhibit Lunch Customers.

Provide the upper bound of the confidence interval that you calculated above (i.e. 96% confidence interval for the population mean).

Answer 1

Solution:
Given that, x̄= 42 minutes,σ =11 minutes,n=68
(1–α)%=88%
α=0.12
α/2=0.06
Zα/2=1.55 ....from standard normal table.
Margin of error=E=Zα/2 ×(σ/√n)
=1.55 ×(12/√68)
=2.2556
Margin of error=E=2.2556
88% confidence interval for the population mean is given as,
x̄± Margin of error=(42-2.2556,42+2.2556)
=(39.7444,44.2556)
Lower bound=39.7444
Upper bound =44.2556
(1–α)%=96%
α=0.04
α/2=0.02
Zα/2=2.05 ....from standard normal table.
Margin of error=E=Zα/2 ×(σ/√n)
=2.05×(12/√68)
=2.9832
Margin of error=E=2.9832
96% confidence interval for the population mean is given as,
x̄± Margin of error=(42-2.9832,42+2.9832)
=(39.0168,44.9832)
Lower bound=39.0168
Upper bound =44.9832