Question

A survey of 100 randomly selected Michigan residents finds that the they spend a mean of $21 per month on going to the movies. An independent study of a 100 randomly selected California residents find that they spend a mean of $46 per month on going to the movies. Assume that the population standard deviation for the Michigan residents is $5 per month and that the population standard deviation for the California residents is $23 per month. Answer the following questions:

(a) If you were asked to construct a confidence interval what are the two pieces of inform you need?

(b) What is the point estimate for the above scenario?

(c) Calculate the Margin of Error at a 90% confidence level.

(d) Construct the Confidence Interval for the true difference in mean amount of money spent on going to the movies per month for the two states.

(e) Give interpretation of the Confidence Interval

Answer 1

Part a)

We need mean ( ) ± Margin of error

Part b)

Part c)

Margin of Error =    = 3.8719

Part d)

Confidence interval :-

Z(α/2) = Z (0.1 /2) = 1.645

Lower Limit =
Lower Limit = -28.8719
Upper Limit =
Upper Limit = -21.1281
90% Confidence interval is ( -28.8719 , -21.1281 )

Part e)

We are 90% confidence that the true population mean difference in amount of money spent on going to the movies per month for the two states. lies within the interval.