Question

In order to determine the average weight of carry-on luggage by passengers in airplanes, a sample of 425 pieces of carry-on luggage was collected and weighed. The average weight was 17 pounds. Assume that we know the standard deviation of the population to be 2.75 pounds.

1. Determine a 90% confidence interval estimate for the mean weight of the carry-on luggage.
2. What wording in the prompt gave away the type of test (z or t) that this should be?

Answer 1

Solution :

Given that,

Sample size = n = 425

Z_/2 = 1.645

Margin of error = E = Z_/2* ( /n)

= 1.645 * (2.75 / 425)

= 0.22

At 90% confidence interval estimate of the population mean is,

- E < < + E

17 - 0.22 < < 17 + 0.22

16.78 < < 17.22

(16.78 , 17.22)

Here confidence population standard deviation is given so we use z confidence interval .