Question

In order to determine the average weight of carry-on luggage by passengers in airplanes, a sample of 100 pieces of carry-on luggage was collected and weighed. The average weight was 34 pounds. Assume that we know the standard deviation of the population to be 5 pounds. a) Determine a 90% confidence interval estimate for the mean weight of the carry-on luggage. b) What determined whether you used a t value or a z value?

Answer 1

Number of observations n = 100

Mean μ = 34

Standard deviation σ = 5

Part a)

90% confidence interval is given as

Here alpha α = 10%

Z­_α/2 = 1.645

Lower limit = 34-1.645*5/10 = 33.1775

Upper limit = 34+1.645*5/10 = 34.8225

90% confidence interval is

(33.1775,34.8225)

Part b)

Since sample size is more than 25 i.e 100 we will use z value instead of t value.

The thumb rule to use t value when

1) sample size is below 30,

2) sample has an unknown population standard deviation.