Question

In order to estimate the Mean weight of carry-on luggage, an airline selected and weighed, a random sample of 25 pieces of carry-on luggage. The sample mean was 32 pounds with a sample standard deviation of 9 pounds. Compute and explain a 90% confidence interval estimate of the population mean weight of all carry-on luggage.

Answer 1

Given that,

= 32

s =9

n = 25

Degrees of freedom = df = n - 1 = 25- 1 = 24

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,24 = 1.711 ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 1.711* ( 9/ 25) = 3.1878

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

- E < < + E

32 - 3.1878 < < 32+ 3.1878

28.8122 < < 35.1878

(28.8122,35.1878 )