Question

The luggage handing branch at the airport wants to estimate the mean weight among all checked suitcases. A random sample of 47 checked suitcases were carefully weighed (to the nearest pound). The sample results found a mean of 40.6 pounds and a standard deviation of 9.5 pounds.

a. Estimate with 90% confidence the mean weight of all checked luggage. (18)

b. Explain fully what is meant when you say, "you have 90% confidence" in your interval in part A.

Answer 1

Given that,

= 40.6

s =9.5

n = 47

Degrees of freedom = df = n - 1 = 47- 1 = 46

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,46 = 1.679 ( using student t table)

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

= 1.679* ( 9.5/ 46) = 2.3518

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

- E < < + E

40.6 - 2.3518 < < 40.6+ 2.3518

38.2482 < < 42.9518

( 38.2482 , 42.9518)