Question

QUESTION 1

1. 
   In order to determine the average weight of carry-on luggage by passengers in airplanes, a sample of 36 pieces of carry-on luggage was weighed. The average weight was 22 pounds. Assume that we know the standard deviation of the population to be 7 pounds.

   a.	Determine a 99% confidence interval estimate for the mean weight of the carry-on luggage.
   b.	Determine a 95% confidence interval estimate for the mean weight of the carry-on luggage.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 22

Population standard deviation =    = 7
Sample size = n =36

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

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

= 2.576* ( 7/ 36)

E= 3.0053

At 99% confidence interval estimate of the population mean is,

- E < < + E

22-3.0053 < < 22+3.0053

18.9947< < 25.0053

(18.9947,< 25.0053)

(B)

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96* ( 7/ 36)

E=2.2867

At 95% confidence interval estimate of the population mean is,

- E < < + E

22-2.2867 < < 22+2.2867

19.7133< < 24.2867

(19.7133,24.2867)