Question

To ensure Fork-lifts are not over-loaded, design engineers have to consider the weights of cargo. Suppose 400 cargo shipments have normally distributed weights with a mean of 500 pounds and a standard deviation of 25 pounds. Given a confidence level of 95%:

a) What is the error of estimate (E)?

b) What does the sample size have to be if we want the error of estimate to be no more than 1.5?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 500 pounds.

Population standard deviation =    = 25 pounds.

Sample size = n = 400

b) At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96

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

= 1.96 * ( 25 /  400 )

= 2.45

Given that,

Population standard deviation = = 25 pounds.

Margin of error = E = 2.45

At 95% confidence level the z is,

= 1 - 95%

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96  

b) sample size = n = [Z/2* / E] 2

n = [1.96 * 25 / 1.5]2

n = 1067.11

Sample size = n = 1068