Question

4. All the savvy business statisticians know that the variance of a population equals 1780. A random sample of 121 observations is going to be taken from the population. a. With a 0.95 probability, what statement can be made about the size of the margin of error? b. With a 0.95 probability, how large of a sample would have to be taken to provide a margin of error of 8 or less?

PLEASE SHOW WORK FOR EACH ANSWER

Answer 1

Solution:
a) Given that, n=121
σ^2= 1780
σ=42.1900
(1–α)%=95%
α=0.05
α/2=0.025
Zα/2=1.96 ....... from standard normal table.
Margin of error=E=Zα/2 ×(σ/√n)
= 1.96×(42.1900/√121)
=7.5175
Margin of error is 7.5175
b) Given that,Margin of error= E=8
σ^2= 1780
σ=42.1900
(1–α)%=95%
α=0.05
α/2=0.025
Zα/2=1.96 ....... from standard normal table.
Sample size = n={(Zα/2 ×σ)/E}^2
={(1.96×42.19)/8}^2   
=106.8443
=107 ... nearest whole number
The sample size required is 107