Question

A population has a standard deviation of 50. A random sample of 100 items from this population is selected. The sample mean is determined to be 600. Calculate the margin of errors for 95%, 90% and 99% confidence interval.

Answer 1

Answer with explanation:

Formula for Margin of error :

`E= z_(\alpha/2)(\sigma)/(√(n))`

, where n= sample size

`\sigma`= sample standard deviation

`z_(\alpha/2)` = two-tailed z-value for confidence level of (
`1-\alpha`).

Given : tex]\sigma=50[/tex]

n= 100

a) Confidence level = 95%

Critical z-value for 95% confidence =
`z_(\alpha/2)=1.96`. (Using z-value table)

Then , Margin of error :
`E= (1.96)(50)/(√(100))`

E=9.8

Hence, Margin of error for 95%= 9.8

b) Confidence level = 90%

Critical z-value for 90% confidence =
`z_(\alpha/2)=1.645`.(Using z-value table)

Then , Margin of error :
`E= (1.645)(50)/(√(100))`

E=8.225

Hence, Margin of error for 90%= 8.225

c) Confidence level = 99%

Critical z-value for 99% confidence =
`z_(\alpha/2)=2.576`.(Using z-value table)

Then , Margin of error :
`E= (2.576)(50)/(√(100))`

E=12.88

Hence, Margin of error for 99%= 12.88