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 :

$E= z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

, 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 : $E= (1.96)\dfrac{50}{\sqrt{100}}$

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 : $E= (1.645)\dfrac{50}{\sqrt{100}}$

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 : $E= (2.576)\dfrac{50}{\sqrt{100}}$

E=12.88

Hence, Margin of error for 99%= 12.88