Apopulation 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.

Formula for Margin of error :

, where n= sample size

= sample standard deviation

= two-tailed z-value for confidence level of ().

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

n= 100

a) Confidence level = 95%

Critical z-value for 95% confidence = . (Using z-value table)

Then , Margin of error :

E=9.8

Hence, Margin of error for 95%= 9.8

b) Confidence level = 90%

Critical z-value for 90% confidence = .(Using z-value table)

Then , Margin of error :

E=8.225

Hence, Margin of error for 90%= 8.225

c) Confidence level = 99%

Critical z-value for 99% confidence = .(Using z-value table)

Then , Margin of error :

E=12.88

Hence, Margin of error for 99%= 12.88

my answer is$ 12

step-by-step explanation:

because 15 -2=12

5/12, or 5: 12

step-by-step explanation: