Question

1) Determine the 95% confidence interval for the average salary in Vineland if a sample of 125 individuals had an average salary of $43,000 and the population standard deviation is $3350

2)The dean is preparing a report for the State of New Jersey and wants to determine a 90% confidence interval for the average income of their student population within $250 dollars of the population average. If he knows the population standard deviation is $2250, how many students will he need to survey?

Answer 1

Solution :

Given that,

1) Z/2 = Z0.025 = 1.96

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

= 1.96 * ( 3350 /  125 )

= 587.28

At 95% confidence interval estimate of the population mean is,

  ± E

= 43000 ± 587.28

= ( $ 42,412.72 to $ 43,587.28 )

2) Z/2 = Z0.05 = 1.645

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

n = [1.645 * 2250 / 250]2

n = 219.18

Sample size = n = 220