Question

Suppose that income X of a randomly selected Illinois adult is a normal random variable. From a sample of 50 such people, suppose the sample mean income is $55,000 and the sample standard deviation is is $10,000. (a) Find a 95 percent confidence interval for the unknown population mean income of Illinois adults. Hint: use the above table of t values. (b) Explain the meaning of the confidence interval you built in part A

pls show work with answer

Answer 1

Given that, sample size ( n ) = 50

sample mean = $55000

sample standard deviation ( s ) = $10000

Confidence level = 95%

significance level = 1 - 0.95 = 0.05

t-critical value at significance level = 0.05 with degrees of freedom = n - 1 = 50 - 1 = 49 is,

The 95% confidence interval for population mean is,

A 95% confidence interval for the unknown population mean income of illinois adults is, ( $52157.43, $57842.57)

Interpretation: we are 95% confident that the mean income of illinois adults is betwewn $52157.43 and $57842.57