A random sample of 45 fresh graduates has a mean starting earnings of $3250. Assume the...

A random sample of 45 fresh graduates has a mean starting earnings of $3250. Assume the population standard deviation is $675. Construct a 90% the confidence interval for the population mean, μ. (Round your answer to the nearest integer)

a-($2575, $3925)

b-($3084, $3416)

c-($3053, $3447)

d-($3221, $3279)

Solutions

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 3250

Population standard deviation = = 675

Sample size n =45

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 * ( 675 / $\sqrt$ 45 )

=166
At 90% confidence interval estimate of the population mean
is,

- E < < + E

3250 - 166 < < 3250+ 166
($3084, $3416)

orchestra answered 1 year ago

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