Question

In: Statistics and Probability

8) An economic surveys 35 adults living is Townsburg about their salaries. The mean annual salaries...

8) An economic surveys 35 adults living is Townsburg about their salaries. The mean annual salaries for those surveyed was found to be x̄ = $48,240, with a standard deviation of $2,794.

a) Find a 90% confidence interval for the true mean annual salary of adults living in Townsburg.

b) Provide the margin of error of the interval as your answer.

Round your answer to the nearest dollar.

Solutions

Expert Solution

Solution :

Given that,

a) Point estimate = sample mean = = $ 48,240

sample standard deviation = s = $ 2,794

sample size = n = 35

Degrees of freedom = df = n - 1 = 35 - 1 = 34

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t_0.05,34 = 1.691

Margin of error = E = t_/2,df * (s / $\sqrt$ n)

= 1.691 * ( 2794 / $\sqrt$ 35)

Margin of error = E = $ 799

The 90% confidence interval estimate of the population mean is,

± E

= $ 48,240 ± $ 799

= ( $ 47,441, $ 49,039 )

b) Margin of error = E = t_/2,df * (s / $\sqrt$ n)

= 1.691 * ( 2794 / $\sqrt$ 35)

Margin of error = E = $ 799

orchestra answered 1 year ago

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