Question

In: Statistics and Probability

An economic surveys 31 adults living is Townsburg about their salaries. The mean annual salaries for...

An economic surveys 31 adults living is Townsburg about their salaries. The mean annual salaries for those surveyed was found to be x̄ = $47,782, with a standard deviation of $2,488.

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.

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 47,782

Population standard deviation = = 2,488
Sample size = n =31

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 * ( 2488 / $\sqrt$ 31 )

= 735.08
At 90% confidence interval
is,

- E < < + E
47782 - 735.08 < <47782 + 735.08

47046.92 < < 48517.08

orchestra answered 2 years ago

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