Question

In: Statistics and Probability

A random sample of professional wrestlers was obtained, and the annual salary (in dollars) for each...

A random sample of professional wrestlers was obtained, and the annual salary (in dollars) for each was recorded. The summary statistics were x = 45,500 and n = 17. Assume the distribution of annual salary is normal with o = 8,500.

(a) Find a 90% confidence interval for the true mean annual salary for all professional wrestlers (in dollars). (Round your answers to four decimal places.)

(______________ , ______________) kg

Expert Solution

Given that,

Point estimate = sample mean = = 45500

Population standard deviation = = 8500

Sample size = n =17

At 90% confidence level

= 1 - 90%

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

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

= 1.645* ( 8500 / $\sqrt$ 17 )

= 3391.2544
At 90% confidence interval estimate of the population mean
is,

- E < < + E

45500- 3391.2544 < < 45500+ 3391.2544

42108.7456 < < 48891.2544

( 42108.7456 , 48891.2544)

orchestra answered 11 months ago

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