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To estimate [tex]\mu[/tex], the mean salary of full professors at American colleges and universities, you obtain the salaries of a random sample of 400 full professors. The sample mean is [tex]\bar{x} = \$73,220[/tex] and the sample standard deviation is [tex]s = \$4,400[/tex].

A 99% confidence interval for [tex]\mu[/tex] is:

A. \$73,220 ± \$1,440

B. \$73,220 ± \$568

C. \$73,220 ± \$431

D. \$73,220 ± \$28.6

Answer :

Answer:

99% confidence interval for [tex]\mu = 73220 \pm 565.4[/tex]

Step-by-step explanation:

Sample mean = [tex]\bar{x} =73220[/tex]

Standard deviation = s = 4400

Z at 99% confidence level = 2.57

Sample = n = 400

Formula of confidence interval :[tex]\bar{x} \pm Z \times \frac{s}{\sqrt{n}}[/tex]

Substitute the values in the formula :

So,99% confidence interval for [tex]\mu = 73220 \pm 2.57 (\frac{4400}{\sqrt{400}})[/tex]

99% confidence interval for [tex]\mu = 73220 \pm 565.4[/tex]

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