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Salaries of 37 college graduates who took a statistics course in college have a mean, x, of $67,500. Assuming a standard devi

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Answer #1

Solution :

Given that,

Point estimate = sample mean = \bar x = $67500

Population standard deviation =   \sigma =$15534
Sample size = n =37

At 99% confidence level the z is ,

\alpha  = 1 - 99% = 1 - 0.99 = 0.01

\alpha / 2 = 0.01 / 2 = 0.005

Z\alpha/2 = Z0.005 = 2.576   ( Using z table )

Margin of error = E = Z\alpha/2* (\sigma /\sqrtn)

= 2.576* ( 15534 / \sqrt 37)

= 6579

At 99% confidence interval estimate of the population mean is,

\bar x- E < \mu < \bar x + E

67500-6579 < \mu < 67500+6579

60921< \mu < 74079

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