A simple random sample of 35 colleges and universities in the United States has a mean...

A simple random sample of 35 colleges and universities in the United States has a mean tuition of 18700 with a standard deviation of 10800. Construct a 99% confidence interval for the mean tuition for all colleges and universities in the United States. Round the answers to the nearest whole number.

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Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 18700

Population standard deviation = = 10800
Sample size = n =35

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

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

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

=2.576 * (10800 / $\sqrt$ 35)

= 4703

At 99% confidence interval is,

- E < < + E

18700 - 4703 < < 18700+4703

13997< < 23403

(18700, 23403)

orchestra answered 1 year ago

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