SOLUTION: The president of a large university wishes to estimate the average age of the students presently enrolled. From past studies, the standard deviation &#963; is known to be 2 years.

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Question 730068: The president of a large university wishes to estimate the average age of the students presently enrolled. From past studies, the standard deviation σ is known to be 2 years. A sample of 50 students is selected and the mean is found to be 23.3 years.
a. Find the 99% confidence interval of the population.
b. Find the 99% confidence interval for the population standard deviation

Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
I'll do the first part to get you started

a)
I'm assuming you want the 99% confidence interval of the population mean. Let me know if it's some other variable.

The sample mean is 23.3

The margin of error is

E = z*sigma/sqrt(n)

E = 2.576*sigma/sqrt(n) ... use a calculator to find the critical value for z (let me know if you need me to show you how to do this on a calculator)

E = 2.576*2/sqrt(50) ... plug in the other given values

E = 0.72860282733461

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The lower bound of the confidence interval is

L = xbar - E

L = 23.3 - 0.72860282733461

L = 22.5713971726653

The upper bound of the confidence interval is

L = xbar + E

L = 23.3 + 0.72860282733461

L = 24.0286028273347

So the 99% confidence interval for the population mean is

(L, U) = (22.5713971726653, 24.0286028273347)

SOLUTION: The president of a large university wishes to estimate the average age of the students presently enrolled. From past studies, the standard deviation &#963; is known to be 2 years.

SOLUTION: The president of a large university wishes to estimate the average age of the students presently enrolled. From past studies, the standard deviation σ is known to be 2 years.