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In: Statistics and Probability

A simple random sample of 40 colleges and universities in the United States has a mean...

A simple random sample of 40 colleges and universities in the United States has a mean tuition of 18,800 with a standard deviation of 10,100. Construct a 98% 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 :

sample size = n = 40

Degrees of freedom = df = n - 1 = 39

t /2,df = 2.426

Margin of error = E = t/2,df * (s /n)

= 2.426 * (10,100 / 40)

Margin of error = E = 3874.2

The 95% confidence interval estimate of the population mean is,

- E < < + E

18,800 - 3,874.2 < < 18,800 + 3,874.2

14,925.8 < < 22,674.2

(14,925.8, 22,674.2)

orchestra answered 2 years ago
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