Question

A simple random sample of 35 colleges and universities in the United States has a mean tuition of $17,700 with a standard deviation of $10,300. Construct a 95% confidence interval for the mean tuition for all colleges and universities in the United States. Round the answers to the nearest whole number.

Answer 1

The 95% confidence interval for the mean tuition for all colleges and universities in the United States is between $14,288 and $21,112.

Explanation:

Sample size greater than 30, which means that we use the normal distribution to find the interval.

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.025 = 0.975`, so
`z = 1.96`

Now, find M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 1.96*(10300)/(√(35)) = 3412`

The lower end of the interval is the sample mean subtracted by M. So it is 17700 - 3412 = $14,288

The upper end of the interval is the sample mean added to M. So it is 17700 + 3412 = $21,112.

The 95% confidence interval for the mean tuition for all colleges and universities in the United States is between $14,288 and $21,112.