Question

4) The Law School Admission Test (LSAT) is an examination for prospective law school students. Scores on the LSAT are known to have a normal distribution and a population standard deviation of σ = 10. A random sample of 250 LSAT takers produced a sample mean of 502. Compute a 99% confidence interval for the population mean LSAT score. Show all work, and round all decimals that you use to three decimal places.

Answer 1

The 99% confidence interval for the population mean LSAT score is between 500.371 and 503.629

Explanation:

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

`\alpha = (1-0.99)/(2) = 0.005`

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
`1-0.005 = 0.995`, so
`z = 2.575`

Now, find the margin of error 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 = 2.575(10)/(√(250)) = 1.629`

The lower end of the interval is the sample mean subtracted by M. So it is 502 - 1.629 = 500.371

The upper end of the interval is the sample mean added to M. So it is 502 + 1.629 = 503.629

The 99% confidence interval for the population mean LSAT score is between 500.371 and 503.629