Question

A social scientist wishes to estimate the mean number of years of education µ for adults in a certain large city. She surveys a random sample of 100 aadults and finds that the sample mean is 12.98 years and sample variance of is 9. canstruct a 99% confidence interval for the mean number of years of education for adults in the city.

Answer 1

The 99% confidence interval for the mean number of years of education for adults in the city is (5.215 years, 20.665 years).

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 M as such

`M = z*s`

In which s is the standard population of the sample. We have a sample variance of 9. The standard deviation is the square root of the variance. So
`s = 3`.

So:

`M = z*s = 3*2.575 = 7.725`

The lower end of the interval is the mean subtracted by M. So it is 12.94 - 7.725 = 5.215 years.

The upper end of the interval is the mean added to M. So it is 12.94 + 7.725 = 20.665 years.

The 99% confidence interval for the mean number of years of education for adults in the city is (5.215 years, 20.665 years).