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

Answer:

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

Step-by-step explanation:

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

$\alpha = \frac{1-0.99}{2} = 0.005$

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).