Asocial 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.

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 level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so .

Now, find M as such

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 .

So:

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).

ca = ec

cpctc makes it so that corresponding parts of the triangle you proved to be equal are equal, the other options aren't part of the part of the triangles you proved to be equal.