A population is normally distributed with a standard deviation of 2.8. A random sample is obtained from this population and the observations are: 8, 9, 10, 13, 14, 16, 17, 20, 21. Construct the 95% confidence interval for the mean of this population. Construct the 99% confidence interval for the mean of this population.

The 95% confidence interval for the mean of this population is between 12.39 and 16.05.

The 99% confidence interval for the mean of this population is between 11.82 and 16.62.

Step-by-step explanation:

The first step is finding the mean of the sample:

There are 9 observations. So

95% confidence interval:

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 is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 14.22 - 1.83 = 12.39

The upper end of the interval is the sample mean added to M. So it is 14.22 + 1.83 = 16.05

The 95% confidence interval for the mean of this population is between 12.39 and 16.05.

99% confidence interval:

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 is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 14.22 - 2.40 = 11.82

The upper end of the interval is the sample mean added to M. So it is 14.22 + 2.40 = 16.62

The 99% confidence interval for the mean of this population is between 11.82 and 16.62.

idk

step-by-step explanation:

answer: ok

step-by-step explanation: