The sick days of employees every two years in a company are normally distributed with a population standard deviation of 7 days and an unknown population mean. If a random sample of 20 employees is taken and results in a sample mean of 21 days, find a 95% confidence interval for the population mean. Round your answer to TWO decimal places. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 You may use a calculator or the common z values above.

The 95% confidence interval for the population mean is between 17.93 days and 24.07 days.

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 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 21 - 3.07 = 17.93 days

The upper end of the interval is the sample mean added to M. So it is 21 + 3.07 = 24.07 days

The 95% confidence interval for the population mean is between 17.93 days and 24.07 days.

what golf ball

step-by-step explanation:

chad should not have included the values (1, 2) and (3, 6) because that would mean he only had half a healing card.