The average zinc concentration recovered from a sample of measurements taken in 36 different locations in a river is found to be 2.6 grams per milliliter. Find the 95% and 99% confidence intervals for the mean zinc concentration in the river. Assume that the population standard deviation is 0.3 gram per milliliter.

The 95% confidence interval for the mean zinc concentration in the river is between 1.75 and 3.45 grams per milliliter.

The 99% confidence interval for the mean zinc concentration in the river is between 1.48 and 3.72 grams per milliliter.

Step-by-step explanation:

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 .

That is z with a pvalue of , so Z = 1.96.

Now, find the margin of error 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 2.6 - 0.85 = 1.75 grams per milliliter.

The upper end of the interval is the sample mean added to M. So it is 2.6 + 0.85 = 3.45 grams per milliliter.

The 95% confidence interval for the mean zinc concentration in the river is between 1.75 and 3.45 grams per milliliter.

99% confidence level:

By the same logic as for the 95% confidence interval, we have that . So

The lower end of the interval is the sample mean subtracted by M. So it is 2.6 - 1.12 = 1.48 grams per milliliter.

The upper end of the interval is the sample mean added to M. So it is 2.6 + 1.12 = 3.72 grams per milliliter.

The 99% confidence interval for the mean zinc concentration in the river is between 1.48 and 3.72 grams per milliliter.

true? is that what you're asking or do you mean something else?

$3,200 x 7.25% = $3,200 x .0725 = $232 interest monthly

$232 monthly x 12 annual = $2, 784

step-by-step explanation: