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GaryK [48]
2 years ago
8

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.
Mathematics
1 answer:
vazorg [7]2 years ago
5 0

Answer:

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

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = z\frac{\sigma}{\sqrt{n}}

M = 1.96\frac{2.6}{\sqrt{36}}

M = 0.85

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 Z = 2.575. So

M = z\frac{\sigma}{\sqrt{n}}

M = 2.575\frac{2.6}{\sqrt{36}}

M = 1.12

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.

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