Can someone explain and give the answer to this question?
2 answers:
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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 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.