Respuesta :
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 [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]M = 1.96\frac{2.6}{\sqrt{36}}[/tex]
[tex]M = 0.85[/tex]
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 [tex]Z = 2.575[/tex]. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]M = 2.575\frac{2.6}{\sqrt{36}}[/tex]
[tex]M = 1.12[/tex]
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.
