A chemist examines 15 sedimentary samples for nitrate concentration. The mean nitrate concentration for the sample data is 0.670 cc/cubic meter with a standard deviation of 0.0616.

a. Determine the 80% confidence interval for the population mean nitrate concentration. Assume the population is approximately normal. Round your answer to three decimal places.
b. Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

(a) The 80% confidence interval for the population mean nitrate concentration is (0.648, 0.692).

(b) The critical value of t that should be used in constructing the 80% confidence interval is 1.345.

Step-by-step explanation:

Let X = nitrate concentration.

The sample mean nitrate concentration is, cc/cubic meter.

The sample standard deviation of the nitrate concentration is, .

It assumed that the population is approximately normal.

And since the population standard deviation is not known, we will use a t-interval.

The (1 - α)% confidence interval for population mean (μ) is:

(a)

The critical value of t for α = 0.20 and degrees of freedom, (n - 1) = 14 is:

*Use a t-table for the critical value.

Compute the 80% confidence interval for the population mean nitrate concentration as follows:

Thus, the 80% confidence interval for the population mean nitrate concentration is (0.648, 0.692).

(b)

The critical value of t for confidence level (1 - α)% and (n - 1) degrees of freedom is:

The value of  is:

α = 0.20

And the degrees of freedom is,

(n - 1) = 15 - 1 = 14

Compute the critical value of t for confidence level 80% and 14 degrees of freedom as follows:

*Use a t-table for the critical value.

Thus, the critical value of t that should be used in constructing the 80% confidence interval is 1.345.

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step-by-step explanation:

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step-by-step explanation:

the text says that there will be 22 children at this party, 1 teacher, and 2 parents. so if you add them all together (22+1+2) it will equal 25. (=25)