A chemist examines 17 sedimentary samples for lead concentration. The mean lead concentration for the sample data is 0.268 cc/cubic meter with a standard deviation of 0.069. Determine the 80% confidence interval for the population mean lead concentration. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

The critical value that should be used in constructing the confidence interval is T = 1.337.

The 80% confidence interval for the population mean lead concentration is between 0.246 and 0.29 cc/cubic meter.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 17 - 1 = 16

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 16 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.337, which is the crical value.

The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 0.268 - 0.022 = 0.246 cc/cubic meter

The upper end of the interval is the sample mean added to M. So it is 0.268 + 0.022 = 0.29 cc/cubic meter

The 80% confidence interval for the population mean lead concentration is between 0.246 and 0.29 cc/cubic meter

is there answer choices?

move it to the other side

step-by-step explanation: