Question

A chemist examines 22 water samples for iron concentration. The mean iron concentration for the sample data is 0.409 cc/cubic meter with a standard deviation of 0.016. Determine the 99% confidence interval for the population mean iron 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.

Answer 1

Answer:

Critical value: T = 2.8314

The 99% confidence interval for the population mean iron concentration is between 0.364 cc/m³ and 0.454 cc/m³

Step-by-step explanation:

We are in posession of the sample's standard deviation. So we use the student 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 = 22 - 1 = 21

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 21 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$. So we have T = 2.8314, which is the critical value that should be used in constructing the interval.

The margin of error is:

M = T*s = 2.8314*0.016 = 0.045

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 0.409 - 0.045 = 0.364 cc/m³

The upper end of the interval is the sample mean added to M. So it is 0.409 + 0.045 = 0.454 cc/m³

The 99% confidence interval for the population mean iron concentration is between 0.364 cc/m³ and 0.454 cc/m³