Question

Noise levels at 5 volcanoes were measured in decibels yielding the following data: 127,174,157,120,161 Construct the 98% confidence interval for the mean noise level at such locations. Assume the population is approximately normal. Step 3 of 4 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

Answer:

The critical value used is T = 3.747.

The 98% confidence interval for the mean noise level at such locations is (108.944, 186.656).

Step-by-step explanation:

Before building the confidence interval, we need to find the sample mean and the sample standard deviation.

Sample mean:

$\overline{x} = \frac{127+174+157+120+161}{5} = 147.8$

Sample standard deviation:

$s = \sqrt{\frac{(127-147.8)^2+(174-147.8)^2+(157-147.8)^2+(120-147.8)^2+(161-147.8)^2}{4}} = 23.188$

Confidence interval:

We have the standard deviation for the sample, which means that the t-distribution is used 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 = 5 - 1 = 4

98% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 4 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.98}{2} = 0.99$. So we have T = 3.747, which is the critical value used.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 3.747\frac{23.188}{\sqrt{5}} = 38.856$

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 147.8 - 38.856 = 108.944  

The upper end of the interval is the sample mean added to M. So it is 147.8 + 38.856 = 186.656.

The 98% confidence interval for the mean noise level at such locations is (108.944, 186.656).