Question

The following measurements (in picocuries per liter) were recorded by a set of argon gas detectors installed in a research facility:
381.3,394.8,396.1,380
Using these measurements, construct a 95% confidence interval for the mean level of argon gas present in the facility. Assume the population is approximately normal.

Answer 1

Answer:

The 95% confidence interval for the mean level of argon gas present in the facility is (374.4, 401.7).

Step-by-step explanation:

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

Sample mean:

$\overline{x} = \frac{381.3+394.8+396.1+380}{4} = 388.05$

Sample standard deviation:

$s = \sqrt{\frac{(381.3-388.05)^2+(394.8-388.05)^2+(396.1-388.05)^2+(380-388.05)^2}{3}} = 8.58$

Confidence interval:

We have the standard deviation for the sample, so 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 = 4 - 1 = 3

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 3 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 3.1824

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 3.1824\frac{8.58}{\sqrt{4}} = 13.65$

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 388.05 - 13.65 = 374.4

The upper end of the interval is the sample mean added to M. So it is 388.05 + 13.65 = 401.7

The 95% confidence interval for the mean level of argon gas present in the facility is (374.4, 401.7).