Answer:
The 90% confidence interval is (493.1903, 550.2097), and the critical value to construct the confidence interval is 2.0150
Step-by-step explanation:
Let X be the random variable that represents a measurement of helium gas detected in the waste disposal facility. We have observed n = 6 values, [tex]\bar{x}[/tex] = 521.7 and S = 31.6368. We will use
[tex]T = \frac{\bar{X}-\mu}{S/\sqrt{n}}[/tex]
as the pivotal quantity. T has a [tex]t[/tex] distribution with 5 degrees of freedom. Then, as we want a 90% confidence interval for the mean level of helium gas present in the facility, we should find the 5th quantile of the t distribution with 5 degrees of freedom, i.e., [tex]t_{5}[/tex], this value is -2.0150. Therefore the 90% confidence interval is given by
[tex]521.7\pm 2.0150(31.6368/\sqrt{5})[/tex], i.e.,
(493.1903, 550.2097)
To find the 5th quantile of the t distribution with 5 degrees of freedom, you can use a table from a book or the next instruction in the R statistical programming language
qt(0.05, df = 5)
