Question

The following measurements (in picocuries per liter) were recorded by a set of helium gas detectors installed in a waste disposal facility: 534.9, 495.3, 556.2, 499.2, 487.4, 557.2 Using these measurements, construct a 90% confidence interval for the mean level of helium gas present in the facility. Assume the population is approximately normal.Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

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, $\bar{x}$ = 521.7 and S = 31.6368. We will use  

$T = \frac{\bar{X}-\mu}{S/\sqrt{n}}$  

as the pivotal quantity. T has a $t$ 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., $t_{5}$, this value is -2.0150. Therefore the 90% confidence interval is given by

$521.7\pm 2.0150(31.6368/\sqrt{5})$, 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)