Question

A sample of 28 elements is selected to estimate a 95% confidence interval for the variance of the population. The chi-square values to be used for this interval estimation are
a. 16.151 and 40.113.

b. 14.573 and 43.195.

c. 15.308 and 44.461.

d. 11.808 and 49.645.

Answer 1

Answer:

$\chi^2_{\alpha/2}=43.195$

$\chi^2_{1- \alpha/2}=14.573$

b. 14.573 and 43.195.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Solution to the problem

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

We need on this case to calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=28-1=27$

Since the Confidence is 0.95 or 95%, the value of $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table for the Chis square distribution with 27 degrees of freedom to find the critical values. We need a value that accumulates 0.025 of the area on the left tail and 0.025 of the area on the right tail.  

The excel commands would be: "=CHISQ.INV(0.025,27)" "=CHISQ.INV(0.975,27)". so for this case the critical values are:

$\chi^2_{\alpha/2}=43.195$

$\chi^2_{1- \alpha/2}=14.573$

b. 14.573 and 43.195.