Use technology to construct the confidence intervals for the population variance sigma σ squared 2 and the population standard deviation sigma σ. Assume the sample is taken from a normally distributed population. c equals = 0.90 0.90, s equals = 31 31, n equals = 17 17 The confidence interval for the population variance is ( 26.30 26.30, 7.96 7.96). (Round to two decimal places as needed.) Enter your answer in the edit fields and then_____________.

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 .

Calculating the confidence interval

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

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

The sample deviation for this case is s=31

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

[tex]df=n-1=17-1=16[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,16)" "=CHISQ.INV(1-0.05,7)". so for this case the critical values are:

[tex]\chi^2_{\alpha/2}=26.30[/tex]

[tex]\chi^2_{1- \alpha/2}=7.96[/tex]

And replacing into the formula for the interval we got:

[tex]\frac{(16)(31)^2}{26.30} \leq \sigma^2 \leq \frac{(16)(31)^2}{7.96}[/tex]

[tex] 584.64 \leq \sigma^2 \leq 1931.66[/tex]

Now we just take square root on both sides of the interval and we got:

[tex] 24.18 \leq \sigma \leq 43.95[/tex]