Consider the following random sample from a normal population: 14, 10, 13, 16, 12, 18, 15, and 11. What is the 95% confidence interval for the population variance?

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 means 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.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Data: 14, 10, 13, 16, 12, 18, 15, 11

We can calculate the sample mean and deviation with the following formulas:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

[tex] s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]

[tex]\bar X=13.625[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

s=2.669 represent the sample standard deviation

n=8 represent the sample size

Calculate the confidence interval

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=8-1=7[/tex]

Since the confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,7)".And we see that [tex]t_{\alpha/2}=2.365[/tex]

Now we have everything in order to replace into formula (1):

[tex]13.625-2.365\frac{2.669}{\sqrt{8}}=11.393[/tex]

[tex]13.625+2.365\frac{2.669}{\sqrt{8}}=15.857[/tex]

So on this case the 95% confidence interval would be given by (11.393;15.857)