Answer to Question #204791 in Statistics and Probability for zaimsquah

Question

If n=25 and the sample standard deviation is 15, construct a 95% confidence interval for the true standard deviation. (Assume the population is approximately Normal).

Answer 1

The confidence interval is used to estimate the population parameter with the help of a sample statistic. The confidence interval of population standard deviation is computed by

$\sqrt{ \frac{(n-1)s^2}{χ^2_{α/2}} } < \sigma < \sqrt{\frac{(n-1)s^2}{χ^2_{1-α/2}} }$

where n is the sample size, and s is the sample standard deviation.

The values $χ^2_{1-α/2}$ and $χ^2_{α/2}$ are critical values corresponding to the left tail and right tail area α/2 respectively. The value of α/2 is computed by

$\frac{α}{2}= \frac{1}{2}(1- \frac{CL}{100})$

The degree of freedom for this distribution is given by

df=n-1

The confidence level is 95%. So, to find the value of α/2 , substitute 95% for CL in

$\frac{α}{2}= \frac{1}{2}(1- \frac{CL}{100}) \\ \frac{α}{2}= \frac{1}{2}(1- \frac{95}{100}) \\ = \frac{1}{2}(1-0.95) \\ = \frac{0.05}{2} \\ = 0.025$

The sample size is 25, so the degree of freedom will be 24.

Using the technology or table of chi-square distribution, the critical value corresponding to the left tail area 0.025 is 12.401 and the critical value corresponding to the right tail area is 39.365.

To find the 95% confidence interval of population standard deviation, substitute

$s= 15 \\ n=25 \\ χ^2_{α/2} = 39.365 \\ χ^2_{1-α/2} = 12.401$

in

$\sqrt{ \frac{(n-1)s^2}{χ^2_{α/2}} } < \sigma < \sqrt{\frac{(n-1)s^2}{χ^2_{1-α/2}} } \\ \sqrt{ \frac{(25-1)15^2}{39.365 }} < \sigma < \sqrt{\frac{(25-1)15^2}{12.401 }} \\ \sqrt{ \frac{24.225}{39.365 }} < \sigma < \sqrt {\frac{24.225}{12.401 }} \\ \sqrt{ 137.177}< \sigma < \sqrt {435.45}\\ 11.71 < \sigma < 20.87$

So, the 95% confidence interval for population standard deviation is (11.71, 20.87).