Answer in Statistics and Probability for zaimsquah #204791

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).

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}{u03c7^2_{u03b1/2}} } < sigma < sqrt{frac{(n-1)s^2}{u03c7^2_{1-u03b1/2}} }”

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

The values “u03c7^2_{1-u03b1/2}” and “u03c7^2_{u03b1/2}” are critical values corresponding to the left tail and right tail area α/2 respectively. The value of α/2 is computed by

“frac{u03b1}{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{u03b1}{2}= frac{1}{2}(1- frac{CL}{100}) \nnfrac{u03b1}{2}= frac{1}{2}(1- frac{95}{100}) \nn= frac{1}{2}(1-0.95) \nn= frac{0.05}{2} \nn= 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 \nnn=25 \nnu03c7^2_{u03b1/2} = 39.365 \nnu03c7^2_{1-u03b1/2} = 12.401”

in

“sqrt{ frac{(n-1)s^2}{u03c7^2_{u03b1/2}} } < sigma < sqrt{frac{(n-1)s^2}{u03c7^2_{1-u03b1/2}} } \nnsqrt{ frac{(25-1)15^2}{39.365 }} < sigma < sqrt{frac{(25-1)15^2}{12.401 }} \nnsqrt{ frac{24.225}{39.365 }} < sigma < sqrt {frac{24.225}{12.401 }} \nnsqrt{ 137.177}< sigma < sqrt {435.45}\nn11.71 < sigma < 20.87”

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