Answer to Question #204791 in Statistics and Probability for zaimsquah

Question #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).


1
Expert's answer
2021-06-13T17:27:34-0400

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}) \\\\\n\n\\frac{\u03b1}{2}= \\frac{1}{2}(1- \\frac{95}{100}) \\\\\n\n= \\frac{1}{2}(1-0.95) \\\\\n\n= \\frac{0.05}{2} \\\\\n\n= 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\nn=25 \\\\\n\n\u03c7^2_{\u03b1\/2} = 39.365 \\\\\n\n\u03c7^2_{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}} } \\\\\n\n\\sqrt{ \\frac{(25-1)15^2}{39.365 }} < \\sigma < \\sqrt{\\frac{(25-1)15^2}{12.401 }} \\\\\n\n\\sqrt{ \\frac{24.225}{39.365 }} < \\sigma < \\sqrt {\\frac{24.225}{12.401 }} \\\\\n\n\\sqrt{ 137.177}< \\sigma < \\sqrt {435.45}\\\\\n\n11.71 < \\sigma < 20.87"

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


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