Question

III 4. The standard deviation of the ages of a random sample of 40 television sets in a neigl forhood 3 years. Find a 95% confidence interval for the standard deviation of the entire population of levisions in this neighborhood. Assume that these ages are randomly distributed.

Answer 1

Given that, sample size (n) = 40 and

sample standard deviation (s) = 3 years

confidence level = 0.95

=> significance level $(\alpha)$ = 1 - 0.95 = 0.05

$\alpha/2 = 0.05/2 = 0.025$ and

$1-\alpha/2 = 1- 0.025 = 0.975$

Degrees of freedom = 40 - 1 = 39

Using Excel we find, the chi-square critical values,

$\chi_{\alpha/2}^2 = CHIINV (0.025, 39) =58.120$

$\chi_{1-\alpha/2}^2 = CHIINV (0.975, 39) =23.654$

The 95% confidence interval for the population standard deviation is,

$\sqrt {\frac {(n-1)*s^2}{\chi_{\alpha/2}^2}} < \sigma < \sqrt {\frac {(n-1)*s^2}{\chi_{1-\alpha/2}^2}}$

$\sqrt {\frac {(40-1)*3^2}{58.120} }< \sigma < \sqrt {\frac {(40-1)*3^2}{23.654} }$

$2.4575 < \sigma < 3.8521$

Therefore, the 95% confidence interval for the population standard deviation is (2.4575, 3.8521)