Question

Assume that the heights of men are normally distributed. a random sample of 16 men have a mean height of 67.5 inches and a standard deviation of 3.2 inches. construct a 99% confidence interval for the population standard deviation, σ.

Answer 1

Solution: The 99% confidence interval for the population standard deviation is given below:

$\sqrt{\frac{(n-1)s^{2}}{chi^{2}_{\frac{\alpha}{2}}}} \leq$σ$\leq \sqrt{\frac{(n-1)s^{2}}{chi^{2}_{1-\frac{\alpha}{2}}}}$

$\sqrt{\frac{15 (3.2)^{2}}{30.578}}\leq$σ$\leq \sqrt{\frac{15(3.2)^{2}}{4.601}}$

$2.2\leq$σ$\leq 5.8$

Answer 2

Answer: $2.16\leq\sigma\leq5.78$

Step-by-step explanation:

The confidence interval for population standard deviation is given by :-

$\sqrt{\dfrac{(n-1)s^2}{\chi^2_{\alpha/2}}}<\sigma<\sqrt{\dfrac{(n-1)s^2}{\chi^2_{1-\alpha/2}}}$

Given : n = 16

Standard deviation : s = 3.2

Significance level : $\alpha: 1-0.99=0.01$

Critical values by using chi-square distribution table :-

$\chi^2_{n-1, \alpha/2}}=\chi^2_{15, 0.005}}=32.80$

$\chi^2_{n-1, 1-\alpha/2}}=\chi^2_{15, 0.995}}=4.6$

The 99% confidence interval for the population standard deviation is given by :-

$\sqrt{\dfrac{(15)(3.2)^2}{32.80}}\leq\sigma\leq\sqrt{\dfrac{(15)(3.2)^2}{4.6}}\\\\\approx2.16\leq\sigma\leq5.78$

Hence,  a 99% confidence interval for the population standard deviation : $2.16\leq\sigma\leq5.78$