Answer:
The 95% confidence interval for the population variance is (8.80, 32.45).
Step-by-step explanation:
The (1 - α)% confidence interval for the population variance is given as follows:
[tex]\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}[/tex]
It is provided that:
n = 20
s = 3.9
Confidence level = 95%
⇒ α = 0.05
Compute the critical values of Chi-square:
[tex]\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907[/tex]
*Use a Chi-square table.
Compute the 95% confidence interval for the population variance as follows:
[tex]\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}[/tex]
[tex]\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45[/tex]
Thus, the 95% confidence interval for the population variance is (8.80, 32.45).
