Answer: [tex]0.3849<\sigma<0.5718[/tex]
Step-by-step explanation:
Given : sample size : n= 51
sample standard deviation : s= 0.46
Significance level : [tex]\alpha=1-0.95=0.05[/tex]
Using chi-square distribution, the critical values will be :
[tex]\chi^2_{\alpha/2, n-1}=\chi_{0.025,\ 50}=71.4202[/tex]
[tex]\chi^2_{1-\alpha/2, n-1}=\chi_{0.975,\ 50}=32.3574[/tex]
Confidence interval for population standard deviation ([tex]\sigma[/tex]) :
[tex]\sqrt{\dfrac{s^2(n-1)}{\chi^2_{\alpha/2}}}<\sigma<\sqrt{\dfrac{s^2(n-1)}{\chi^2_{1-\alpha/2}}}[/tex]
Substitute all the values , we get
[tex]\sqrt{\dfrac{(0.46)^2(50)}{71.4202}}<\sigma<\sqrt{\dfrac{(0.46)^2(50)}{32.3574}}\\\\ \sqrt{0.148137361699}<\sigma<\sqrt{0.326973118977}\\\\\approx0.3849<\sigma<0.5718 [/tex]
Hence, the 95% two-sided confidence interval for σ : [tex]0.3849<\sigma<0.5718[/tex]
