Answer: [tex]0.12< \sigma<0.42[/tex]
Step-by-step explanation:
Confidence interval for standard deviation is given by :-
[tex]\sqrt{\dfrac{s^2(n-1)}{\chi^2_{\alpha/2}}}< \sigma<\sqrt{\dfrac{s^2(n-1)}{\chi^2_{1-\alpha/2}}}[/tex]
Given : Confidence level : [tex]1-\alpha=0.95[/tex]
⇒[tex]\alpha=0.05[/tex]
Sample size : n= 7
Degree of freedom = 6 Â Â (df= n-1)
sample standard deviation : s= 0.19
Critical values by using chi-square distribution table :
[tex]\chi^2_{\alpha/2, df}}=\chi^2_{0.025, 6}}=14.4494\\\\\chi^2_{1-\alpha/2, df}}=\chi^2_{0.975, 6}}=1.2373[/tex]
Confidence interval for standard deviation of the weights of the packages prepared by the machine is given by :-
[tex]\sqrt{\dfrac{ 0.19^2(6)}{14.4494}}< \sigma<\sqrt{\dfrac{ 0.19^2(6)}{1.2373}}[/tex]
[tex]\Rightarrow0.12243< \sigma<0.418400[/tex]
[tex]\approx0.12< \sigma<0.42[/tex]
Hence, the 95% confidence interval to estimate the standard deviation of the weights of the packages prepared by the machine. Â :
[tex]0.12< \sigma<0.42[/tex]
