Answer:
The correct option is 1.148 < σ < 6.015
Explanation:
The 99% confidence interval for the standard deviation is given below:
[tex]\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\frac{0.01}{2} }} } <\sigma<\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{1-\frac{0.01}{2} }} }[/tex]
Where:
[tex]\chi^{2}_{\frac{0.01}{2} }=18.548[/tex]
[tex]\chi^{2}_{1-\frac{0.01}{2}}=0.676[/tex]
Therefore, the 99% confidence interval is:
[tex]\sqrt{\frac{(7-1)2.019^{2}}{18.548} } <\sigma<\sqrt{\frac{(7-1)2.019^{2}}{0.676} }[/tex]
[tex]1.148<\sigma<6.015[/tex]
Therefore, the option 1.148 < σ < 6.015 is correct
