You have to calculate the confidence interval for the population variance and population standard deviation of carbon monoxide deaths.
To calculate the confidence interval for the population variance you have to use the following formula, which is derived from the Chi-Square distribution:
[tex]\lbrack\frac{(n-1)S^2}{\chi^2_{n-1;1-\frac{\alpha}{2}}};\frac{(n-1)S^2}{\chi^2_{n-1;\frac{\alpha}{2}}}\rbrack[/tex]For a sample of n=6 years, the sample standard deviation is S=4.1 deaths per year
To calculate the interval, the first step is to determine the values under the Chi-Square distribution with n-1 degrees of freedom and a confidence level of 0.99
The degrees of freedom for this particular distribution is:
[tex]n-1=6-1=5[/tex]The confidence level is 0.99, calculate the value of α:
[tex]\begin{gathered} 1-\alpha=0.99 \\ \alpha=1-0.99 \\ \alpha=0.01 \end{gathered}[/tex]For the chi-Square value of the left bound, the value of probability is
[tex]1-\alpha/2=1-0.01/2=1-0.005=0.995[/tex]The Chi-Square value for the left bound of the interval corresponds to a distribution with 5 degrees of freedom and 0.995 of accumulated probability
[tex]\chi^2_{5;0.995}=16.8[/tex]So the value of the distribution for the left bound is χ²left= 16.8
For the right bound of the interval, the accumulated probability under the distribution is α/2
[tex]\frac{\alpha}{2}=\frac{0.01}{2}=0.005[/tex]The Chi-Square value for the right bound of the interval is for distribution with 5 degrees of freedom and 0.005 accumulated probability:
[tex]\chi^2_{5;0.005}=0.412[/tex]The value of the distribution for the right bound is χ²right= 0.412
The sample variance is equal to the square of the sample standard deviation so that:
[tex]\begin{gathered} S^2=4.1^2 \\ S^2=16.81 \end{gathered}[/tex]Now you can calculate the confidence interval as follows:
Left bound of the interval
[tex]\frac{(n-1)S^2}{\chi^2_{5;0.995}}=\frac{(6-1)16.81}{16.8}=5.00[/tex]Right bound of the interval
[tex]\frac{(n-1)S^2}{\chi^2_{5;0.005}}=\frac{(6-1)16.81}{0.412}=204.00[/tex]The calculation of the interval can be written as follows:
[tex]\frac{(6-1)4.1^2}{16.8}\leq\sigma^2\leq\frac{(6-1)4.1^2}{0.412}[/tex]So, using a 99% confidence level, the interval for the population variance of the carbon monoxide deaths is [5.00;204.00] deaths per year²
Finally, apply the square root to both bounds of the interval to determine the confidence interval for the population standard deviation:
[tex]\begin{gathered} \sqrt[]{\frac{(6-1)4.1^2}{16.8}}\leq\sigma\leq\sqrt[]{\frac{(6-1)4.1^2}{0.412}} \\ \sqrt[]{\frac{5\cdot16.81}{16.8}}\leq\sigma\leq\sqrt[]{\frac{5\cdot16.81}{0.412}} \\ 2.236\leq\sigma\leq14.283 \\ \end{gathered}[/tex]Rounding to the nearest tenth the interval can be expressed as follows
[tex]2.2\leq\sigma\leq14.3[/tex]Using the same confidence level the confidence interval for the population standard deviation is [2.2;14.3] deaths per year.
