Respuesta :
Answer:
a.
[tex]\begin{gathered} 0.000041\leq\sigma^2\leq0.00015 \\ 0.0064\leq\sigma\leq0.0125 \end{gathered}[/tex]b. The standard deviation is at an acceptable level
Explanation:
The confidence interval for the variance can be calculated as:
[tex]\frac{(n-1)s^2}{\chi^2_{\frac{\propto}{2}}}\leq\sigma^2\leq\frac{(n-1)s^2}{\chi^2_{1-\frac{\propto}{2}}}[/tex]Where n is the size of the sample, s² is the variance of the sample, (1-∝) is the level of confidence, and χ² is the value of chi-square with n-1 degrees of freedom.
In this case, we get that n is 14 tablets, s² is 0.000071, and 1-∝ = 90%, so ∝ is 10%.
Then, the values of chi-square with 13 (14 -1) degrees of freedom is:
[tex]\begin{gathered} \chi^2_{\frac{\propto}{2}}=22.36 \\ \chi^2_{1-\frac{\propto}{2}}=5.89 \end{gathered}[/tex]Therefore, the interval of confidence for the population variance is:
[tex]\begin{gathered} \frac{(14-1)(0.000071)}{22.36}\leq\sigma^2\leq\frac{(14-1)(0.000071)}{5.89} \\ 0.000041\leq\sigma^2\leq0.00015 \end{gathered}[/tex]Then, the standard deviation is the square root of the variance, so the interval of confidence for the population standard deviation is:
[tex]\begin{gathered} \sqrt[]{0.000041}\leq\sigma\leq\sqrt[]{0.00015} \\ 0.0064\leq\sigma\leq0.0125 \end{gathered}[/tex]Finally, since the upper limit of the interval for the standard deviation is less than 0.15 milligrams, we can say that the population standard deviation is at an acceptable level with a confidence level of 90%
