# You randomly select and weigh 27 samples of an allergy medication. The sample standard deviation is 1.22 milligrams. Assuming the weights are normally distributed, construct 90% confidence intervals for the population variance and standard deviation. Please interpret your result.

You randomly select and weigh 27 samples of an allergy medication. The sample standard deviation is 1.22 milligrams. Assuming the weights are normally distributed, construct $90\mathrm{%}$ confidence intervals for the population variance and standard deviation. Please interpret your result.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Demi-Leigh Barrera

Step 1
From the given information, $S=1.22$ milligrams and $n=27$.
The level of confidence is 0.90.
$1-\alpha =0.90$
$\alpha =1-0.90$
$\alpha =0.10$
Step 2
Then, a $90\mathrm{%}$ interval for the population variance is
$CI=\left(\frac{\left(n-1\right){S}^{2}}{{X}_{1-\frac{a}{n},n-1}^{2}}<{\sigma }^{2}<\frac{\left(n-1\right){S}^{2}}{{X}_{\frac{a}{n},n-1}^{2}}\right)$
$=\left(\frac{\left(27-1\right){1.22}^{2}}{{X}_{1-\frac{0.10}{2},27-1}^{2}}<{\sigma }^{2}<\frac{\left(27-1\right){1.22}^{2}}{{X}_{\frac{0.10}{2},27-1}^{2}}\right)$
$=\left(\frac{26×{1.22}^{2}}{{X}_{0.95,26}^{2}}<{\sigma }^{2}<\frac{26×{1.22}^{2}}{{X}_{0.5,26}^{2}}\right)$
$=\left(\frac{26×{1.22}^{2}}{38.855}<{\sigma }^{2}<\frac{26×{1.22}^{2}}{15.379}\right)$
$=\left(0.9952<{\sigma }^{2}<2.5163\right)$
Interpretation:
There is $90\mathrm{%}$ confidence that the population variance lies between 0.9952 and 2.5163.
Step 3
The $90\mathrm{%}$ confidence interval for the population standard deviation is
$CI=\left(\frac{\left(n-1\right){S}^{2}}{{X}_{1-\frac{a}{n},n-1}^{2}}<\sigma <\frac{\left(n-1\right){S}^{2}}{{X}_{\frac{a}{n},n-1}^{2}}\right)$
$=\left(\sqrt{\frac{\left(27-1\right){1.22}^{2}}{{X}_{1-\frac{0.10}{2},27-1}}}<\sigma <\sqrt{\frac{\left(27-1\right){1.22}^{2}}{{X}_{\frac{0.10}{2},27-1}^{2}}}\right)$
$=\left(\sqrt{\frac{26×{1.22}^{2}}{{X}_{0.95,26}^{2}}}<\sigma <\sqrt{\frac{26×{1.22}^{2}}{{X}_{0.05,26}^{2}}}\right)$
$\left(\sqrt{\frac{26×{1.22}^{2}}{38.885}}<\sigma <\sqrt{\frac{26×{1.22}^{2}}{15.379}}\right)$ [(Using excel function), (=CHISQ.INV(0.95,26)),