Question

you randomly select and measure the contents of 10 bottles of cough syrup. the results (in fluid ounces) are shown to the right. 4.218 4.298 4.253 4.242 4.188 4.244 4.265 4.246 4.221 4.236
assume the sample is taken from a normally distributed population. construct 80% confidence intervals for (a) the population variance ( sigma^{2} ) and (b) the population standard deviation ( sigma ). interpret the results.
(a) the confidence interval for the population variance is (). (round to six decimal places as needed.)

Explanation:

Step1: Calculate sample mean $\bar{x}$

$$\bar{x}=\frac{4.218 + 4.298+4.253+4.242+4.188+4.244+4.265+4.246+4.221+4.236}{10}$$
$$\bar{x}=\frac{42.411}{10}=4.2411$$

Step2: Calculate sample variance $s^{2}$

$$s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$$
$$(x_1-\bar{x})=(4.218 - 4.2411)=- 0.0231,(x_2-\bar{x})=(4.298 - 4.2411)=0.0569,(x_3-\bar{x})=(4.253 - 4.2411)=0.0119,(x_4-\bar{x})=(4.242 - 4.2411)=0.0009,(x_5-\bar{x})=(4.188 - 4.2411)=-0.0531,(x_6-\bar{x})=(4.244 - 4.2411)=0.0029,(x_7-\bar{x})=(4.265 - 4.2411)=0.0239,(x_8-\bar{x})=(4.246 - 4.2411)=0.0049,(x_9-\bar{x})=(4.221 - 4.2411)=-0.0201,(x_{10}-\bar{x})=(4.236 - 4.2411)=-0.0051$$
$$\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=(-0.0231)^{2}+(0.0569)^{2}+(0.0119)^{2}+(0.0009)^{2}+(-0.0531)^{2}+(0.0029)^{2}+(0.0239)^{2}+(0.0049)^{2}+(-0.0201)^{2}+(-0.0051)^{2}$$
$$=0.00053361+0.00323761+0.00014161+0.00000081+0.00281961+0.00000841+0.00057121+0.00002401+0.00040401+0.00002601$$
$$=0.0077669$$
$$s^{2}=\frac{0.0077669}{9}\approx0.000863$$

Step3: Determine degrees of freedom and $\chi^{2}$ - values

Degrees of freedom $df=n - 1=10 - 1 = 9$
For an $80\%$ confidence interval, $\alpha=1 - 0.80 = 0.20$, $\frac{\alpha}{2}=0.10$
From the $\chi^{2}$ - distribution table, $\chi_{L}^{2}=\chi_{1-\frac{\alpha}{2},df}^{2}=\chi_{0.90,9}^{2}=4.168$, $\chi_{R}^{2}=\chi_{\frac{\alpha}{2},df}^{2}=\chi_{0.10,9}^{2}=14.684$

Step4: Construct the confidence interval for $\sigma^{2}$

The formula for the confidence interval of the population variance is $\frac{(n - 1)s^{2}}{\chi_{R}^{2}}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{L}^{2}}$
Substitute $n = 10$, $s^{2}=0.000863$, $\chi_{L}^{2}=4.168$, $\chi_{R}^{2}=14.684$
$$\frac{(10 - 1)\times0.000863}{14.684}\leq\sigma^{2}\leq\frac{(10 - 1)\times0.000863}{4.168}$$
$$\frac{0.007767}{14.684}\leq\sigma^{2}\leq\frac{0.007767}{4.168}$$
$$0.00053\leq\sigma^{2}\leq0.00186$$

Answer:

$(0.000530,0.001860)$