Question

you randomly select and measure the contents of 10 bottles of cough 4.218 4.298 4.253 4.242 4.188 syrup. the results (in fluid ounces) are shown to the right. 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 and sample variance

First, calculate the sample mean \(\bar{x}\) using the formula \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
\(\sum_{i=1}^{10}x_{i}=4.218 + 4.298+4.253+4.242+4.188+4.244+4.265+4.246+4.221+4.236 = 42.411\)
\(\bar{x}=\frac{42.411}{10}=4.2411\)

Then, calculate the sample variance \(s^{2}\) using the formula \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\)
\(\sum_{i=1}^{10}(x_{i}-\bar{x})^{2}=(4.218 - 4.2411)^{2}+(4.298 - 4.2411)^{2}+(4.253 - 4.2411)^{2}+(4.242 - 4.2411)^{2}+(4.188 - 4.2411)^{2}+(4.244 - 4.2411)^{2}+(4.265 - 4.2411)^{2}+(4.246 - 4.2411)^{2}+(4.221 - 4.2411)^{2}+(4.236 - 4.2411)^{2}\)
\(\sum_{i=1}^{10}(x_{i}-\bar{x})^{2}=0.00053361+0.00323761+0.00014161+0.00000081+0.00281961+0.00000841+0.00057121+0.00002401+0.00040401+0.00002601=0.0077669\)
\(s^{2}=\frac{0.0077669}{9}\approx0.000863\)

Step2: Determine the critical values

For a \(80\%\) confidence interval and \(n - 1=9\) degrees of freedom, \(\alpha=1 - 0.80 = 0.20\), \(\frac{\alpha}{2}=0.10\)

From the \(\chi^{2}\) - distribution table, \(\chi_{L}^{2}=\chi_{1-\frac{\alpha}{2},n - 1}^{2}=\chi_{0.90,9}^{2}=4.168\), \(\chi_{R}^{2}=\chi_{\frac{\alpha}{2},n - 1}^{2}=\chi_{0.10,9}^{2}=14.684\)

Step3: Calculate the confidence interval for the population variance

The formula for the confidence interval for the population variance \(\sigma^{2}\) 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:

\((0.000530,0.001860)\)