Question

1. the daily intakes of milk (in ounces) for ten randomly selected people were:

23.3 28.4 10.5 16.4 26.4
18.1 20.4 17.3 27.4 13.2
find a 99% confidence interval for the population standard deviation σ.
a) 3.78 oz < σ < 13.95 oz b) 3.66 oz < σ < 12.53 oz
c) 0.88 oz < σ < 3.38 oz d) 3.78 oz < σ < 12.53 oz

Explanation:

Step1: Calculate the sample variance \(s^{2}\)

First, find the sample mean \(\bar{x}=\frac{23.3 + 28.4+10.5 + 16.4+26.4+18.1+20.4+17.3+27.4+13.2}{10}=\frac{201.4}{10} = 20.14\)

Then, \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\)

\((23.3-20.14)^{2}=9.8596\), \((28.4 - 20.14)^{2}=68.2276\), \((10.5-20.14)^{2}=92.9296\), \((16.4-20.14)^{2}=13.9876\), \((26.4-20.14)^{2}=39.1876\), \((18.1-20.14)^{2}=4.1616\), \((20.4-20.14)^{2}=0.0676\), \((17.3-20.14)^{2}=8.0656\), \((27.4-20.14)^{2}=52.7076\), \((13.2-20.14)^{2}=48.1636\)

\(\sum_{i=1}^{10}(x_{i}-\bar{x})^{2}=9.8596+68.2276 + 92.9296+13.9876+39.1876+4.1616+0.0676+8.0656+52.7076+48.1636=337.358\)

\(s^{2}=\frac{337.358}{9}\approx37.4842\)

Step2: Determine the critical values

For a \(99\%\) confidence interval and \(n = 10\) (so \(df=n - 1=9\)), \(\alpha=0.01\), \(\frac{\alpha}{2}=0.005\) and \(1-\frac{\alpha}{2}=0.995\)

From the \(\chi^{2}\) - distribution table, \(\chi_{L}^{2}=\chi_{0.005,9}^{2}=23.589\) and \(\chi_{U}^{2}=\chi_{0.995,9}^{2}=1.735\)

Step3: Calculate the confidence interval for \(\sigma\)

The formula for the confidence interval for \(\sigma\) is \(\sqrt{\frac{(n - 1)s^{2}}{\chi_{U}^{2}}}<\sigma<\sqrt{\frac{(n - 1)s^{2}}{\chi_{L}^{2}}}\)

Substitute \(n - 1 = 9\), \(s^{2}\approx37.4842\), \(\chi_{U}^{2}=1.735\) and \(\chi_{L}^{2}=23.589\)

\(\sqrt{\frac{9\times37.4842}{23.589}}<\sigma<\sqrt{\frac{9\times37.4842}{1.735}}\)

\(\sqrt{\frac{337.3578}{23.589}}<\sigma<\sqrt{\frac{337.3578}{1.735}}\)

\(\sqrt{14.3}<\sigma<\sqrt{194.44}\)

\(3.78<\sigma<13.95\)

Answer:

A. \(3.78\space oz<\sigma<13.95\space oz\)