Question

the grade point averages for 10 randomly selected students are listed below. construct a 90% confidence interval for the population standard deviation, σ. assume the sample is from a normally distributed population.
2.0 3.2 1.8 2.9 0.9 4.0 3.3 2.9 3.6 0.8
a. (0.32, 0.85)
b. (0.81, 1.83)
c. (0.53, 1.01)
d. (1.10, 2.01)

Explanation:

Step1: Calculate the sample standard deviation \(s\)

First, find the sample mean \(\bar{x}=\frac{2.0 + 3.2+1.8+2.9+0.9+4.0+3.3+2.9+3.6+0.8}{10}=\frac{25.4}{10} = 2.54\)

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

\(\sum_{i=1}^{10}(x_{i}-\bar{x})^{2}=(2.0 - 2.54)^{2}+(3.2-2.54)^{2}+(1.8 - 2.54)^{2}+(2.9-2.54)^{2}+(0.9-2.54)^{2}+(4.0-2.54)^{2}+(3.3-2.54)^{2}+(2.9-2.54)^{2}+(3.6-2.54)^{2}+(0.8-2.54)^{2}\)

\(=(- 0.54)^{2}+(0.66)^{2}+(-0.74)^{2}+(0.36)^{2}+(-1.64)^{2}+(1.46)^{2}+(0.76)^{2}+(0.36)^{2}+(1.06)^{2}+(-1.74)^{2}\)

\(=0.2916+0.4356 + 0.5476+0.1296+2.6896+2.1316+0.5776+0.1296+1.1236+3.0276\)

\(=11.184\)

\(s=\sqrt{\frac{11.184}{9}}\approx1.11\)

Step2: Determine the critical values

For a \(90\%\) confidence interval and \(n = 10\) (degrees of freedom \(df=n - 1=9\)), the critical values for the \(\chi^{2}\) distribution are \(\chi_{1-\alpha/2,df}^{2}=\chi_{0.95,9}^{2}=3.325\) and \(\chi_{\alpha/2,df}^{2}=\chi_{0.05,9}^{2}=16.919\)

Step3: Calculate the confidence interval

The formula for the confidence interval for the population standard deviation \(\sigma\) is \(\sqrt{\frac{(n - 1)s^{2}}{\chi_{\alpha/2,df}^{2}}}<\sigma<\sqrt{\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2,df}^{2}}}\)

Substitute \(n = 10\), \(s\approx1.11\), \(\chi_{0.05,9}^{2}=16.919\), \(\chi_{0.95,9}^{2}=3.325\)

\(\sqrt{\frac{(10 - 1)\times(1.11)^{2}}{16.919}}<\sigma<\sqrt{\frac{(10 - 1)\times(1.11)^{2}}{3.325}}\)

First, \((10 - 1)\times(1.11)^{2}=9\times1.2321 = 11.0889\)

\(\sqrt{\frac{11.0889}{16.919}}\approx\sqrt{0.655}\approx0.81\)

\(\sqrt{\frac{11.0889}{3.325}}\approx\sqrt{3.335}\approx1.83\)

Answer:

B. \((0.81,1.83)\)