Question

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

Explanation:

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

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, calculate \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\).
\(\sum_{i=1}^{10}(x_{i}-\bar{x})^{2}=(2.0 - 2.54)^{2}+(3.2-2.54)^{2}+(1.8 - 2.54)^{2}+(2.9-2.54)^{2}+(0.9-2.54)^{2}+(4.0 - 2.54)^{2}+(3.3-2.54)^{2}+(2.9-2.54)^{2}+(3.6-2.54)^{2}+(0.8-2.54)^{2}\)
\(=(- 0.54)^{2}+0.66^{2}+(-0.74)^{2}+0.36^{2}+(-1.64)^{2}+1.46^{2}+0.76^{2}+0.36^{2}+1.06^{2}+(-1.74)^{2}\)
\(=0.2916 + 0.4356+0.5476+0.1296+2.6896+2.1316+0.5776+0.1296+1.1236+3.0276=11.084\)
\(s^{2}=\frac{11.084}{9}\approx1.2316\)

Step2: Determine \(\chi^{2}\) values

For a \(90\%\) confidence interval and \(n = 10\) (degrees of freedom \(df=n - 1=9\)), \(\alpha=1 - 0.90 = 0.10\), \(\frac{\alpha}{2}=0.05\), \(1-\frac{\alpha}{2}=0.95\).
From the \(\chi^{2}\) - distribution table, \(\chi_{1-\frac{\alpha}{2},df}^{2}=\chi_{0.95,9}^{2}=3.325\), \(\chi_{\frac{\alpha}{2},df}^{2}=\chi_{0.05,9}^{2}=16.919\)

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

The formula for the confidence interval for the population standard deviation \(\sigma\) is \(\sqrt{\frac{(n - 1)s^{2}}{\chi_{\frac{\alpha}{2},df}^{2}}}<\sigma<\sqrt{\frac{(n - 1)s^{2}}{\chi_{1-\frac{\alpha}{2},df}^{2}}}\)
Substitute \(n = 10\), \(s^{2}\approx1.2316\), \(\chi_{0.05,9}^{2}=16.919\), \(\chi_{0.95,9}^{2}=3.325\)
\(\sqrt{\frac{9\times1.2316}{16.919}}<\sigma<\sqrt{\frac{9\times1.2316}{3.325}}\)
\(\sqrt{\frac{11.0844}{16.919}}<\sigma<\sqrt{\frac{11.0844}{3.325}}\)
\(\sqrt{0.655}<\sigma<\sqrt{3.334}\)
\(0.81<\sigma<1.83\)

Answer:

C. \((0.81,1.83)\)