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. (1.10,2.01) b. (0.32,0.85) c. (0.53,1.01) d. (0.81,1.83)

Explanation:

Step1: Calculate sample variance

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

Step2: Determine the critical values

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

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

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

Answer:

D. \((0.81,1.83)\)