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

Explanation:

Step1: Calculate sample standard deviation

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

The formula for the confidence interval for the population standard deviation \(\sigma\) is \(\sqrt{\frac{(n - 1)s^{2}}{\chi_{R}^{2}}}<\sigma<\sqrt{\frac{(n - 1)s^{2}}{\chi_{L}^{2}}}\)
Substitute \(n = 10\), \(s\approx1.11\), \(\chi_{L}^{2}=3.325\), \(\chi_{R}^{2}=16.919\)
\(\sqrt{\frac{9\times(1.11)^{2}}{16.919}}<\sigma<\sqrt{\frac{9\times(1.11)^{2}}{3.325}}\)
\(\sqrt{\frac{9\times1.2321}{16.919}}<\sigma<\sqrt{\frac{9\times1.2321}{3.325}}\)
\(\sqrt{\frac{11.0889}{16.919}}<\sigma<\sqrt{\frac{11.0889}{3.325}}\)
\(\sqrt{0.6554}<\sigma<\sqrt{3.335}\)
\(0.81<\sigma<1.83\)

Answer:

B. \((0.81, 1.83)\)