# Construct 98% confidence intervals for the a) population variance and the b)population standard deviation sigma The number of hours of reserve capacity of 10 randomly selected automotive batteries. 1.71, 1.87, 1.58, 1.61, 1.78, 1.98, 1.36, 1.58, 1.47, 2.05, Assume the sample is taken from a normally distributed population.

Question
Confidence intervals
Construct 98% confidence intervals for the a) population variance and the b)population standard deviation $$\displaystyle\sigma$$ The number of hours of reserve capacity of 10 randomly selected automotive batteries. 1.71, 1.87, 1.58, 1.61, 1.78, 1.98, 1.36, 1.58, 1.47, 2.05, Assume the sample is taken from a normally distributed population.

2020-11-09
Use STDEV function in excel to find the standard deviation. Then, $$\displaystyle{s}={0.2215}.$$ And the variance is $$\displaystyle{s}^{{{2}}}={\left({0.2215}\right)}^{{{2}}}$$
$$\displaystyle={0.04906}$$ Step 2 At 98% level, $$\displaystyle\alpha={1}-{98}\%$$
$$\displaystyle={1}-{0.98}$$
$$\displaystyle={0.02}$$ And The degree of freedom $$\displaystyle={n}-{1}$$
$$\displaystyle={10}-{1}$$
$$\displaystyle={9}$$ Step 3 The critical values are $$\displaystyle{X}^{{{2}}}_{\left\lbrace\alpha,{n}-{1}\right\rbrace}={X}^{{{2}}}_{\left\lbrace{0.02},{9}\right\rbrace}$$
$$\displaystyle={19.679}$$ And $$\displaystyle{X}^{{{2}}}_{\left\lbrace{1}-\alpha,{n}-{1}\right\rbrace}={X}^{{{2}}}_{\left\lbrace{0.98},{9}\right\rbrace}$$
$$\displaystyle={2.5324}$$ Step 4 The 98% confidence interval for variance is $$\displaystyle{\left[{\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{{x}^{{{2}}}_{\left\lbrace\alpha,{n}-{1}\right\rbrace}}}}\leq\sigma^{{{2}}}\leq{\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{{x}^{{{2}}}_{\left\lbrace{1}-\alpha,{n}-{1}\right\rbrace}}}}\right]}={\frac{{{9}{\left({0.04906}\right)}}}{{{19.679}}}}\leq\sigma^{{{2}}}\leq{\frac{{{9}{\left({0.04906}\right)}}}{{{2.5324}}}}$$
$$\displaystyle={\frac{{{0.44154}}}{{{19.679}}}}\leq\sigma^{{{2}}}\leq{\frac{{{0.44154}}}{{{2.5324}}}}$$
$$\displaystyle={0.022}\leq\sigma^{{{2}}}\leq{0.174}$$ The 98% confidence interval for standard deviation is $$\displaystyle{\left[\sqrt{{{\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{{X}^{{{2}}}_{\left\lbrace\alpha,{n}-{1}\right\rbrace}}}}}}\leq\sigma\leq\sqrt{{{\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{'{X}^{{{2}}}_{\left\lbrace{1}-\alpha,{n}-{1}\right\rbrace}}}}}}\right]}=\sqrt{{{0.022}}}\leq\sigma\leq\sqrt{{{0.174}}}$$
$$\displaystyle={0.150}\leq\sigma\leq{0.417}$$

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