# Construct 98% confidence intervals for the a) population variance and the b)population standard deviation sigma The number of hours of reserve capacit

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