# Use technology to construct the confidence intervals for the population variance sigma^{2} and the population standard deviation sigma. Assume the sam

Use technology to construct the confidence intervals for the population variance ${\sigma }^{2}$ and the population standard deviation $\sigma$. Assume the sample is taken from a normally distributed population. $c=0.99,s=37,n=20$ The confidence interval for the population variance is (?, ?). The confidence interval for the population standard deviation is (?, ?)
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Step 1 Solution: We need to construct the 99% confidence interval for the population variance. We have been provided with the following information about the sample variance and sample size: ${s}^{2}=1369$
$n=20$ Step 2 The critical values for $\alpha =0.01$ and $df=19$ degrees of freedom are ${X}_{L}^{2}={X}_{1-\frac{\alpha }{2},n-1}^{2}=6.844$ and ${X}_{U}^{2}={X}_{1-\frac{\alpha }{2},n-1}^{2}=38.5823$. The corresponding confidence interval is computed as shown below: $CI\left(Variance\right)=\left(\frac{\left(n-1\right){s}^{2}}{{X}_{\frac{\alpha }{2},n-1}^{2}},\frac{\left(n-1\right){s}^{2}}{{X}_{1-\frac{\alpha }{2},n-1}^{2}}\right)$
$\left(\frac{\left(20-1\right)×1369}{38.5823},\frac{\left(20-1\right)×1369}{6.844}$
$=\left(674.17,3800.5711\right)$ Now that we have the limits for the confidence interval, the limits for the 99% confidence interval for the population standard deviation are obtained by simply taking the squared root of the limits of the confidence interval for the variance, so then: CI(Standard Deviation) $=\left(\sqrt{674.17},\sqrt{3800.5711}\right)=\left(25.9648,61.6488\right)$ Therefore, based on the data provided, the 99% confidence interval for the population variance is $674.17<{\sigma }^{2}<3800.5711$, and the 99% confidence interval for the population standard deviation is $25.9648<\sigma <61.6488$. Step 3 Ci for population variance (674.17 , 3800.57) Ci for population standard deviation ( 25.96 , 61.65)