Question

Question:In Exercises 5–8, use the given information to find the number of degrees of freedom, the critical values X2 L and X2R, and the confidence interval estimate of $\sigma$. The samples are from Appendix B and it is reasonable to assume that a simple random sample has been selected from a population with a normal distribution.

Heights of Men 99% confidence;n= 153,s= 7.10 cm.

Short answer

The degrees of freedom is 152.

The critical values are ${\chi }_L^2=$110.8458 and ${\chi }_R^2=$200.6568.

The 99% confidence interval estimate is 56.9<$\sigma$ <76.6.

Step-by-step solution

Given information

The size of the sample is $n=153$.

The sample standard deviation is $s=7.10$.

The level of confidence is 99%.

Compute the degrees of freedom, critical values, and confidence interval estimate of  σ

The degrees of freedom is computed as,

$$\begin{gathered} df=n-1 \\ =153-1 \\ =152 \end{gathered}$$

The level of confidence is 99%, which implies that the level of significance is 0.01.

Using the Chi-square table, the critical values at 0.01 level of significance and at 152 degrees of freedom are ${\chi }_L^2=110.8458$ and${\chi }_R^2=200.6568$.

The 95% confidence interval estimate of $\sigma$ is computed as,

$$\begin{gathered} \sqrt{\frac{\left(n-1\right)s^2}{{\chi }_R^2}}<\sigma <\sqrt{\frac{\left(n-1\right)s^2}{{\chi }_L^2}} \\ \sqrt{\frac{\left(153-1\right){65.4}^2}{200.6568}}<\sigma <\sqrt{\frac{\left(153-1\right){65.4}^2}{110.8458}} \\ 56.9<\sigma <76.6 \end{gathered}$$

Therefore, the 99% confidence interval estimate of role="math" localid="1648107586512" $\sigma$ is role="math" localid="1648107574037" $56.9<\sigma <76.6$.