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 thedegrees of freedom, critical values, and confidence interval estimate of \({\bf{\sigma }}\)

The degrees of freedom is computed as,

\(\begin{array}{c}df = n - 1\\ = 153 - 1\\ = 152\end{array}\)

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,

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

Therefore, the 99% confidence interval estimate of\(\sigma \)is\(56.9 < \sigma < 76.6\).