Chapter 7: Q8 (page 340)

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 $σ$ . 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

Expert verified

The degrees of freedom is 152.

The critical values are $χ_{L}^{2} =$ 110.8458 and $χ_{R}^{2} =$ 200.6568.

The 99% confidence interval estimate is 56.9< $σ$ <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,

$d f = n - 1 = 153 - 1 = 152$

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 $χ_{L}^{2} = 110.8458$ and $χ_{R}^{2} = 200.6568$ .

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

$\sqrt{\frac{(n - 1) s^{2}}{χ_{R}^{2}}} < σ < \sqrt{\frac{(n - 1) s^{2}}{χ_{L}^{2}}} \sqrt{\frac{(153 - 1) {65.4}^{2}}{200.6568}} < σ < \sqrt{\frac{(153 - 1) {65.4}^{2}}{110.8458}} 56.9 < σ < 76.6$

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

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Step by step solution

Given information

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

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