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.

Platelet Counts of Women 99% confidence;n= 147,s= 65.4.

Short answer

The degrees of freedom is 146.

The critical values are ${\chi }_L^2=$105.7411 and ${\chi }_R^2=$193.7611.

The 99% confidence interval estimate is 56.8<$\sigma$<76.8.

Step-by-step solution

Given information

The size of the sample is $n=147$.

The sample standard deviation is $s=65.4$.

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,

$$\begin{gathered} df=n-1 \\ =147-1 \\ =146 \end{gathered}$$

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 146 degrees of freedom are ${\chi }_L^2=105.7411$and ${\chi }_R^2=193.7611$.

The 95% confidence interval estimate of is computed as,

$$\begin{gathered} \sqrt{\frac{\left(n-1\right)s^2}{{\chi }_R^2}}<\sigma <\sqrt{\frac{\left(n-1\right)s^2}{{\chi }_L^2}} \\ \sqrt{\frac{\left(147-1\right){65.4}^2}{193.7611}}<\sigma <\sqrt{\frac{\left(147-1\right){65.4}^2}{105.7411}} \\ 56.8<\sigma <76.8 \end{gathered}$$

Therefore, the 99% confidence interval estimate of $\sigma$is $56.8<\sigma <76.8$.