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.

Nicotine in Menthol Cigarettes 95% confidence;n= 25,s= 0.24 mg.

Short answer

The degrees of freedom is 24.

The critical values are ${\chi }_L^2=$12.401 and ${\chi }_R^2=$39.364.

The 95% confidence interval estimate of is $0.19\text{mg}<\sigma <0.33\text{mg}$.

Step-by-step solution

Given information

The size of the sample is $n=25$.

The sample standard deviation is $s=0.24$.

The level of confidence is 95%.

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

The degrees of freedom (df) is computed as,

$$\begin{gathered} df=n-1 \\ =25-1 \\ =24 \end{gathered}$$

Using the Chi-square table, the critical values are ${\chi }_L^2=12.401$and ${\chi }_R^2=39.364$.

The 95% confidence interval estimate of is computed as,

role="math" localid="1648052944759" $$\begin{gathered} \sqrt{\frac{\left(n-1\right)s^2}{{\chi }_R^2}}<\sigma <\sqrt{\frac{\left(n-1\right)s^2}{{\chi }_L^2}} \\ \sqrt{\frac{\left(25-1\right){0.24}^2}{39.364}}<\sigma <\sqrt{\frac{\left(25-1\right){0.24}^2}{12.401}} \\ 0.19<\sigma <0.33 \end{gathered}$$

Therefore, the 95% confidence interval estimate of role="math" localid="1648052981275" $\sigma$isrole="math" localid="1648052969717" $0.19mg<\sigma <0.33mg$.