Chapter 11.2, Problem 15E

a.

To determine

Conduct the required hypothesis test and state the conclusion using p-value approach.

Answer to Problem 15E

There is enough evidence to infer that the two variances are equal. That is, ${\sigma }_1^2={\sigma }_2^2$.

Explanation of Solution

Calculation:

The given information is that the sample variance for first sample is $s_{{}_1}^2=8.2$, the sample variance for second sample is $s_{{}_2}^2=4$, the sample size for first sample is $n_1=21$, and the sample size for second sample is $n_2=26$. The significance level is $\alpha =0.05$.

The hypotheses are given below:

Null hypothesis:

$H_0:{\sigma }_1^2={\sigma }_2^2$

Alternative hypothesis:

$H_a:{\sigma }_1^2\ne {\sigma }_2^2$

Test statistic:

The formula for test statistic is given below:

$F=\frac{s_1^2}{s_2^2}$ with $n_1-1$ degrees of freedom for the numerator and $n_2-1$ degrees of freedom for the denominator.

Where $s_1^2$ is the sample variance for population 1, and $s_2^2$ is the sample variance for population 2.

Substitute $s_1^2=8.2$ and $s_2^2=4$ in test statistic formula.

$\begin{array}{c}F=\frac{s_1^2}{s_2^2}\\ =\frac{8.2}{4}\\ =\text{2}\text{.05}\end{array}$

Thus, the test statistic is 2.05.

Degrees of freedom for numerator:

$\begin{array}{c}{n}_1-1=21-1\\ =20\end{array}$

Degrees of freedom for denominator:

$\begin{array}{c}{n}_2-1=26-1\\ =25\end{array}$

p-value:

Software procedure:

Step-by-step procedure to obtain the probability value using Excel:

• Open an EXCEL sheet and select the cell A1.
• Enter the formula =F.DIST.RT(2.05,20,25) in the cell A1.
• Press Enter.

Output obtained using EXCEL software is given below:

From the output, the p-value for the upper tail is 0.0452.

For a two-tailed test, the p-value is two times the upper-tail area.

From the output:

$\begin{array}{c}p\text{-value}=2\left(0.0452\right)\\ =0.0904\end{array}$

Thus, the p-value is 0.0904.

p-value approach:

Rejection rule:

If $p\text{-value}\le \alpha ,$, reject the null hypothesis $H_0$.

If $p\text{-value}\le \alpha ,$ do not reject the null hypothesis $H_0$.

Conclusion:

Here, the p-value is greater than the level of significance.

That is, $p\text{-value}\left(=0.0904\right)>\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is not rejected.

There is enough evidence to infer that the two variances are equal. That is, ${\sigma }_1^2={\sigma }_2^2$.

b.

To determine

Conduct the required hypothesis test and state the conclusions using critical value approach.

Answer to Problem 15E

There is enough evidence to infer that the two variances are equal. That is, ${\sigma }_1^2={\sigma }_2^2$.

Explanation of Solution

Calculation:

In Part (a), the test statistic is 2.05.

Critical value:

Step-by-step procedure to obtain the value of $F_{0.025}$ using Table 4 of Appendix B:

• Locate the value 20 and $\alpha =0.025$ in the left column of Table 4.
• Go through the row corresponding to the value 20 and the column corresponding to the value 25 of the table.
• The intersecting value that corresponds to the degrees of freedom 20 and 25 with level of significance 0.025 is 2.30.

The value of $F_{0.025}$ with degrees of freedom 20 and 25 is 2.30.

Rejection rule:

If $F\ge F_{\frac{\alpha }{2}},$ reject the null hypothesis $H_0$.

If $F<F_{\frac{\alpha }{2}},$ do not reject the null hypothesis $H_0$.

Conclusion:

Here, the value of test statistic is lesser than the critical value.

That is, $\text{test}\text{statistic}\left(=2.05\right)<F_{\frac{\alpha }{2}}\left(=2.30\right)$.

Therefore, the null hypothesis is not rejected.

There is enough evidence to infer that the two variances are equal. That is, ${\sigma }_1^2={\sigma }_2^2$.