Chapter 10.3, Problem 22E

a.

To determine

Provide the appropriate alternative and null hypotheses.

Answer to Problem 22E

Null hypothesis: $H_0:{\mu }_1={\mu }_2$.

Alternative hypothesis: $H_a:{\mu }_1\ne {\mu }_2$.

Explanation of Solution

It may be expected that there is no difference in the mean pre-test scores for students who were taught using the traditional method and activity-oriented method.

Therefore, the test hypotheses are given below:

Denote ${\mu }_1$ as the mean pre-test scores for students who were taught using the traditional method and ${\mu }_2$ as the mean pre-test scores for students who were taught using activity-oriented method.

Null hypothesis:

 $H_0:{\mu }_1={\mu }_2$.

That is, there is no difference in the mean pre-test scores for students who taught using the traditional method and activity-oriented method.

Alternative hypothesis: $H_a:{\mu }_1\ne {\mu }_2$.

That is, there is difference in the mean pre-test scores for students who were taught using the traditional method and activity-oriented method.

b.

To determine

State whether the alternative hypothesis in Part (a) imply a one-tailed or two-tailed test.

Answer to Problem 22E

Two-tailed test.

Explanation of Solution

From Part (a), it is observed that the alternative hypothesis is $H_a:{\mu }_1\ne {\mu }_2$. This implies a two-tailed test.

c.

To determine

State whether the data provide the evidence to conclude that there is no difference in the mean pre-test scores for students who taught using the traditional method and activity-oriented method.

Answer to Problem 22E

There is no sufficient evidence that there is difference in the mean pre-test scores for students who were taught using the traditional method and activity-oriented method.

Explanation of Solution

Test statistic:

For large sample n, the sample mean ${\overline{Y}}_1\text{and}{\overline{Y}}_2$ is a point estimate of ${\mu }_1\text{ and }{\mu }_2$. Thus, the test statistic is given below:

$Z=\frac{\left({\overline{Y}}_1-{\overline{Y}}_2\right)-D_0}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}}$

Where, $D_0={\mu }_1-{\mu }_2$ is the difference, $n_1\text{and }{n}_2$ is the sample sizes of two population and ${\sigma }_1^2\text{and}{\sigma }_2^2$ is the population variances.

For large samples $(n_i>30)$, the population variance is not known. The sample variances provide a good estimate of the population.

Substitute ${\overline{Y}}_1$ as 14.06, ${\overline{Y}}_2$ as 13.38, $D_0$ as 0, $s_1$ as 5.45, $s_2$ as 5.59, $n_1$ as 368, and $n_2$ as 372.

$\begin{array}{c}Z=\frac{\left({\overline{Y}}_1-{\overline{Y}}_2\right)-D_0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{14.06-13.38}{\sqrt{\frac{{5.45}^2}{368}+\frac{{5.59}^2}{372}}}\\ =\frac{0.68}{0.4058}\\ =1.675\end{array}$

Rejection region:

Here, the alternative is two-tailed test. In context, the level of the test, $\alpha$ is 0.01.

If $z\le -z_{\frac{\alpha }{2}}\text{or }z\ge z_{\frac{\alpha }{2}}$, reject the null hypothesis.

The critical value has to be obtained for $z_{\frac{0.01}{2}}=z_{0.005}$.

Critical value:

From the “Appendix 3, Table 4 Normal curve areas”, the critical value for the level of significance 0.005 is 2.58.

Thus, the rejection region under level of significance of 0.01 is $z\le -2.58\text{or }z\ge 2.58$.

Decision rule:

• If $z\le -2.58\text{or }z\ge 2.58$, reject the null hypothesis.
• Otherwise, fail to reject the null hypothesis.

Conclusion:

Here, the test statistic, Z does not fall in the rejection region.

Therefore, by the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence to conclude that there is difference in the mean pre-test scores for students who were taught using the traditional method and activity-oriented method.