Chapter 10.8, Problem 77E

a.

To determine

State whether there is sufficient evidence to conclude that there is a difference in the mean verbal SAT scores for high school students intending to major in engineering and in language at the $\alpha =0.05$.

Obtain the associated p-value.

Answer to Problem 77E

Yes, there is sufficient evidence to conclude that there is a difference in the mean verbal SAT scores for high school students intending to major in engineering and in language at the $\alpha =0.05$.

The attained significance level is approximately 0.

Explanation of Solution

The test hypotheses are given as follows:

Denote ${\mu }_1$ as the mean verbal SAT scores for high school students intending to major in engineering and ${\mu }_2$ as the mean verbal SAT scores for high school students intending to major in language.

Null hypothesis:

$H_0:{\mu }_1={\mu }_2$

That is, there is no difference in the mean verbal SAT scores for high school students intending to major in engineering and in language.

Alternative hypothesis:

$H_a:{\mu }_1\ne {\mu }_2$

That is, there is a difference in the mean verbal SAT scores for high school students intending to major in engineering and in language.

Assumptions:

It is assumed that $Y_1,Y_2,\mathrm{...}{Y}_n$ is an independent random sample from a normal population with ${\sigma }_1^2={\sigma }_2^2$.

Test statistic:

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

$T=\frac{\left({\overline{Y}}_1-{\overline{Y}}_2\right)-D_0}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

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

Here,

 $S_p=\sqrt{\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}}$.

The pooled estimate of ${\sigma }^2$ is obtained as given below:

Substitute $s_1$ as 42, $s_2$ as 45, $n_1$ as 15, and $n_2$ as 15.

$\begin{array}{c}{S}_p=\sqrt{\frac{\left(15-1\right){\left(42\right)}^2+\left(15-1\right){\left(45\right)}^2}{15+15-2}}\\ =\sqrt{\frac{24,696+28,350}{28}}\\ =\sqrt{1,894.5}\\ =43.53\end{array}$

The test statistic is obtained as given below:

Substitute ${\overline{Y}}_1$ as 446, ${\overline{Y}}_2$ as 534, $D_0$ as 0, $S_p$ as 43.53, $n_1$ as 15, and $n_2$ as 15.

$\begin{array}{c}T=\frac{\left(446-534\right)-0}{\left(43.53\right)\sqrt{\frac{1}{15}+\frac{1}{15}}}\\ =\frac{-88}{15.895}\\ =-5.54\end{array}$

Degrees of freedom:

$\begin{array}{c}v=n_1+n_2-2\\ =15+15-2\\ =28\end{array}$

Here, the probability of the test statistic $2\left[P\left(T>5.54\right)\right]$ is obtained as given below:

Step-by-step procedure to obtain the critical value using Applet:

• Go to Applets/Simulations under Book Resources.
• Select Student’s t Probabilities and Quantiles.
• In df, enter 28.
• In x:, enter 5.54.

The output obtained is as follows:

From the output, it is clear that the value must be in the range [–4,4]. Here, the value of x is out of the range. Therefore, the probability of $2\left[P\left(T>5.54\right)\right]$ will be approximately 0.

Decision rule:

• If $p\text{-value}\le \alpha$, reject the null hypothesis.
• Otherwise, fail to reject the null hypothesis.

Conclusion:

Here, the p-value is less than the significance level of 0.05.

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

Therefore, the data provide the evidence to support the claim that there is a difference in the mean verbal SAT scores for high school students intending to major in engineering and in language at $\alpha =0.05$.

b.

To determine

State whether the results in Part (a) are consistent with Exercise 8.90(a).

Explanation of Solution

On observing Exercise 8.90(a), the confidence interval of difference in the mean verbal SAT score for high school students intending to major in engineering and in language is (–120.55, –55.45).

Since 0 does not lie in the confidence interval, the null hypothesis is rejected and supports the conclusion that is same as in Part (a).

Therefore, the results in Part (a) are consistent with Exercise 8.90(a).

c.

To determine

State whether there is sufficient evidence to conclude that there is a difference in the mean math SAT scores for high school students intending to major in engineering and in language at the $\alpha =0.05$.

Obtain the associated p-value.

Answer to Problem 77E

Yes, there is sufficient evidence to conclude that there is a difference in the mean math SAT scores for high school students intending to major in engineering and in language at the $\alpha =0.05$.

The attained significance level is approximately 0.

Explanation of Solution

The test hypotheses are given as follows:

Denote ${\mu }_1$ as the mean math SAT scores for high school students intending to major in engineering and ${\mu }_2$ as the mean math SAT scores for high school students intending to major in language.

Null hypothesis:

$H_0:{\mu }_1={\mu }_2$

That is, there is no difference in mean math SAT scores for high school students intending to major in engineering and in language.

Alternative hypothesis:

$H_a:{\mu }_1\ne {\mu }_2$

That is, there is a difference in mean math SAT scores for high school students intending to major in engineering and in language.

Assumptions:

It is assumed that $Y_1,Y_2,\mathrm{...}{Y}_n$ is a independent random sample from a normal population with ${\sigma }_1^2={\sigma }_2^2$.

Test statistic:

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

$T=\frac{\left({\overline{Y}}_1-{\overline{Y}}_2\right)-D_0}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

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

Here,

 $S_p=\sqrt{\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}}$.

The pooled estimate of ${\sigma }^2$ is obtained as given below:

Substitute $s_1$ as 57, $s_2$ as 52, $n_1$ as 15, and $n_2$ as 15.

$\begin{array}{c}{S}_p=\sqrt{\frac{\left(15-1\right){\left(57\right)}^2+\left(15-1\right){\left(52\right)}^2}{15+15-2}}\\ =\sqrt{\frac{45,486+37,856}{28}}\\ =\sqrt{2,976.5}\\ =54.56\end{array}$

The test statistic is obtained as given below:

Substitute ${\overline{Y}}_1$ as 548, ${\overline{Y}}_2$ as 517, $D_0$ as 0, $S_p$ as 54.56, $n_1$ as 15, and $n_2$ as 15.

$\begin{array}{c}T=\frac{\left(548-517\right)-0}{\left(54.56\right)\sqrt{\frac{1}{15}+\frac{1}{15}}}\\ =\frac{31}{19.9225}\\ =1.56\end{array}$

Degrees of freedom:

$\begin{array}{c}v=n_1+n_2-2\\ =15+15-2\\ =28\end{array}$

Here, the probability of test statistic $2\left[P\left(T>1.56\right)\right]$ is obtained as given below:

Step-by-step procedure to obtain the critical value using Applet:

• Go to Applets/Simulations under Book Resources.
• Select Student’s t Probabilities and Quantiles.
• In df, enter 28.
• In x:, enter 1.56.

The output obtained is as follows:

From the output, the probability of $P\left(T>1.56\right)$ will be 0.65.

Therefore, the probability of $2P\left(T>1.56\right)$ is $\underset{\_}{0.13\left(2\times 0.065\right)}$.

Conclusion:

Here, the p-value is greater than the significance level of 0.05.

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

Therefore, the data do not provide the evidence to support the claim that there is difference in mean math SAT scores for high school students intending to major in engineering and in language at $\alpha =0.05$.

d.

To determine

State whether the results in Part (c) are consistent with Exercise 8.90(b).

Explanation of Solution

On observing Exercise 8.90(b), the confidence interval of difference in the mean math SAT score for high school students intending to major in engineering and in language is (-9.799, 71.799).

Since 0 lie in the confidence interval, the null hypothesis is not rejected and supports the conclusion that is same as in Part (c).

Therefore, the results in Part (c) are consistent with Exercise 8.90(b).