Chapter 10, Problem R10.6RE

(a)

To determine

To explain why a two sample t test and not a paired t test is the appropriate inference procedure in this setting.

Explanation of Solution

A two sample t test and not a paired t test is the appropriate inference procedure in this setting because the treatment group and the control group have no subjects in common, which implies that the treatment group and the control group are independent and then it is more appropriate to use a two-sample t test.

(b)

To determine

To write a few sentences comparing the performance of these two groups.

Explanation of Solution

The center the treatment group is higher than the treatment group for the control group, because the median vertical line in the box of the boxplot lies more to the right of the box of the boxplot.

The spread for both groups seems to be about the same, because the width between the whiskers of the boxplots are the roughly same for both boxplots.

The distribution of the control group seems to be right-skewed, while the distribution of the treatment group seems symmetric.

(c)

To determine

To find out do the data provide convincing evidence at the $\alpha =0.01$ significance level that subliminal messages help students like the ones in the study learn math, on average or not.

Answer to Problem R10.6RE

There is sufficient evidence that the subliminal messages help students learn math at the $5\%$ of the significance level.

Explanation of Solution

It is given in the question that:

  $\begin{array}{l}{n}_1=10\\ n_2=8\\ \alpha =0.05\end{array}$

The mean is the sum of all the values divided by the number of values:

  $\begin{array}{l}{\overline{x}}_1=\frac{6+7+12+11+15+16+11+13+13+10}{10}\\ =11.4\\ {\overline{x}}_2=\frac{11+5+4+8+14+5+7+12}{8}\\ =8.25\end{array}$

The standard deviation is as follows:

  $\begin{array}{l}{s}_1=\sqrt{\frac{\begin{array}{}$ {(6-11.4)}^2+{(7-11.4)}^2+{(12-11.4)}^2+{(11-11.4)}^2+{(15-11.4)}^2+{(16-11.4)}^2$$ +{(11-11.4)}^2+{(13-11.4)}^2+{(13-11.4)}^2+{(10-11.4)}^2$\end{array}}{10-1}}\\ =3.1693\\ s_2=\sqrt{\frac{\begin{array}{}$ {(11-8.25)}^2+{(5-8.25)}^2+{(4-8.25)}^2+{(8-8.25)}^2+$$ {(14-8.25)}^2+{(5-8.25)}^2+{(7-8.25)}^2+{(12-8.25)}^2$\end{array}}{8-1}}\\ =3.6936\end{array}$

Now, we will conduct the hypothesis test:

Claim: Treatment group has higher mean then the control group.

The claim is either the null hypothesis or the alternative hypothesis.

Thus, define hypothesis:

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_a:{\mu }_1>{\mu }_2\end{array}$

Now to determine the value of test statistics:

  $\begin{array}{l}t=\frac{{\overline{x}}_1-{\overline{x}}_2-({\mu }_1-{\mu }_2)}{\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}}\\ =\frac{11.4-8.25-0}{\sqrt{\frac{{3.1693}^2}{10}+\frac{{3.6936}^2}{8}}}\\ =1.914\end{array}$

And calculate the degree of freedom:

  $\begin{array}{l}df=\min (n_1-1,n_2-1)\\ =\min (10-1,8-1)\\ =7\end{array}$

The P-value is the probability of obtaining the value of test statistics when the null hypothesis is true. The P-value will be:

  $0.025<P<0.05$

If the P-value is less than or equal to the significance level then the null hypothesis is rejected:

  $P<0.05\Rightarrow \text{Reject }{H}_0$

So there is sufficient evidence that the subliminal messages help students learn math at the $5\%$ of the significance level.

(d)

To determine

To explain why or why not can we generate these results to the population of all students who failed the mathematics part of the city university of New York skills assessment test.

Answer to Problem R10.6RE

No.

Explanation of Solution

No, we cannot generate these results to the population of all students who failed the mathematics part of the city university of New York skills assessment testbecause the students agreed to participate in the study and thus the sample is not a random sample. Since the random requirement is not satisfied, it is not possible to generalize to the entire population.