Chapter 10.2, Problem 63E

(a)

To determine

To write a few sentences comparing the DRP scores for the two groups.

Explanation of Solution

The conclusion the educator make at the $\frac{305}{1526}=0.200=20.0\%,\alpha =0.05$ significance level will be as:

Shape: The distribution of activities appears to be left skewed because the box in the boxplot lies to the right between the whiskers. The distribution of control appears to right skewed because the boxplot lies to the left between the whickers.

Center: The center for the activities seems to be higher than the center for control because the boxplot for the activities lies more to the right of the boxplot for control.

Spread: The spread for the control group seems to be more than the spread for the activities group because the width between the whickers of the boxplot for the control group is greater than the width between the whickers for the activities group.

(b)

To determine

To find out what conclusion should the educator make at the $\frac{305}{1526}=0.200=20.0\%,\alpha =0.05$ significance level.

Answer to Problem 63E

There is convincing evidence that the true mean DRP score of third grades who do the new reading activities is higher than the true mean DRP score of third grades who follow the same curriculum without activities.

Explanation of Solution

It is given that:

  $\begin{array}{l}{\overline{x}}_1=51.5\\ {\overline{x}}_2=41.5\\ n_1=21\\ n_2=23\\ s_1=11.0\\ s_2=17.1\end{array}$

  $\alpha =0.05$

Therefore, the appropriate hypotheses for this is:

  $\begin{array}{l}{H}_0:{\mu }_A={\mu }_C\\ H_a:{\mu }_A>{\mu }_C\end{array}$

Now, find the test statistics:

  $\begin{array}{l}t=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{51.5-41.5-0}{\sqrt{\frac{{11}^2}{21}+\frac{{17.1}^2}{23}}}\\ =2.327\end{array}$

Now, the degree of freedom will be:

  $\begin{array}{l}df=\min (n_1-1,n_2-1)\\ =\min (21-1,23-1)\\ =20\end{array}$

So the P -value will be:

  $0.01<P<0.02$

On the other hand by using the calculator command: $2\times \text{tcdf(2}\text{.327,1E99,20)}$ which results in the P -values as $0.03058$ .

And we know that if the P -value is less than or equal to the significance level then the null hypothesis is rejected, then,

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

Thus, we conclude that there is convincing evidence that the true mean DRP score of third grades who do the new reading activities is higher than the true mean DRP score of third grades who follow the same curriculum without activities.

(c)

To determine

To explain can we conclude that the new reading activities caused an increase in the mean DRP score or not.

Answer to Problem 63E

Yes, we conclude that the new reading activities caused an increase in the mean DRP score.

Explanation of Solution

We conclude in part (b) that,

There is convincing evidence that the true mean DRP score of third grades who do the new reading activities is higher than the true mean DRP score of third grades who follow the same curriculum without activities.

A completely randomized experiment randomly assigns all subjects to a group.

Since the students were randomly assigned to a treatment group the experiment is completely randomized experiment and thus the different groups were not as similar as possible prior to the treatment.

This then implies that the difference among the groups after the treatments need to be due to the treatments themselves and thus we can conclude that the new reading activities caused an increase I n the mean DRP score.

(d)

To determine

To explain what type of error could you have made- Type I error or Type II error, based on the conclusion in part (b).

Answer to Problem 63E

Type I error.

Explanation of Solution

We conclude in part (b) that,

There is convincing evidence that the true mean DRP score of third grades who do the new reading activities is higher than the true mean DRP score of third grades who follow the same curriculum without activities.

A type I error occurs if we reject a null hypothesis when the null hypothesis is true. And the Type II error occurs if we fails to reject the null hypothesis when the null hypothesis is false.

Thus, in this case we reject the null hypothesis then it is a Type I error.