91Ó°ÊÓ

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

Because the box in the boxplot lies to the right between the whiskers, the distribution of activities appears to be skewed. Because the boxplot lies to the left between the whickers, the control distribution appears to be skewed to the right.

Because the boxplot for the activities is more to the right of the boxplot for control, the center for the activities appears to be higher than the center for control.

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, the spread for the control group appears to be greater than the spread for the activities group.

$$\begin{gathered} {\stackrel{¯}{x}}_1=51.5 \\ {\stackrel{¯}{x}}_2=41.5 \\ {n}_1=21 \\ {n}_2=23 \\ {s}_1=11.0 \\ {s}_2=17.1 \\ \mathrm{Î\pm }=0.05 \end{gathered}$$

The appropriate hypotheses for this are:

$$\begin{gathered} H_0:{\mathrm{μ}}_A={\mathrm{μ}}_c \\ {H}_0:{\mathrm{μ}}_A>{\mathrm{μ}}_C \end{gathered}$$

Locate the following test statistics:

$$\begin{gathered} =\frac{51.5-41.5-0}{\sqrt{\frac{112}{21}+\frac{17.12}{23}}} \\ =2.327 \end{gathered}$$

The degree of liberty will now be:

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

So the P-value will be:

$0.01<P<0.02$

On the other hand by using the calculator command: $2×tcdf(2.327,1\mathrm{E}99,20)$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,

$P<0.05⇒\text{Reject}{H}_0$

Therefore, we conclude that the true mean DRP score of third-graders who participate in the new reading activities is higher than the true mean DRP score of third-graders who do not participate in the new reading activities.

We conclude in part (b) that,

There is compelling evidence that the true mean DRP score of third-grade students who participate in the new reading activities is higher than the true mean DRP score of third-grade students who do not participate in the new reading activities.

In a completely randomized experiment, all subjects are assigned to a group at random.

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

As a result, the difference between the groups after the treatments must be due to the treatments, and we can conclude that the new reading activities resulted in an increase I $n$the mean DRP score.

We conclude in part (b) that,

There is compelling evidence that the true mean DRP score of third-grade students who participate in the new reading activities is higher than the true mean DRP score of third-grade students who do not participate in the new reading activities.

When we reject a null hypothesis when the null hypothesis is true, we make a type I error. When we fail to reject the null hypothesis when the null hypothesis is false, we make a Type II error.

As a result, if we reject the null hypothesis, we have made a Type I error.