Question

Teaching reading An educator believes that new reading activities in the classroom will help elementary school pupils improve their reading ability. She recruits $44$third-grade students and randomly assigns them into two groups. One group of $21$students does these new activities for an $8$week period. A control group of $23$third-graders follows the same curriculum without the activities. At the end of the $8$weeks, all students are given the Degree of Reading Power (DRP) test, which measures the aspects of reading ability that the treatment is designed to improve. Comparative boxplots and summary statistics for the data from Fathom are shown below.

(a) Based on the graph and numerical summaries, write a few sentences comparing the DRP scores for the two groups.

(b) Is the mean DRP score significantly higher for the students who did the reading activities? Carry out an appropriate test to support your answer.

(c) Can we conclude that the new reading activities caused an increase in the mean DRP score? Explain.

(d) Construct and interpret a $95\%$confidence interval for the difference in mean DRP scores. Explain how this interval provides more information than the significance test in part (b).

Short answer

(a) The Activity Centre appears to be larger than the Control Centre, indicating a higher mean for activities and a more accurate boxplot

(b) Yes, the students who completed the reading activities had a significantly higher mean DRP score.

(c) Yes, the increased reading activities did result in an increase in the mean DRP score.

(d) $\left(0.9688,18.9402\right)$

Step-by-step solution

Part (a) Step 1: Given Information

Given in the question that

We have to write few sentences comparing the DRP scores for the two groups.

Part(a) Step 2: Explanation

The Activity Centre appears to be larger than the Control Centre, indicating a higher mean for activities and a more accurate boxplot. Because the standard deviation is higher and the distance between the boxplot whiskers is bigger, the spread seems to be wider for the Activities group than for the Control group. Because the median (line in the box of the boxplot) tends to be more to the right in the boxplot, both distributions are left-skewed.

Part (b) Step 1: Given Information

Given in the questionthat

$$\begin{gathered} {\overline{x}}_1=51.4762 \\ {\overline{x}}_2=41.5217 \\ {s}_1=11.0074 \\ {s}_2=17.1487 \\ {n}_1=21 \\ {n}_2=23 \end{gathered}$$

we have to determine Is the mean DRP score significantly higher for the students who did the reading activities .

Part(b) Step 2: Explanation

The formula for calculating test statistic is:

$$\begin{gathered} t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{x_1}{n_1}+\frac{x_2^2}{n_2}}} \\ =\frac{51.4762-41.5217}{\sqrt{\frac{11.{0074}^2}{21}+\frac{17.14872}{23}}} \\ =2.311 \end{gathered}$$

The degree of freedom:

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

The P value is

$$\begin{gathered} 0.01<P<0.02 \\ P<0.05\Rightarrow reject{H}_0 \end{gathered}$$

So, the students who completed the reading activities had a significantly higher mean DRP score .

Part (c) Step 1: Given Information 

We have to explain Is it possible to infer that recent reading practices have led to an increase in the average DRP score, yes or no?

Part(c) Step 2: Explanation 

Yes, the increased reading activities did result in an increase in the mean DRP score.

Part (d) Step 1: Given Information

Given in the question that,

$$\begin{gathered} \overline{X_1}=51.476 \\ \overline{X_2}=41.5217 \\ {S}_1=11.0074 \\ {S}_2=17.1487 \\ {n}_1=21 \\ {n}_2=23 \end{gathered}$$

We have to construct and interpret a $95\%$confidence interval for the difference in mean DRP score, and to explain how this interval provides more information than the significance test in part (b).

Part(a) Step 2: Explanation 

The formula of confidence interval is

$\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

The degree of freedom is $20$

Finding $t^{*}$by using $df=20andc=95\%$

$t^{*}=t_{\alpha /2}=2.086$

The Confidence interval is

$$\begin{gathered} \left({\overline{x}}_1-{\overline{x}}_2\right)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=(51.4762-41.5217)-2.086 \\ \times \sqrt{\frac{11.{0074}^2}{21}+\frac{17.{1487}^2}{23}} \\ =0.9688 \end{gathered}$$

$$\begin{gathered} \left({\overline{x}}_1-{\overline{x}}_2\right)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=(51.4762-41.5217)+2.086 \\ \times \sqrt{\frac{11.{0074}^2}{21}+\frac{17.{1487}^2}{23}} \\ =18.9402 \end{gathered}$$