91影视

Given in the question that

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

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.

Given in the questionthat

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

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

The formula for calculating test statistic is:

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

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

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

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

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?

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

Given in the question that,

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

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).

The formula of confidence interval is

$\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)卤t_{伪/2}脳\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

The degree of freedom is $20$

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

$t^{*}=t_{伪/2}=2.086$

The Confidence interval is

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

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