Chapter 10.2, Problem 51E

(a)

To determine

To Write: a few lines comparing, on the basis of graph and numerical summaries, the DRP scores for the two groups.

Explanation of Solution

Given:

  

The Activity Centre seems to be greater than the control Centre, so there is a higher mean for activities and the boxplot is more correct. The spread appears to be more than the spread for the Activities group for the Control group, since it has a higher standard deviation and the distance is greater between the boxplot whiskers. Both distributions tend to be left-skewed, since in the boxplot, the median (line in the box of the boxplot) tends to be more to the right.

(b)

To determine

To Explain: the finding that an improvement in the mean DRP score was prompted by new reading practises.

Answer to Problem 51E

Yes

Explanation of Solution

Given:

  $\begin{array}{l}{\overline{x}}_1=51.4762\\ {\overline{x}}_2=41.5217\\ s_1=11.0074\\ s_2=17.1487\\ n_1=21\\ n_2=23\end{array}$

Formula used:

  $t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

Calculation:

The test statistic:

  $\begin{array}{c}t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{51.4762-41.5217}{\sqrt{\frac{{11.0074}^2}{21}+\frac{{17.1487}^2}{23}}}\\ =2.311\end{array}$

Degrees of freedom:

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

The P-value is

  $0.01<P<0.02$

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

Enough evidence exists to support the statement that the mean DRP score for the children who took the reading activities is significantly greater.

(c)

To determine

To Explain: that it can be inferred that the latest reading practises have contributed to an improvement in the average DRP score, yes or no.

Answer to Problem 51E

Yes

Explanation of Solution

Yes, since the data was obtained by using a randomised trial and a randomised trial can show causation.

(d)

To determine

To construct: and interpret the difference in mean DRP scores for a 95 percent confidence interval and clarify that this interval presents more details than part (b) of the significance test.

Answer to Problem 51E

(0.9688, 18.9402)

Explanation of Solution

Given:

  $\begin{array}{c}{\overline{x}}_1=51.4762\\ {\overline{x}}_2=41.5217\\ s_1=11.0074\\ s_2=17.1487\\ n_1=21\\ n_2=23\end{array}$

Formula used:

For the confidence interval

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

Calculation:

Degrees of freedom is

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

Finding the $t^{\ast }$ with $df=20$ and $c=95\%$

  $t^{\ast }=t_{\alpha /2}=2.086$

The confidence interval for ${\mu }_1-{\mu }_2$ are:

  $\begin{array}{c}\left({\overline{x}}_1-{\overline{x}}_2\right)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=\left(51.4762-41.5217\right)-2.086\times \sqrt{\frac{{11.0074}^2}{21}+\frac{{17.1487}^2}{23}}=0.9688\\ \left({\overline{x}}_1-{\overline{x}}_2\right)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=\left(51.4762-41.5217\right)+2.086\times \sqrt{\frac{{11.0074}^2}{21}+\frac{{17.1487}^2}{23}}=18.9402\end{array}$

There are 95% sufficient confidence that the mean difference is between 0.9688 and 18.9402.