Chapter 10, Problem 16CAP

(a)

To determine

Find the mean difference $\left(M_D\right)$, standard deviation $\left(s_D\right)$, and standard error for the difference scores $\left(s_{MD}\right)$.

Answer to Problem 16CAP

The mean difference for the difference scores is –0.8.

The standard deviation for the difference scores is 2.8.

The standard error for the difference scores is 1.3.

Explanation of Solution

Calculations:

From the Question 15, the difference scores are –4, 1, –2, –2, and 3.

The formula for sample mean difference score is, $M_D=\frac{{\displaystyle \sum D}}{n_D}$.

The formula for sample standard deviation difference score is, $s_D=\sqrt{\frac{{\displaystyle \sum {\left(D-M_D\right)}^2}}{n_D-1}}$.

The formula for sample standard error of difference score is, $s_{MD}=\frac{s_D}{\sqrt{n_D}}$.

Mean for the difference:

Substitute, the values for difference scores, –4, 1, –2, –2, 3 and $n_D=5$ in mean for the difference scores formula.

$\begin{array}{c}{M}_D=\frac{-4+1-2-2+3}{5}\\ =\frac{-4}{5}\\ =-0.8\end{array}$

Thus, the mean for the difference scores is –0.8.

Standard deviation for the difference scores:

Substitute, the values for difference scores, –4, 1, –2, –2, 3, $M_D=-0.8$ and $n_D=5$ in standard deviation for the difference scores formula.

$\begin{array}{c}{s}_D=\sqrt{\frac{{\left(-4+0.8\right)}^2+{\left(1+0.8\right)}^2+{\left(-2+0.8\right)}^2+{\left(-2+0.8\right)}^2+{\left(3+0.8\right)}^2}{5-1}}\\ =\sqrt{\frac{10.24+3.24+1.44+1.44+14.44}{4}}\\ =\sqrt{\frac{30.8}{4}}\\ =\sqrt{7.7}\end{array}$

$=2.8$

Thus, the standard deviation for the difference scores is 2.8.

Standard error for the difference scores:

Substitute, $s_D=2.8,n_D=8$ in standard error for the difference scores formula.

$\begin{array}{c}{s}_{MD}=\frac{2.8}{\sqrt{5}}\\ =\frac{2.8}{2.23}\\ =1.3\end{array}$

Thus, the standard error for the difference scores is 1.3.

(b)

To determine

Sketch the shape of the distribution of mean difference scores $M_D\pm s_D$.

Answer to Problem 16CAP

The shape of the distribution of mean difference scores $M_D\pm s_D$ is,

Explanation of Solution

Calculation:

The value of $M_D$ is -0.8 and the value of $s_D$ is 2.8.

For 1 standard deviation:

$\begin{array}{c}{M}_D-s_D=-0.8-2.8\\ =-3.6\end{array}$

$\begin{array}{c}{M}_D+s_D=-0.8+2.8\\ =2.0\end{array}$

For two standard deviation:

$\begin{array}{c}{M}_D-2s_D=-0.8-\left(2\times 2.8\right)\\ =-0.8-5.6\\ =-6.4\end{array}$

$\begin{array}{c}{M}_D+2s_D=-0.8+\left(2\times 2.8\right)\\ =-0.8+5.6\\ =4.8\end{array}$

For three standard deviation:

$\begin{array}{c}{M}_D-3s_D=-0.8-\left(3\times 2.8\right)\\ =-0.8-8.4\\ =-9.2\end{array}$

$\begin{array}{c}{M}_D+3s_D=-0.8+\left(3\times 2.8\right)\\ =-0.8+8.4\\ =7.6\end{array}$

Thus, the standard deviation limits for the distribution are $\left(-3.6,2.0\right),\left(-6.4,4.8\right)$ and $\left(-9.2,7.6\right)$.

Sketch of Graph:

Step by step procedure for sketching the graph:

• The mean of the distribution is taken in the centre of the graph.
• The one standard deviation limits are taken on either side of the mean.
• The two standard deviations limits are taken on either side of one standard deviation limits.
• The three standard deviations limits are taken on either side of two standard deviations limits.

(c)

To determine

Sketch the shape of the distribution of mean difference scores $M_D\pm s_{MD}$.

Answer to Problem 16CAP

The shape of the distribution of mean difference scores $M_D\pm s_{MD}$ is,

Explanation of Solution

Calculation:

The value of $M_D$ is –0.8 and the value of $s_{MD}$ is 1.3.

For 1 standard deviation:

$\begin{array}{c}{M}_D-s_{MD}=-0.8-1.3\\ =-2.1\end{array}$

$\begin{array}{c}{M}_D+s_{MD}=-0.8+1.3\\ =0.5\end{array}$

For two standard deviation:

$\begin{array}{c}{M}_D-2s_{MD}=-0.8-\left(2\times 1.3\right)\\ =-0.8-2.6\\ =-3.4\end{array}$

$\begin{array}{c}{M}_D+2s_{MD}=-0.8+\left(2\times 1.3\right)\\ =-0.8+2.6\\ =1.8\end{array}$

For three standard deviation:

$\begin{array}{c}{M}_D-3s_{MD}=-0.8-\left(3\times 1.3\right)\\ =-0.8-3.9\\ =-4.7\end{array}$

$\begin{array}{c}{M}_D+3s_{MD}=-0.8+\left(3\times 1.3\right)\\ =-0.8+3.9\\ =3.1\end{array}$

Thus, the standard deviation limits for the distribution are $\left(-2.1,0.5\right),\left(-3.4,1.8\right)$ and $\left(-4.7,3.1\right)$.

Sketch of Graph:

Step by step procedure for sketching the graph:

• The mean of the distribution is taken in the centre of the graph.
• The one standard deviation limits are taken on either side of the mean.
• The two standard deviations limits are taken on either side of one standard deviation limits.
• The three standard deviations limits are taken on either side of two standard deviations limits.