Chapter 4, Problem 17SP

To determine

The measure that you should compute to summarize the experiment. If the normally distributed ratio scores from an experiment

Condition A	Condition B	Condition C
12	33	47
11	33	48
11	34	49
10	31	48

Explanation of Solution

By computing the mean and standard deviation in each case we usually summarize the experiment. When the standard deviation is relatively small, then the scores are similar in situations, and therefore, there is a more consistent-strong relationship.

To determine

The appropriate descriptive statistics and summarize the relationship in the sampledata. If the normally distributed ratio scores from an experiment

Condition A	Condition B	Condition C
12	33	47
11	33	48
11	34	49
10	31	48

Explanation of Solution

Need to summarize the given experiment of three conditions. The procedure is specified in part (a)

 First we need to calculate the mean and the standard deviation

Condition A

Mean for first sample

$\begin{array}{l}\overline{X}=\frac{{\displaystyle \sum X_1}}{n}\\ =\frac{44}{4}\\ =11\end{array}$

Standard deviation

$\begin{array}{l}\overline{X}=\sqrt{{\displaystyle \sum X^2}-\frac{{\left({\displaystyle \sum X_1}\right)}^2}{N}}\\ =\sqrt{\frac{{12}^2+{11}^2+{11}^2+{10}^2+\frac{{\left(44\right)}^2}{4}}{4}}\\ =\sqrt{0.50}\\ =0.7071\end{array}$

Condition B

Mean of the second sample

$\begin{array}{l}\overline{X}=\frac{{\displaystyle \sum X_1}}{n}\\ =\frac{131}{4}\\ =32.75\end{array}$

Standard deviation

$\begin{array}{l}\overline{X}=\sqrt{{\displaystyle \sum X^2}-\frac{{\left({\displaystyle \sum X_1}\right)}^2}{N}}\\ =\sqrt{\frac{{33}^2+{33}^2+{34}^2+{31}^2+\frac{{\left(131\right)}^2}{4}}{4}}\\ =\sqrt{1.1875}\\ =1.09\end{array}$

Condition C

Mean of the first sample is

$\begin{array}{l}\overline{X}=\frac{{\displaystyle \sum X_1}}{n}\\ =\frac{192}{4}\\ =48\end{array}$

Standard deviation

$\begin{array}{l}\overline{X}=\sqrt{{\displaystyle \sum X^2}-\frac{{\left({\displaystyle \sum X_1}\right)}^2}{N}}\\ =\sqrt{\frac{{47}^2+{48}^2+{49}^2+{48}^2+\frac{{\left(192\right)}^2}{4}}{4}}\\ =\sqrt{0.5}\\ =0.07071\end{array}$

To determine

The participants appeared consistent were in each condition. If the normally distributed ratio scores from an experiment

Condition A	Condition B	Condition C
12	33	47
11	33	48
11	34	49
10	31	48

Explanation of Solution

We need to verify the stability of all the participants in each situation.

The three conditions are relatively smaller than the standard deviation. This indicates that scores are relatively close to each other in each case. Therefore, the participants scored consistently in each situation.

To determine

This relationship account for much of the variance in the scores. If the normally distributed ratio scores from an experiment

Condition A	Condition B	Condition C
12	33	47
11	33	48
11	34	49
10	31	48

Explanation of Solution

Yes, this relation is very high because the standard deviation in A, B and C is 0.7071, 1.089 and 0.7071, from the above calculation.