Chapter 4, Problem 21P

21 . Within a population. the differences that exist from one person to another are often called diversity.

Researchers comparing cognitive kills for younger adults and older adults typically find greater difference. (greater diversity ,in the older population (Morse, 1993). Following are typical data showing problem-solving scores for two groups of participants.

Older Adults (average age 72)

738

845

266

Younger Adults (average age 31)

678

668

869

1. Compute the me.an, variance. and standard deviation for each group.
2. Is one group of scores 111.oticeabty more variable (more diverse) than the other?

To determine

Mean, variance and standard deviation for the two groups given also compare the variability of these two groups.

Answer to Problem 21P

Solution:

1. Mean=5.467, variance=4.695, SD=2.167
2. There is a significant difference between the SD of the two groups

Explanation of Solution

Here two groups are given we need to find the sample mean variance and SD and hence we need to compare that there is significant difference in SD`s. To do the problem will follow the steps that first mean then variance followed by SD hence we can compare these two SD.

Given:

We have provided with scores of two groups younger and olders. So, for this available data we will proceed toward solving the problem

Formula Used:

For sample of size n, we have below formulas

Mean= $\overline{x}$ = $\frac{1}{n}{\displaystyle \sum _{i=1}^n{x}_i}$ SS= ${\displaystyle \sum _{i=1}^n{(x_i-\overline{x})}^2}$

Variance= $\frac{1}{n-1}{\displaystyle \sum _{i=1}^n{(x_i-\overline{x})}^2}$

Standard Deviation (SD)= $\sqrt{\frac{1}{n-1}{\displaystyle \sum _{i=1}^n{(x_i-\overline{x})}^2}}$

Calculation:

For older

Here n=15, samples are given as also the calculation in the below table

X	(x-$\overline{x}$ )2
9	12.4821
4	2.1521
7	2.3501
3	6.0861
8	6.4161
6	0.2841
2	12.0201
8	6.4161
4	2.1521
5	0.2181
7	2.3501
5	0.2181
2	12.0201
6	0.2841
6	0.2841
Total= ${\displaystyle \sum _{i=1}^{15}{x}_i}$ =82	${\displaystyle \sum _{i=1}^{15}{(x_i-\overline{x})}^2}$ = 65.7333

$\overline{x}$= $\frac{1}{15}{\displaystyle \sum _{i=1}^{15}{x}_i}$=5.467

Variance = $\frac{1}{15-1}{\displaystyle \sum _{i=1}^{15}{(x_i-\overline{x})}^2}$=65.7333/14=4.695

Standard deviation= $\sqrt{\mathrm{var}iance}$= $\sqrt{4.695}$=2.167

For Younger

Here n=15, samples are given as also the calculation in the below table

X	(x-$\overline{x}$ )2
7	0.07129
9	3.00329
6	1.60529
7	0.07129
8	0.53729
6	1.60529
7	0.07129
6	1.60529
6	1.60529
8	0.53729
9	3.00329
7	0.07129
8	0.53729
6	1.60529
9	3.00329
Total= ${\displaystyle \sum _{i=1}^{15}{x}_i}$ =109	${\displaystyle \sum _{i=1}^{15}{(x_i-\overline{x})}^2}$ = 18.9333

$\overline{x}$= $\frac{1}{15}{\displaystyle \sum _{i=1}^{15}{x}_i}$=7.267

Variance = $\frac{1}{15-1}{\displaystyle \sum _{i=1}^{15}{(x_i-\overline{x})}^2}$=18.9333/14=1.352

Standard deviation= $\sqrt{\mathrm{var}iance}$= $\sqrt{1.352}$=1.163

Here there is a significant difference between the SD of the two groups for younger groups we get the less variability than the older groups. Hence, we can comment that there is less variability in the younger than the older groups.

Conclusion:

Here we can note that older group is more variable than the younger group, so we may comment that older are more dispersed than the younger.