Chapter 4, Problem 22P

1. Wegesin and Stern (2004) found greater consistency (less variability) in the memory performance scores for younger women than for older women. The following data represent memory scores obtained for two women, one older and one younger, over a series of memory trials.

Younger	Older
8	7
6	5
6	8
7	5
8	7
7	6
8	8
8	5

1. Calculate the variance of the scores for each woman.

Are the scores for the younger woman more consistent (less variable)?

To determine

Variance for the two groups given also compare the variability of these two groups.

Answer to Problem 22P

Solution:

1. Variance=0.786
2. younger women groups we get the less variability than the older women 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:

a)For younger

Here n=8, samples are given as also the calculation in the below table

X	(x-$\overline{x}$ )2
8	0.5625
6	1.5625
6	1.5625
7	0.0625
8	0.5625
7	0.0625
8	0.5625
8	0.5625
Total= ${\displaystyle \sum _{i=1}^8{x}_i}$ =58	${\displaystyle \sum _{i=1}^8{(x_i-\overline{x})}^2}$ = 5.5

$\overline{x}$= $\frac{1}{8}{\displaystyle \sum _{i=1}^8{x}_i}$=7.25

Variance = $\frac{1}{8-1}{\displaystyle \sum _{i=1}^8{(x_i-\overline{x})}^2}$=5.5/7=0.786

Standard deviation= $\sqrt{\mathrm{var}iance}$= $\sqrt{0.786}$=0.886

For older

Here n=8, samples are given as also the calculation in the below table

X	(x-$\overline{x}$ )2
7	0.39063
5	1.89063
8	2.64063
5	1.89063
7	0.39063
6	0.14063
8	2.64063
5	1.89063
Total= ${\displaystyle \sum _{i=1}^8{x}_i}$ =51	${\displaystyle \sum _{i=1}^8{(x_i-\overline{x})}^2}$ = 11.875

$\overline{x}$= $\frac{1}{8}{\displaystyle \sum _{i=1}^8{x}_i}$=6.375

Variance = $\frac{1}{8-1}{\displaystyle \sum _{i=1}^8{(x_i-\overline{x})}^2}$= 11.875/7=1.696

Standard deviation= $\sqrt{\mathrm{var}iance}$= $\sqrt{1.696}$=1.302

b) 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.