Chapter 15, Problem 14P

14. Identifying individuals with a high risk of Alzheimer’s disease usually involves a long series of cognitive tests. However, researchers have developed a 7-Minute Screen, which is a quick and easy way to accomplish the same goal. The question is whether the 7-Minute Screen is as effective as the complete series of tests. To address this question, Ijuin et al. (2008) administered both tests to a group of patients and compared the results. The following data represent results similar to those obtained in the study.

Patient

7-Minutes

Screen

Cognitive

Series

A

3

1

B

8

19

C

10

22

D

8

20

E

4

14

F

7

13

G

4

9

H

5

20

I

14

25

a. Compute the Pearson correlation to measure the degree of relationship between the two test scores.

b. Is the correlation statistically significant? Use a two-tailed test with $\alpha =.01$ .

c. What percentage of variance for the cognitive scores is predicted from the 7-Minute Screen scores? (Compute the value of $r^2$ .)

To determine

1. The Pearson's correlation to measure the degree of relationship between the two test scores.
2. If the correlation is statistically significant using a two tailed test with $\alpha =.01$.
3. What percentage of variance for the cognitive scores is predicted from the 7-Minute screen scores? (Compute the value of $r^2$)

Answer to Problem 14P

Solution:

1. Patient	7-Minute Screen $(X)$	Cognitive Series $(Y)$	$XY$	$X^2$	$Y^2$
   A	3	11	33	9	121
   B	8	19	152	64	361
   C	10	22	220	100	484
   D	8	20	160	64	400
   E	4	14	56	16	196
   F	7	13	91	49	169
   G	4	9	36	16	81
   H	5	20	100	25	400
   I	14	25	350	196	625
   	${\displaystyle \sum _{}^{}{X}_{}}=$ 63	${\displaystyle \sum _{}^{}Y=}$ 153	${\displaystyle \sum _{}^{}XY=}$ 1198	${\displaystyle \sum _{}^{}{X}_{}^2}=$ 539	${\displaystyle \sum _{}^{}{Y}^2=}$ 2837

   The Pearson's correlation between the 7-Minute Screen $(X)$and the Cognitive Series $(Y)$ is $0.8351$. Therefore, the 7-Minute Screen $(X)$and the Cognitive Series $(Y)$ have strong positive correlation.
2. Since the test statistic is beyond the right tail of the critical value $3.499$, therefore we reject H₀. That is, the sample of $n=9$pairs of observation indicates that the correlation is statistically significant.
3. The percentage of variance for the cognitive scores which is predicted from the 7-Minute screen scores is $69.74\%$.

Explanation of Solution

Given: The given data:

Patient	7-Minute Screen	Cognitive Series
A	3	11
B	8	19
C	10	22
D	8	20
E	4	14
F	7	13
G	4	9
H	5	20
I	14	25

Formula Used:

$r=\frac{N{\displaystyle \sum _{}^{}{X}_{}{Y}_{}-\left({\displaystyle \sum _{}^{}X}\right)}\left({\displaystyle \sum _{}^{}Y}\right)}{\sqrt{\left[N{\displaystyle \sum _{}^{}{X}_{}^2-{\left({\displaystyle \sum _{}^{}X}\right)}^2}\right]}\left[N{\displaystyle \sum _{}^{}{Y}_{}^2-{\left({\displaystyle \sum _{}^{}Y}\right)}^2}\right]}$

Calculations:

1. Now to compute the Pearson's correlation between the 7-Minute Screen and the Cognitive Series, let's denote the 7-Minute Screen by $X$ and the Cognitive Series by $Y$. Also we know the formula for Pearson's correlation and below is given same;

   $r=\frac{N{\displaystyle \sum _{}^{}{X}_{}{Y}_{}-\left({\displaystyle \sum _{}^{}X}\right)}\left({\displaystyle \sum _{}^{}Y}\right)}{\sqrt{\left[N{\displaystyle \sum _{}^{}{X}_{}^2-{\left({\displaystyle \sum _{}^{}X}\right)}^2}\right]}\left[N{\displaystyle \sum _{}^{}{Y}_{}^2-{\left({\displaystyle \sum _{}^{}Y}\right)}^2}\right]}$

   Where:

   $N$ = Number of pairs of observations.

   ${\displaystyle \sum _{}^{}X}=\text{ Sum of X observations}$ ${\displaystyle \sum _{}^{}Y=\text{Sum of Y observations}}$ ${\displaystyle \sum _{}^{}XY}=\text{Sum of the products of paired observations }$ ${\displaystyle \sum _{}^{}{X}_{}^2}=\text{Sum of squared X observations}$ ${\displaystyle \sum _{}^{}{Y}_{}^2}=\text{Sum of squared Y observations}$

   Patient	7-Minute Screen $(X)$	Cognitive Series $(Y)$	$XY$	$X^2$	$Y^2$
   A	3	11	33	9	121
   B	8	19	152	64	361
   C	10	22	220	100	484
   D	8	20	160	64	400
   E	4	14	56	16	196
   F	7	13	91	49	169
   G	4	9	36	16	81
   H	5	20	100	25	400
   I	14	25	350	196	625
   	${\displaystyle \sum _{}^{}{X}_{}}=$ 63	${\displaystyle \sum _{}^{}Y=}$ 153	${\displaystyle \sum _{}^{}XY=}$ 1198	${\displaystyle \sum _{}^{}{X}_{}^2}=$ 539	${\displaystyle \sum _{}^{}{Y}^2=}$ 2837

   $r=\frac{9\times 1198-63\times 153}{\sqrt{\left[9\times 539-{63}^2\right]\left[9\times 2837-{153}^2\right]}}$ $=\frac{10782-9639}{\sqrt{(4851-3969)(25533-23409)}}=\frac{1143}{\sqrt{882\times 2124}}=\frac{1143}{1368.71}=0.8351$

   The Pearson's correlation between the 7-Minute Screen $(X)$and the Cognitive Series $(Y)$ is $0.8351$. Therefore, the 7-Minute Screen $(X)$and the Cognitive Series $(Y)$ have strong positive correlation.

2. Now to find if the correlation is statistically significant, we will follow the 4-step procedure of hypothesis testing as:
Step 1:

The first step is to set up the hypothesis and determine the significance level

The null and alternative hypotheses are given below:

$H_0:\rho =0$i.e., the observed sample correlation coefficient is not significant of any correlation in the population.

$H_a:\rho \ne 0$ i.e., the observed sample correlation coefficient is significant of correlation in the population.

Also the significance level is $\alpha =.01$.

Step 2:

The second step is to set the criteria for a decision.

Here $n=9$ and the two tailed critical value for $n-2=9-2=7df$ and $.01\text{ significance level is}$ $t_{0.01,7}=\pm 3.499$The Critical $t-value$ $=\pm 3.499$ and the criteria for decision is, if the test statistic is less than the critical value then we fail to reject the null hypothesis else we reject the null hypothesis.

Step 3:

The third step is to compute the test statistic.

Now we have to calculate the test statistic. The test statistic is given below:

$t=\frac{r\sqrt{(n-2)}}{\sqrt{(1-r^2)}}$ $=\frac{0.8351\sqrt{(9-2)}}{\sqrt{(1-{(0.8351)}^2)}}=\frac{0.8351\times 2.6458}{\sqrt{1-0.6974}}=\frac{2.2095}{0.5501}=4.02\text{ rounded to two decimals}$

Step 4:

The fourth step is to make a decision.

Since the test statistic is beyond the right tail of the critical value $3.499$, therefore we reject H₀. That is, the sample of $n=9$ pairs of observation indicates that the correlation is statistically significant.

3. The percentage of variance for the cognitive scores which is predicted from the 7-Minute screen scores can be calculated using Square of Correlation $r$ i.e., $r^2$. Therefore we have:

   $r^2={0.8351}^2=0.6974\text{ or 69}\text{.74\%}$.

   The percentage of variance for the cognitive scores which is predicted from the 7-Minute screen scores is $69.74\%$ .

Conclusion:

1. The Pearson's correlation between the 7-Minute Screen $(X)$ and the Cognitive Series $(Y)$ is $0.8351$. Therefore, the 7-Minute Screen $(X)$ and the Cognitive Series $(Y)$ have strong positive correlation.
2. Using a two-tailed test with $\alpha =.01$, we clearly see that the correlation is statistically significant.
3. The percentage of variance for the cognitive scores which is predicted from the 7-Minute screen scores is $69.74\%$ .