Chapter 28, Problem 1EQ

Here are data for height and weight among 10 male college students.

Height (cm)	Weight (kg)
159	48
162	50
161	52
175	60
174	64
198	81
172	58
180	74
161	50
173	54

A. Calculate the correlation coefficient for height and weight for this group.

B. Is the correlation coefficient statistically significant? Explain.

Summary Introduction

To review:

The following on the basis of the data for height and weight of 10 male college students:

The correlation coefficient for the height and weight of a given group.

Whether the correlation coefficient is significant or not.

Introduction:

The strength of the association between two variables can be calculated by the correlation coefficient. It ranges from -1 to +1 and indicates how the two factors or variable differ from each other. The positive value indicates that the two variable varies in a similar way relative to each other, while the negative value indicates that they vary inopposite directions, that is, if one variable increases, the other will decrease.

Explanation of Solution

The correlation coefficient (r) can be calculated by the following formula:

${\text{r}}_{\left(X,Y\right)}=\frac{{\text{CoV}}_{\left(X,Y\right)}}{{\text{SD}}_X{\text{SD}}_Y}$

Where CoV is covariance of the two variables,

SD_X is the standard deviation of variable x

SD_Y is the standard deviation of variable y

The height and weight of 10 college students is listed below:

Height in centimeter (X)	Weight in kilogram (Y)
159	48
162	50
161	52
175	60
174	64
198	81
172	58
180	74
161	50
173	54

Mean (Ax)of the given data for height can be calculated by ${\displaystyle \sum \frac{\text{X}}{\text{N}}}$.

${\displaystyle \sum \text{X}}$ is the sum of height of 10 students and ${\displaystyle \sum \text{N}}$ is the total number of students, that is, 10. Thus, from the given data,

$\begin{array}{c}{A}_X=\frac{159+162+161+175+174+198+172+180+161+173}{10}\\ {\text{A}}_X=\frac{\text{159}+\text{162}+\text{161}+\text{175}+\text{174}+\text{198}+\text{172}+\text{180}+\text{161}+\text{173}}{\text{10}}\\ {\text{A}}_X=\text{175}\text{.1}\end{array}$

Similarly, mean (Ay) for weight = ${\displaystyle \sum \frac{Y}{N}}$

$\begin{array}{c}{\text{A}}_Y=\frac{48+50+52+60+64+81+58+74+50+54}{\text{10}}\\ {\text{A}}_Y=\text{59}\text{.1}\end{array}$

The differnet values of (X-A_X), (X-A_X)², (Y-A_Y), (Y-A_Y)², and (X-A_X)(Y-A_Y) are listed below:

Height (X)	X-AX	(X-AX)2	Weight (Y)	Y-AY	(Y-AY)2	(X-AX) (Y-AY)
159	-16.1	259.21	48	-11.1	123.21	178.71
162	-13.1	171.61	50	-9.1	82.81	119.21
161	-14.1	198.81	52	-7.1	50.41	100.11
175	-0.1	0.01	60	0.9	0.81	-0.09
174	-1.1	1.21	64	4.9	24.01	-5.39
198	22.9	524.41	81	21.9	479.61	501.51
172	-3.1	9.61	58	-1.1	1.21	3.41
180	4.9	24.01	74	14.9	222.01	73.01
161	-14.1	198.81	50	-9.1	82.81	128.31
173	-2.1	4.41	54	-5.1	26.01	10.71

The variance of the data is calculated by:

$\begin{array}{c}\text{Vx}={\displaystyle \sum {\frac{\left(\text{X}-\text{Ax}\right)}{\text{N}-\text{1}}}^2}\\ {\text{V}}_{\text{x}}=\frac{\text{1392}\text{.1}}{\text{9}}\\ {\text{V}}_{\text{x}}=\text{154}\text{.67}\end{array}$

$$

$\begin{array}{c}{\text{V}}_{\text{Y}}=\frac{\text{1092}\text{.9}}{\text{9}}\\ {\text{V}}_{\text{Y}}=\text{121}\text{.43}\end{array}$

Standard deviation,

$\begin{array}{c}{\text{SD}}_{\text{x}}=\sqrt{{\text{V}}_{\text{x}}}\\ {\text{SD}}_{\text{x}}=12.43\\ {\text{SD}}_Y=\sqrt{{\text{V}}_Y}\\ {\text{SD}}_Y=11.01\end{array}$

The covariance of height and weight,

$\begin{array}{c}{\text{CoV}}_{\left(X,Y\right)}={\displaystyle \sum \frac{\left(\text{X}-{\text{A}}_X\right)\left(\text{Y}-{\text{A}}_Y\right)}{N-1}}\\ {\text{CoV}}_{\left(X,Y\right)}=\frac{\text{1109}\text{.5}}{\text{10}-\text{1}}\\ {\text{CoV}}_{\left(X,Y\right)}=\text{123}\text{.27}\end{array}$

Correlation coefficient,

$\begin{array}{c}{\text{r}}_{\left(X,Y\right)}=\frac{{\text{CoV}}_{\left(X,Y\right)}}{{\text{SD}}_X{\text{SD}}_Y}\\ {\text{r}}_{\left(X,Y\right)}=\frac{\text{123}\text{.27}}{\text{12}\text{.43}\times \text{11}\text{.01}}\\ {\text{r}}_{\left(X,Y\right)}=\frac{\text{123}\text{.27}}{\text{137}\text{.97}}\\ {\text{r}}_{\left(X,Y\right)}=\text{0}\text{.893}\end{array}$

The significance of the ‘r’ can be explained bynull hypothesis. It says that the observed r value differs from 0 because of a random sampling of error. It can be calculated using the degree of freedom (df). According to df, null hypothesis is rejected if the value of r is less than 0.05 (5%). According to the significance level of r value, df will be N-2, that is, 10-2 = 8.

Thus, according to the 5% significant level the value of r must be 0.632 or greater, while the calculated value is 0.892. Thus, we can reject the null hypothesis and can depict that the two variables are in positive correlation with each other. It can be ascertained that the association is not just due to randon sampling error, but it may or may not imply cause and effect.

Conclusion

Therefore, it can be concluded that the value of r is 0.892 and the height and weight are in positive correlation with each other. The correlation is due to the reason other than random sampling error.