Chapter 24.2, Problem 25AQP

(a)

Summary Introduction

To calculate:

Correlation coefficient for height of mother and daughter.

Introduction:

Correlation coefficient is denoted by the symbol “r”. It can be calculated by dividing covariance of two variables by the product of their standard deviation. Mathematically, it can be expressed as follows:

$r=\frac{{\mathrm{cov}}_{xy}}{s_x{s}_y}$

Here, r is the correlation of two variables.

x and y are the variables.

cov_xy refers to the covariance of x and y variables.

S_xs_y is the product of standard deviations of x and y.

Explanation of Solution

In the given scenario, the height of various mothers (x_i) and daughters (y_i) are given. Mean is calculated by calculating the sum of all the possible values and dividing that sum by total number of values. Mean of mother’s heights $\left(\overline{x}\right)$ and daughter’s heights $\left(\overline{y}\right)$ can be calculated as follows:

$\begin{array}{l}\text{Mean}\text{=}\frac{\text{Sum}\text{}\text{of}\text{observed}\text{values}}{\text{Total}\text{number}\text{of}\text{values}}\\ \text{Mean}\left(\overline{x}\right)=\frac{64+65+66+64+63+63+59+62+61+60}{10}\\ \text{Mean}\left(\overline{x}\right)=\frac{627}{10}\\ \text{Mean}\left(\overline{x}\right)=62.7\\ \text{Mean}\left(\overline{y}\right)=\frac{66+66+68+65+65+62+62+64+63+62}{10}\\ \text{Mean}\left(\overline{y}\right)=\frac{643}{10}\\ \text{Mean}\left(\overline{y}\right)=64.3\end{array}$

Variance is determined as the average of the squared deviation from the mean. Variance of a distribution is calculated by finding the mean, then subtracting the mean from all the values of the data and then again finding the average of their squared differences. Variance is denoted by “s²”.

$s^2=\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}$

Here,

s² represents variance.

$n$ represents the number of original measurements.

$\overline{x}$ represents the mean.

$x_i$ represents the observations of the data.

xi (wet weight in g)	yi (length in mm)	$\left(x_i-\overline{x}\right)$	$\left(y_i-\overline{y}\right)$	${\left(x_i-\overline{x}\right)}^2$	${\left(y_i-\overline{y}\right)}^2$	$\left(x_i-\overline{x}\right)\times \left(y_i-\overline{y}\right)$
64	66	1.3	1.7	1.69	2.89	2.21
65	66	2.3	1.7	5.29	2.89	3.91
66	68	3.3	3.7	10.89	13.69	12.12
64	65	1.3	0.7	1.69	0.49	0.91
63	65	0.3	0.7	0.09	0.49	0.21
63	62	0.3	-2.3	0.09	5.29	-0.69
59	62	-3.7	-2.3	13.69	5.29	8.51
62	64	-0.7	-0.3	0.49	0.09	0.21
61	63	-1.7	-1.3	2.89	1.69	2.21
60	62	-2.7	-2.3	7.29	5.29	0.21
Total				${\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}$= 44.1	${\displaystyle \sum {\left(y_i-\overline{y}\right)}^2}$= 38.1	${\displaystyle \sum \left(x_i-\overline{x}\right)\left(y_i-\overline{y}\right)}$= 35.9

$\begin{array}{l}{s}_x{}^2=\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}\\ s_x{}^2=\frac{44.1}{10-1}\\ s_x{}^2=\frac{44.1}{9}\\ s_x{}^2=4.9\end{array}$

$\begin{array}{l}{s}_y{}^2=\frac{{\displaystyle \sum {\left(y_i-\overline{y}\right)}^2}}{n-1}\\ s_y{}^2=\frac{38.1}{10-1}\\ s_y{}^2=\frac{38.1}{9}\\ s_y{}^2=4.2\end{array}$

Standard deviation quantifies the amount of variation in a given set of data. It is calculated by taking the square root of the variance. So, standard deviations of x and y can be calculated as follows:

$\begin{array}{c}\text{Standard}\text{deviation}\text{=}\sqrt{\text{variance}}\text{}\\ s_x=\sqrt{s^2}\\ s_x=\sqrt{4.9}\\ s_x=2.21\\ s_y=\sqrt{s^2}\\ s_y=\sqrt{4.2}\\ s_y=2.04\end{array}$

Covariance is calculated as follows:

$\begin{array}{l}{\mathrm{cov}}_{xy}=\frac{{\displaystyle \sum \left(x_i-\overline{x}\right)\left(y_i-\overline{y}\right)}}{n-1}\\ {\mathrm{cov}}_{xy}=\frac{35.9}{10-1}\\ {\mathrm{cov}}_{xy}=\frac{35.9}{9}\\ {\mathrm{cov}}_{xy}=3.99\end{array}$

Now, correlation of wet weight and length can be calculated as follows:

$\begin{array}{l}r=\frac{{\mathrm{cov}}_{xy}}{s_x{s}_y}\\ r=\frac{3.99}{2.21\times 2.04}\\ r=0.885\approx 0.89\end{array}$

(b)

Summary Introduction

To predict:

The expected height of a daughter whose mother is 67 inches tall using regression.

Introduction:

Correlation coefficient is denoted by the symbol “r”. It can be calculated by dividing covariance of two variables by the product of their standard deviation. Mathematically, it can be expressed as follows:

$r=\frac{{\mathrm{cov}}_{xy}}{s_x{s}_y}$

Here, r is the correlation of two variables.

x and y are the variables.

cov_xy refers to the covariance of x and y variables.

S_xs_y is the product of standard deviations of x and y.

Explanation of Solution

The regression line that best suits is as follows:

$y=a+bx$

Here, y is the daughter’s height.

x is the mother’s height.

b is the slope of regression line (regression coefficient).

a  is the intercept of the line.

b and a can be calculated as follows:

$\begin{array}{l}b=\frac{{\mathrm{cov}}_{xy}}{s_x^2}\\ b=\frac{3.99}{4.9}\\ b=0.814\end{array}$

$\begin{array}{l}a=\overline{y}-b\overline{x}\\ a=64.3-\left(0.814\right)\text{}\left(62.7\right)\\ a=13.26\end{array}$

Putting the values of a and b in the equation as follows:

$\begin{array}{l}y=a+bx\\ y=13.26+\left(0.814\right)\left(67\right)\\ y=67.798\end{array}$

Height of the daughter would be 67.79 inches if the height of mother is 67 inches.

Conclusion

Correlation coefficient of mother’s and daughter’s heights is 0.89. Height of daughter would be 67.79 inches if mother’s height is 67 inches.