Answer to Question #254481 in Statistics and Probability for Goldie

Question

CORRELATION.

1. A graduate student in developmental psychology believes there may be a relationship between birth weight and subsequent IQ. She randomly samples seven psychology majors at her university and gives them an IQ test. Next, she obtains the weight at birth of the seven majors from the appropriate hospitals (after obtaining permission from the students, of course). The data are shown in the following table.

Student: 1 2 3 4 5 6 7 

Birth Weight: 5.8 6.5 8.0 5.9 8.5 7.2 9.0 

IQ: 122 120 129 112 127 116 130 

a. Construct a scatter plot of the data, plotting birth weight on the X axis and IQ on the Y axis. Does the relationship appear to be linear?

b. Assume the relationship is linear and compute the value of Pearson r.

Answer 1

a.

We can consider that the relationship appears to be linear.

b.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & X & Y & XY & X^2 & Y^2 \\ \hline & 5.8 & 122 & 707.6 & 33.64 & 14884 \\ \hdashline & 6.5 & 120 & 780 & 42.25 & 14400 \\ \hdashline & 8.0 & 129 & 1032 & 64 & 16641 \\ \hdashline & 5.9 & 112 & 660.8 & 34.81 & 12544 \\ \hdashline & 8.5 & 127 & 1079.5 & 72.25 & 16129 \\ \hdashline & 7.2 & 116 & 835.2 & 51.84 & 13456 \\ \hdashline & 9.0 & 130 & 1170 & 81 & 16900 \\ \hdashline Sum= & 50.9 & 856 &6265.1 & 379.79 & 104954 \\ \end{array}$$

$$\bar X = \frac{1}{n} \sum_{i=1}^{n} X_i = \frac{ 50.9}{ 7} = 7.27142857$$

$$\bar Y = \frac{1}{n} \sum_{i=1}^{n} Y_i = \frac{ 856}{ 7} =122.28571429$$

$$SS_{XX} = \sum_{i=1}^{n} X_i^2 - \frac{1}{n} (\sum_{i=1}^{n} X_i)^2$$

$$=379.79-\frac{ 50.9^2}{ 7} =9.67428571$$

$$SS_{YY} = \sum_{i=1}^{n} Y_i^2 - \frac{1}{n} (\sum_{i=1}^{n} Y_i)^2$$

$$=104954-\frac{ 856^2}{ 7} =277.42857143$$

$$SS_{XY} = \sum_{i=1}^{n} X_iY_i - \frac{1}{n} (\sum_{i=1}^{n} X_i) (\sum_{i=1}^{n} Y_i)$$

$$=6265.1-\frac{50.9(856)}{ 7} =40.75714286$$

Therefore, based on the above calculations, the regression coefficients (the slope $m,$ and the $y$ -intercept $n$) are obtained as follows:

$$m=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{40.75714286}{9.67428571}=4.21293562$$

$$n=\bar{Y}-m\cdot\bar{X}$$

$$=122.28571429-4.21293562\cdot7.27142857$$

$$=91.65165386$$

We find that the regression equation is:

$$Y=91.65165386+4.21293562X$$

Correlation coefficient (Pearson $r$)

$$r=\dfrac{SS_{XY}}{\sqrt{SS_{XX}}\sqrt{SS_{YY}}}$$

$$=\dfrac{40.75714286}{\sqrt{9.67428571}\sqrt{277.42857143}}=0.78671727$$

$$|r|=0.78671727\geq0.7$$

Strong correlation.