Question

It is commonly recognized that a poor environment (for example, poverty, low educational level of parents) and the presence of stressors (for example divorce of parents, abuse) are associated with lower IQs in children. Recent research indicates that a child's IQ is related to the number of such risk factors in the children's background, regardless of which specific factors are present. Below are the numbers of risk factors and the IQs of 12 children.

Child

Risk factors (x)

IQ (y)

A

3

112

B

6

82

C

0

105

D

5

102

E

1

115

F

1

101

G

5

94

H

4

89

I

3

98

J

3

109

K

4

91

L

2

107

A. Draw a scatterplot of the data. What does the plot tell you? Determine the regression equation. Be sure to include an interpretation of each parameter in the regression equation.

B. What IQ do you predict for James, who has 5 risk factors, versus Elizabeth, who has only 2?

Answer 1

A) The scatterplot of the data is as follows:What can we interpret from the scatterplot of the given data?We can observe that the higher the number of risk factors, the lower the IQ of the children. This fact is supported by the trend line that shows a negative correlation. This is what the plot is telling us.A regression equation is given by the formula: $y=\beta_0+\beta_1x$Where, $y$ is the IQ of the child, $x$ is the number of risk factors, $\beta_0$ is the y-intercept and $\beta_1$ is the slope of the line.Both $\beta_0$ and $\beta_1$ can be calculated using the following formulae: $$\beta_1=r_{xy}\frac{s_y}{s_x}$$$$\beta_0=\bar{y}-\beta_1\bar{x}$$Where, $r_{xy}$ is the correlation coefficient between $x$ and $y$, $s_x$ and $s_y$ are the standard deviations of $x$ and $y$ respectively, and $\bar{x}$ and $\bar{y}$ are the means of $x$ and $y$ respectively.From the scatterplot, we can observe that $r_{xy}$ is negative. Therefore, there exists a negative correlation between $x$ and $y$.Let us calculate the values of $\beta_1$, $\beta_0$ using the above formulae.$$r_{xy}=\frac{\sum(x-\bar{x})(y-\bar{y})}{\sqrt{\sum(x-\bar{x})^2\sum(y-\bar{y})^2}}$$$$r_{xy}=\frac{(3)(112)+(6)(82)+(0)(105)+(5)(102)+(1)(115)+(1)(101)+(5)(94)+(4)(89)+(3)(98)+(3)(109)+(4)(91)+(2)(107)}{\sqrt{(3^2+6^2+0^2+5^2+1^2+1^2+5^2+4^2+3^2+3^2+4^2+2^2)(112^2+82^2+105^2+102^2+115^2+101^2+94^2+89^2+98^2+109^2+91^2+107^2)}}$$$$r_{xy}=-0.835$$Now, let's calculate the standard deviations of $x$ and $y$:$$s_x=\sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}$$$$s_y=\sqrt{\frac{\sum(y-\bar{y})^2}{n-1}}$$$$s_x=\sqrt{\frac{(3-2)^2+(6-2)^2+(0-2)^2+(5-2)^2+(1-2)^2+(1-2)^2+(5-2)^2+(4-2)^2+(3-2)^2+(3-2)^2+(4-2)^2+(2-2)^2}{12-1}}$$$$s_x=\sqrt{\frac{44}{11}}=2$$$$s_y=\sqrt{\frac{\sum(y-\bar{y})^2}{n-1}}$$$$s_y=\sqrt{\frac{(112-99.17)^2+(82-99.17)^2+(105-99.17)^2+(102-99.17)^2+(115-99.17)^2+(101-99.17)^2+(94-99.17)^2+(89-99.17)^2+(98-99.17)^2+(109-99.17)^2+(91-99.17)^2+(107-99.17)^2}{12-1}}$$$$s_y=\sqrt{\frac{5626.1}{11}}=9.025$$Finally, let's calculate $\beta_1$ and $\beta_0$:$$\beta_1=r_{xy}\frac{s_y}{s_x}$$$$\beta_1=-0.835\frac{9.025}{2}=-3.762$$$$\beta_0=\bar{y}-\beta_1\bar{x}$$$$\beta_0=99.17-(-3.762)(3.08)=112.91$$Therefore, the regression equation is $y=112.91-3.762x$.Here, $\beta_0=112.91$ represents the expected IQ of a child with zero risk factors, while $\beta_1=-3.762$ represents the decrease in IQ per risk factor.B) James has 5 risk factors. Substituting $x=5$ in the regression equation, we get:$$y=112.91-3.762x$$$$y=112.91-3.762(5)$$$$y=93.13$$Therefore, the IQ predicted for James is $93.13$.Similarly, Elizabeth has 2 risk factors. Substituting $x=2$ in the regression equation, we get:$$y=112.91-3.762x$$$$y=112.91-3.762(2)$$$$y=105.39$$Therefore, the IQ predicted for Elizabeth is $105.39$.