Chapter 3: Q 69. (page 208)

Managing diabetes People with diabetes measure their fasting plasma glucose (FPG, measured in milligrams per milliliter) after fasting for at least 8 hours. Another measurement, made at regular medical checkups, is called HbA. This is roughly the percent of red blood cells that have a glucose molecule attached. It measures average exposure to glucose over a period of several months. The table gives data on both HbA and FPG for 18 diabetics five months after they had completed a diabetes education class.
a. Make a scatterplot with HbA as the explanatory variable. Describe what you see.
b. Subject 18 is an outlier in the x-direction. What effect do you think this subject has on the correlation? What effect do you think this subject has on the equation of the least-squares regression line? Calculate the correlation and equation of the least-squares regression line with and without this subject to confirm your answer.
c. Subject 15 is an outlier in the y-direction. What effect do you think this subject has on the correlation? What effect do you think this subject has on the equation of the least-squares regression line? Calculate the correlation and equation of the least-squares regression line with and without this subject to confirm your answer.

Short Answer

Expert verified

Part (a) the scatterplot confirms a weak relationship because the points seem to lie far apart.

Part (b) No, they do not affect.

Part (c) It makes the regression line steeper.

Step by step solution

Part (a) Step 1: Given information

Part (a) Step 2: Explanation

The scatterplot with HbA as the explanatory variable is as:

Because the scatterplot slopes upwards, we can conclude that the scatterplot confirms a positive linear connection. Because the points appear to be widely apart, the scatterplot indicates a weak association.

Part (b) Step 1: Calculation

Now we must use the excel function to calculate the correlation:

First, we'll enter the data into an excel file, and then we'll utilize the correlation function, which is,

CORREL function returns the correlation coefficient of the $a r r a y 1 a n d a r r a y 2$ cell ranges. Thus, the syntax is as:

$C O R R E L (a r r a y 1, a r r a y 2)$

AVERAGE function returns the average of the $a r r a y 1 a n d a r r a y 2$ cell ranges. The syntax is as:

$A V E R A G E (a r r a y 1, a r r a y 2)$

For the case with outlier:

Thus, the calculation will be as:

$C o r r e l a t i o n = C O R R E L (H 1 : H 18, I 1 : I 18)$

And the result will be as:

$C o r r e l a t i o n = 0.4506$

Thus, the slope will be,

$b = r \frac{s_{y}}{s_{x}} = 0.4506 \times \frac{81.4827}{3.2619} = 11.2569$

And the $y -$ intercept will be,

$a = y - b x = 172.3846 - 11.2569 \times 10.3769 = 55.5726$

The regression line will be as:

$\overset{⏜}{y} = 55.5726 + 11.2569 x$

For the case without outlier:

Thus, the calculation will be as:

$C o r r e l a t i o n = C O R R E L (H 1 : H 17, I 1 : I 17)$

And the result will be as:

$C o r r e l a t i o n 0.3837$

Thus, the slope will be,

$b = r \frac{s_{y}}{s_{x}} = 0.3837 \times \frac{68.9020}{2.1821} = 12.1158$

And the $y -$ intercept will be,

$a = y - b x = 157.8824 - 12.1156 \times 8.7176 = 52.2615$

Part (b) Step 2: Calculation

The regression line will be as:

$\overset{⏜}{y} = 55.5726 + 12.1158 x$

As a result, we can see that the correlation coefficient with the outlier is greater than the correlation coefficient without it. Due to the fact that subject $18$ follows the same linear trend as the other points in the scatterplot, the outlier enhances the correlation. We then notice that the two regression lines in the scatterplot are nearly identical, implying that the outlier has little effect on the regression line.

Part (c) Step 1: Explanation

Now we must use the excel function to calculate the correlation:

First, we'll enter the data into an excel file, and then we'll utilize the correlation function, which is,

CORREL function returns the correlation coefficient of the $a r r a y 1$ and $a r r a y 2$ cell ranges. Thus, the syntax is as:

$C O R R E L (a r r a y 1, a r r a y 2)$

AVERAGE function returns the average of the $a r r a y 1$ and $a r r a y 2$ cell ranges. The syntax is as:

AVERAGE $(a r r a y 1, a r r a y 2)$

For the case with outlier:

Thus, the calculation will be as:

$C o r r e l a t i o n = C O R R E L (H 1 : H 18, I 1 : I 18)$

And the result will be as:

$C o r r e l a t i o n 0.4506$

Thus, the slope will be,

$b = r \frac{s_{y}}{s_{x}} = 0.4506 \times \frac{81.4827}{3.2619} = 11.2569$

And the $y -$ intercept will be,

$a = y - b x = 172.3846 - 11.2569 \times 10.3769 = 55.5726$

The regression line will be as:

$\overset{⏜}{y} = 55.5726 + 11.2569 x$

For the case without outlier:

Thus, the calculation will be as:

$C o r r e l a t i o n = C O R R E L (H 1 : H 17, I 1 : I 17)$

And the result will be as:

$C o r r e l a t i o n 0.5684$

Thus, the slope will be,

$b = r \frac{s_{y}}{s_{x}} = 0.5684 \times \frac{52.6231}{3.3531} = 8.9204$

And the $y -$ intercept will be,

$a = y - b x = 151.7647 - 8.9204 \times 9.2235 = 69.4872$

Part (c) Step 2: Explanation

The regression line will be

$\overset{⏜}{y} = 69.4872 + 8.9204 x$

As a result, we can see that the correlation coefficient with the outlier is lower than without the outlier. We then notice that the outlier reduces the correlation since subject $15$ deviates from the general linear pattern in the other scatterplot points. Then we see that the regression line with the outlier is steeper than the regression line without the outlier, implying that the outlier causes the regression line to be steeper.