Chapter 3.2, Problem 69E

(a)

To determine

To make a scatterplot with HbA as the explanatory variable and describe what you see.

Explanation of Solution

The scatterplot with HbA as the explanatory variable is as:

  

From the scatterplot we can say that, the scatterplot confirms a positive linear relationship, because the scatterplot slopes upwards. The scatterplot confirms a weak relationship, because the points seem to lie far apart.

(b)

To determine

To explain what effect do you think this subject has on the correlation and on the equation of the least squares regression line and calculate the correlation and equation of the least squares regression line with and without this subject to confirm your answer.

Answer to Problem 69E

No, they do not affect.

Explanation of Solution

Now, we have to calculate the correlation by using the excel function as:

First we will put the data in the excel file and then we will use the function for the correlation, that is,

CORREL function returns the correlation coefficient of the $\text{array1 and array2}$ cell ranges. Thus, the syntax is as:

  $\text{CORREL}\left(\text{array1, array2}\right)$ .

AVERAGE function returns the average of the $\text{array1 and array2}$ cell ranges. The syntax is as:

  $\text{AVERAGE}\left(\text{array1, array2}\right)$ .

For the case with outlier:

Thus, the calculation will be as:

Correlation

=CORREL(H1:H18,I1:I18)

And the result will be as:

Correlation

0.4506

Thus, the slope will be,

  $\begin{array}{l}b=r\frac{s_y}{s_x}\\ =0.4506\times \frac{81.4827}{3.2619}\\ =11.2569\end{array}$

And the y intercept will be,

  $\begin{array}{l}a=\overline{y}-b\overline{x}\\ =172.3846-11.2569\times 10.3769\\ =55.5726\end{array}$

The regression line will be as:

  $\widehat{y}=55.5726+11.2569x$

For the case without outlier:

Thus, the calculation will be as:

Correlation

=CORREL(H1:H17,I1:I17)

And the result will be as:

Correlation

0.3837

Thus, the slope will be,

  $\begin{array}{l}b=r\frac{s_y}{s_x}\\ =0.3837\times \frac{68.9020}{2.1821}\\ =12.1158\end{array}$

And the y intercept will be,

  $\begin{array}{l}a=\overline{y}-b\overline{x}\\ =157.8824-12.1156\times 8.7176\\ =52.2615\end{array}$

The regression line will be as:

  $\widehat{y}=55.5726+12.1158x$

  

Thus, we note that the correlation coefficient with the outlier is more than the correlation coefficient without outlier. We then note that the outlier increases the correlation due to the fact that subject $18$ lies in the same linear pattern as the other points in the scatterplot. We then note that the two regression lines in the scatterplot are roughly the same and thus the outlier does not seem to affect the regression line very much.

(c)

To determine

To explain what effect do you think this subject has on the correlation and on the equation of the least squares regression line and calculate the correlation and equation of the least squares regression line with and without this subject to confirm your answer.

Answer to Problem 69E

It makes the regression line steeper.

Explanation of Solution

Now, we have to calculate the correlation by using the excel function as:

First we will put the data in the excel file and then we will use the function for the correlation, that is,

CORREL function returns the correlation coefficient of the $\text{array1 and array2}$ cell ranges. Thus, the syntax is as:

  $\text{CORREL}\left(\text{array1, array2}\right)$ .

AVERAGE function returns the average of the $\text{array1 and array2}$ cell ranges. The syntax is as:

  $\text{AVERAGE}\left(\text{array1, array2}\right)$ .

For the case with outlier:

Thus, the calculation will be as:

Correlation

=CORREL(H1:H18,I1:I18)

And the result will be as:

Correlation

0.4506

Thus, the slope will be,

  $\begin{array}{l}b=r\frac{s_y}{s_x}\\ =0.4506\times \frac{81.4827}{3.2619}\\ =11.2569\end{array}$

And the y intercept will be,

  $\begin{array}{l}a=\overline{y}-b\overline{x}\\ =172.3846-11.2569\times 10.3769\\ =55.5726\end{array}$

The regression line will be as:

  $\widehat{y}=55.5726+11.2569x$

For the case without outlier:

Thus, the calculation will be as:

Correlation

=CORREL(H1:H17,I1:I17)

And the result will be as:

Correlation

0.5684

Thus, the slope will be,

  $\begin{array}{l}b=r\frac{s_y}{s_x}\\ =0.5684\times \frac{52.6231}{3.3531}\\ =8.9204\end{array}$

And the y intercept will be,

  $\begin{array}{l}a=\overline{y}-b\overline{x}\\ =151.7647-8.9204\times 9.2235\\ =69.4872\end{array}$

The regression line will be as:

  $\widehat{y}=69.4872+8.9204x$

  

Thus, we note that the correlation coefficient with the outlier is less than the correlation coefficient without outlier. We then note that the outlier decreases the correlation due to the fact that subject $15$ deviates from the general linear pattern in the other points in the scatterplot. We then note that the regression line with the outlier is steeper than the regression line without outlier and thus the outlier makes the regression line steeper.