91Ó°ÊÓ

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.

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 $array1andarray2$cell ranges. Thus, the syntax is as:

$CORREL(array1,array2)$

AVERAGE function returns the average of the $array1andarray2$cell ranges. The syntax is as:

$AVERAGE(array1,array2)$

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,

$$\begin{gathered} b=r\frac{s_y}{s_x} \\ =0.4506×\frac{81.4827}{3.2619} \\ =11.2569 \end{gathered}$$

And the $y-$intercept will be,

$$\begin{gathered} a=\overline{y}−b\overline{x} \\ =172.3846−11.2569×10.3769 \\ =55.5726 \end{gathered}$$

The regression line will be as:

$\stackrel{Áåœ}{y}=55.5726+11.2569x$

For the case without outlier:

Thus, the calculation will be as:

$Correlation=CORREL(H1:H17,I1:I17)$

And the result will be as:

$Correlation0.3837$

Thus, the slope will be,

$$\begin{gathered} b=r\frac{s_y}{s_x} \\ =0.3837×\frac{68.9020}{2.1821} \\ =12.1158 \end{gathered}$$

And the$y-$intercept will be,

$$\begin{gathered} a=\overline{y}−b\overline{x} \\ =157.8824−12.1156×8.7176 \\ =52.2615 \end{gathered}$$

The regression line will be as:

$\stackrel{Áåœ}{y}=55.5726+12.1158x$

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.

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 $array1$and $array2$cell ranges. Thus, the syntax is as:

$CORREL(array1,array2)$

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

AVERAGE $(array1,array2)$

For the case with outlier:

Thus, the calculation will be as:

$Correlation=CORREL(H1:H18,I1:I18)$

And the result will be as:

$Correlation0.4506$

Thus, the slope will be,

$$\begin{gathered} b=r\frac{s_y}{s_x} \\ =0.4506×\frac{81.4827}{3.2619} \\ =11.2569 \end{gathered}$$

And the $y-$intercept will be,

$$\begin{gathered} a=\overline{y}−b\overline{x} \\ =172.3846−11.2569×10.3769 \\ =55.5726 \end{gathered}$$

The regression line will be as:

$\stackrel{Áåœ}{y}=55.5726+11.2569x$

For the case without outlier:

Thus, the calculation will be as:

$Correlation=CORREL(H1:H17,I1:I17)$

And the result will be as:

$Correlation0.5684$

Thus, the slope will be,

$$\begin{gathered} b=r\frac{s_y}{s_x} \\ =0.5684×\frac{52.6231}{3.3531} \\ =8.9204 \end{gathered}$$

And the $y-$intercept will be,

$$\begin{gathered} a=\overline{y}−b\overline{x} \\ =151.7647−8.9204×9.2235 \\ =69.4872 \end{gathered}$$

The regression line will be

$\stackrel{Áåœ}{y}=69.4872+8.9204x$

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.