Use the given data to find the equation of the regression line. x 44 44 11 11 55 y 66 55 negative 1−1 negative 3−3 88 ModifyingAbove y with caretyequals=nothingplus+nothingx​ (Round to two decimal places as​ needed.)

Respuesta :

Answer:

[tex]y=1.98 x -24.34[/tex]

Step-by-step explanation:

Assuming the following data

X: 44, 44, 11, 11, 55

Y: 66, 55, -1, -3, 88

We want to find a linear model [tex] Y= mx +b[/tex]

For this case we need to calculate the slope with the following formula:

[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]

Where:

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]

So we can find the sums like this:

[tex]\sum_{i=1}^n x_i =44+44+11+11+55=165[/tex]

[tex]\sum_{i=1}^n y_i =66+55-1-3+88=205[/tex]

[tex]\sum_{i=1}^n x^2_i =7139[/tex]

[tex]\sum_{i=1}^n y^2_i =15135[/tex]

[tex]\sum_{i=1}^n x_i y_i =10120[/tex]

With these we can find the sums:

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=7139-\frac{165^2}{5}=1694[/tex]

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=10120-\frac{165*205}{5}=3355[/tex]

And the slope would be:

[tex]m=\frac{3355}{1694}=1.98[/tex]

Nowe we can find the means for x and y like this:

[tex]\bar x= \frac{\sum x_i}{n}=\frac{165}{5}=33[/tex]

[tex]\bar y= \frac{\sum y_i}{n}=\frac{205}{5}=41[/tex]

And we can find the intercept using this:

[tex]b=\bar y -m \bar x=41-(1.98*33)=-24.34[/tex]

So the line would be given by:

[tex]y=1.98 x -24.34[/tex]

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