Question

What is the equation of the line of best fit for the following data? Round the

Answer 1

Answer:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=438-\frac{44^2}{5}=50.8$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=415-\frac{44*42}{5}=45.4$

And the slope would be:

$m=\frac{45.4}{50.8}=0.8937 \approx 0.894$

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

$\bar x= \frac{\sum x_i}{n}=\frac{44}{5}=8.8$

$\bar y= \frac{\sum y_i}{n}=\frac{42}{5}=8.4$

And we can find the intercept using this:

$b=\bar y -m \bar x=8.4-(0.894*8.8)=0.535$

So the line would be given by:

$y=0.894 x +0.535$

And the best option is:

A. y = 0.894x + 0.535

Step-by-step explanation:

We have the following dataset given

x: 5,6,9,10,14

y: 4,6,9,11,12

We want to find the least-squares line appropriate for this data given by this general expresion:

$y = mx +b$

Where m is the slope and b the intercept

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

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

So we can find the sums like this:

$\sum_{i=1}^n x_i = 44$

$\sum_{i=1}^n y_i =42$

$\sum_{i=1}^n x^2_i =438$

$\sum_{i=1}^n y^2_i =398$

$\sum_{i=1}^n x_i y_i =415$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=438-\frac{44^2}{5}=50.8$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=415-\frac{44*42}{5}=45.4$

And the slope would be:

$m=\frac{45.4}{50.8}=0.8937 \approx 0.894$

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

$\bar x= \frac{\sum x_i}{n}=\frac{44}{5}=8.8$

$\bar y= \frac{\sum y_i}{n}=\frac{42}{5}=8.4$

And we can find the intercept using this:

$b=\bar y -m \bar x=8.4-(0.894*8.8)=0.535$

So the line would be given by:

$y=0.894 x +0.535$

And the best option is:

A. y = 0.894x + 0.535