Question

Use the given data to find a regression line that best fits the price-demand data for price p in dollars as a function of the demand x widgets. Here, price is the dependent variable, and demand is the independent variable. Find the regression function for price, and write it as p ( x )

Answer 1

Answer:

$m=-\frac{7600}{8250}=-0.921$

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

$\bar x= \frac{\sum x_i}{n}=\frac{550}{10}=55$

$\bar y= \frac{\sum y_i}{n}=\frac{1042}{10}=104.2$

And we can find the intercept using this:

$b=\bar y -m \bar x=104.2-(-0.921*55)=104.707$

So the line would be given by:

$y=-0.921 x +104.707$

Step-by-step explanation:

For this case we have the following data given:

Demand (x): 10,20,30,40,50,60,70,80,90,100

Price (y): 141 , 133,126, 128,113,97, 90, 82,79,53

We want to construct a linear model like this:

$y = mx +b$

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:

$\sum_{i=1}^n x_i =550$

$\sum_{i=1}^n y_i =1042$

$\sum_{i=1}^n x^2_i =38500$

$\sum_{i=1}^n y^2_i =115882$

$\sum_{i=1}^n x_i y_i =49710$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=38500-\frac{550^2}{10}=8250$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=49710-\frac{550*1042}{10}=-7600$

And the slope would be:

$m=-\frac{7600}{8250}=-0.921$

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

$\bar x= \frac{\sum x_i}{n}=\frac{550}{10}=55$

$\bar y= \frac{\sum y_i}{n}=\frac{1042}{10}=104.2$

And we can find the intercept using this:

$b=\bar y -m \bar x=104.2-(-0.921*55)=104.707$

So the line would be given by:

$p(x)=-0.921 x +104.707$