Assume you have noted the following prices for books and the number of pages that each book contains.Book Pages (x) Price (y)A 5

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Assume you have noted the following prices for books and the number of pages that each book contains.Book Pages (x) Price (y)A 5

00 $7.00B 700 7.50C 750 9.00D 590 6.50E 540 7.50F 650 7.00G 480 4.50a. Develop a least-squares estimated regression line.b. Compute the coefficient of determination and explain its meaning.c. Compute the correlation coefficient between the price and the number of pages. Test to see if x and y are related. Use ? = 0.10.

Business

1 answer:

tigry1 [53] · Answer 1 · 2024-06-11T07:56:49+03:00

$a) $y=0.00991 x+1.042$b) $r^2=0.7503^2=0.563$\\C) $r=\frac{7(30095)-(4210)(49)}{\sqrt{\left[7(2595100)-(4210)^2\right]\left[7(354)-(49)^2\right]}}=0.7503$$

x: 500, 700, 750, 590 , 540, 650, 480

y: 7.00, 7.50 , 9.00, 6.5, 7.50 , 7.0, 4.50

We want to create a linear model like this :

$y=m x+b$

Where

$m=\frac{S_{x y}}{S_{x x}}$

And:

$\begin{aligned}& S_{x y}=\sum_{i=1}^n x_i y_i-\frac{\left(\sum_{i=1}^n x_i\right)\left(\sum_{i=1}^n y_i\right)}{n} \\& S_{x x}=\sum_{i=1}^n x_i^2-\frac{\left(\sum_{i=1}^n x_i\right)^2}{n}\end{aligned}$

With these we can find the sums:

$\begin{aligned}& S_{x x}=\sum_{i=1}^n x_i^2-\frac{\left(\sum_{i=1}^n x_i\right)^2}{n}=2595100-\frac{4210^2}{7}=63085.714 \\& S_{x y}=\sum_{i=1}^n x_i y_i-\frac{\left(\sum_{i=1}^n x_i\right)\left(\sum_{i=1}^n y_i\right) n}{=} 30095-\frac{4210 * 49}{7}=625\end{aligned}$

And the slope would be:

$m=\frac{625}{63085.714}=0.00991$

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

$\begin{aligned}& \bar{x}=\frac{\sum x_i}{n}=\frac{4210}{7}=601.429 \\& \bar{y}=\frac{\sum y_i}{n}=\frac{49}{7}=7\end{aligned}$

And we can find the intercept using this:

$b=\bar{y}-m \bar{x}=7-(0.00991 * 601.429)=1.042$

And the line would be:

$y=0.00991 x+1.042$

Part b

The correlation coefficient is given by:

$r=\frac{n\left(\sum x y\right)-\left(\sum x\right)\left(\sum y\right)}{\sqrt{\left[n \sum x^2-\left(\sum x\right)^2\right]\left[n \sum y^2-\left(\sum y\right)^2\right]}}$

For our case we have this:

$\begin{aligned}& \mathrm{n}=7 \sum x=4210, \sum y=49, \sum x y=30095, \sum x^2=2595100, \sum y^2=354 \\& r=\frac{7(30095)-(4210)(49)}{\sqrt{\left[7(2595100)-(4210)^2\right]\left[7(354)-(49)^2\right]}}=0.7503\end{aligned}$