Write 0.45 as a fraction in simplest form

Table below shows the age in years and the retail value in thousands of dollars of a random sample of ten automobiles of the sam

horrorfan [7]

Answer:

Step-by-step explanation:

Hello!

Given the study variables

X: Age of an automobile (years)

Y: Retail value of an automobile (thousands of dollars)

a) Construct a scatter diagram.

(attached image)

b) Compute the linear correlation coefficient r. Interpret its value in the context of the problem.

The formula for the correlation coefficient (Person) is:

$r= \frac{sum(XY)-\frac{sum(X)*sum(Y)}{n} }{[sum(X^2)-\frac{(sum(X))^2}{n}] * [sum(Y^2)-\frac{(sum(Y))^2}{n} ] }$

First you have to do some intermediate summatories:

∑x= 40

∑x²= 174

∑y= 246.30

∑y²= 6154.15

∑xy= 956.5

$r= \frac{956.5-\frac{40*246.3}{10} }{[147-\frac{(40)^2}{10}] * [6154.15-\frac{(246.3)^2}{10} ] }$

r= -0.818 ≅ 0.82

-0.82 indicates there is a strong negative linear correlation between the age of the automobiles and the retail price. Wich means that the older the car, the less it costs.

c) Compute the least squares regression line. Plot it on the scatter diagram.

Interpret the meaning of the slope of the least-squares regression line in the context of the problem.

The least squares regression line is Y= a + bX

To calculate the regression line you need to calculate the values of "a" and "b"

$a= (\frac{sum(Y)}{n} ) -b(\frac{sum(X)}{n} )$

$a= (\frac{246.3}{10} ) -(-2.05)(\frac{40}{10} )$

a= 32.83

$b= \frac{sum(XY)-\frac{sum(X)*sum(Y)}{n} }{sum(X^2)-(\frac{sum(X)^2}{n})}$

$b= \frac{956.5-\frac{40*246.3}{10} }{174-(\frac{40^2}{10})}$

b= -2.05

Y= 32.83 - 2.05X

Check attachment for graphic.

-2.05 $(\frac{years}{thousanfs of dollars})$ is the modification of the estimated mean of the retail value of automobiles every time their age increases one year. In other words, every time the automobiles increase their age in one year, the price decreases 2.05 thousands of dollars.

e) Suppose a 20 -year-old automobile of this make and model is selected at random. Use the regression equation to predict its retail value. Interpret the result.

You need to calculate the value of Y for a value of X of 20 years. To do so you replace the values in the regression equation:

Y= 32.83 - 2.05(20)

Y= -8.17 thousands of dollars.

Under the model Y= 32.83 - 2.05X, a 20-year-old automobile has a retail price of -8.17 thousands of dollars.

This is mathematically correct even if it doesn't make sense in context.

f) Comment on the validity of using the regression equation to predict the price of a brand new automobile of this make and model.

You can only use the regression equation to predict possible values of Y for values of X that are between the range of definition of the variable. In this case, the variable is defined in numbers from 2 to 6 years.

You can predict values outside the range of definition of X but the further these values are from the definition range, the less accurate and imprecise the prediction is.

For example in question e) I've predicted the value of Y for X= 20 years, although I've obtained a result truth is that this prediction is not accurate nor reliable.

I hope it helps!