Respuesta :
Answer:
a) [tex]y = 3.8 x +100[/tex]
b) [tex]Abs. change= |168.4-167.5|=0.9[/tex]
So the calculated value is 0.9 points above the actual value.
[tex]Relative. Change =\frac{|168.4 -167.5|}{167.5}x100 =0.537[/tex]%
And the calculated value it's 0.537% higher than the actual value.
c) For this case we can use the slope obtained from the linear model to answer this question, and we can conclude that the CPI is increasing at approximate 3.8 units per year.
Step-by-step explanation:
Data given
1982 , CPI=100
1986, CPI = 191.2
Notation
Let CPI the dependent variable y. And the time th independent variable x.
For this case we want to adjust a linear model givn by the following expression:
[tex]y=mx+b[/tex]
Solution to the problem
Part a
For this case we can find the slope with the following formula:
[tex] m =\frac{CPI_{2006}-CPI_{1982}}{2006-1982}[/tex]
And if we replace we got:
[tex] m =\frac{191.2-100}{2006-1982}=3.8[/tex]
Let X represent the number of years after. Then for 1982 t = 0, and if we replace we can find b:
[tex] 100 = 3.8(0)+b[/tex]
And then [tex]b=100[/tex]
So then our linear model is given by:
[tex]y = 3.8 x +100[/tex]
Part b
For this case we need to find the years since 1982 and we got x = 2000-1982=18, and if we rpelace this into our linear model we got:
[tex]y = 3.8(18) +100=168.4[/tex]
And the actual value is 167.5 we can compare the result using absolute change or relative change like this:
[tex]Abs. change= |168.4-167.5|=0.9[/tex]
So the calculated value is 0.9 points above the actual value.
And we can find also the relative change like this:
[tex]Relative. Change =\frac{|Calculated -Real|}{Real}x100[/tex]
And if we replace we got:
[tex]Relative. Change =\frac{|168.4 -167.5|}{167.5}x100 =0.537[/tex]%
And the calculated value it's 0.537% higher than the actual value.
Part c
For this case we can use the slope obtained from the linear model to answer this question, and we can conclude that the CPI is increasing at approximate 3.8 units per year.