Question

Profit The net profits P (in millions of dollars) of Medco Health solutions from 2000 through 2009 are shown in the table.

Answer 1

Answer:

a) For this case we have the linear model given:

$P(t) = 223.89 e^{0.1979t}$

And we want to see the fit to the model so we can calculate the value for the model and the difference respect to the observed value.

$t=0, P(0)= 223.89 e^{0.1979*0}= 223.89 , e= |216.8-223.89|=7.09$

$t=1, P(1)= 223.89 e^{0.1979*1}= 272.89 , e= |256.6-272.89|=16.29$

$t=2, P(2)= 223.89 e^{0.1979*2}=332.60 , e= |361.6-332.60|=29.00$

$t=3, P(3)= 223.89 e^{0.1979*3}= 405.39 , e= |425.8-405.39|=20.41$

$t=4, P(4)= 223.89 e^{0.1979*4}= 494.11 , e= |481.6-494.11|=12.51$

As we can see the model not perfect fits to the data but the residuals are not to higher compared to the real values.

b) $y=116.85 x +111.15$

c) $t=15, P(15)= 223.89 e^{0.1979*15}= 4357.505$

$y=116.85(15) +111.15=1863.9$

As we can see the difference between the two models is higher.

Step-by-step explanation:

For this case we have the following data:

x=t      y=P(t)

0       216.8

1        256.6

2       361.6

3       425.8

4       481.6

5       602.0

6       729.8

7       912.0

8       1102.9

9       1280.3

Where x represent  the year, with t = 0 corresponding to 2000

Part a

For this case we have the linear model given:

$P(t) = 223.89 e^{0.1979t}$

And we want to see the fit to the model so we can calculate the value for the model and the difference respect to the observed value.

$t=0, P(0)= 223.89 e^{0.1979*0}= 223.89 , e= |216.8-223.89|=7.09$

$t=1, P(1)= 223.89 e^{0.1979*1}= 272.89 , e= |256.6-272.89|=16.29$

$t=2, P(2)= 223.89 e^{0.1979*2}=332.60 , e= |361.6-332.60|=29.00$

$t=3, P(3)= 223.89 e^{0.1979*3}= 405.39 , e= |425.8-405.39|=20.41$

$t=4, P(4)= 223.89 e^{0.1979*4}= 494.11 , e= |481.6-494.11|=12.51$

As we can see the model not perfect fits to the data but the residuals are not to higher compared to the real values.

Part 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:

$\sum_{i=1}^n x_i =45$

$\sum_{i=1}^n y_i =6369.4$

$\sum_{i=1}^n x^2_i =285$

$\sum_{i=1}^n y^2_i =5239157$

$\sum_{i=1}^n x_i y_i =38302.3$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=285-\frac{45^2}{10}=82.5$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=38302.3-\frac{45*6369.4}{10}=964-$

And the slope would be:

$m=\frac{9640}{82.5}=116.85$

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

$\bar x= \frac{\sum x_i}{n}=\frac{45}{10}=4.5$

$\bar y= \frac{\sum y_i}{n}=\frac{6369.4}{10}=636.94$

And we can find the intercept using this:

$b=\bar y -m \bar x=636.94-(116.85*4.5)=111.15$

So the line would be given by:

$y=116.85 x +111.15$

Part c

$t=15, P(15)= 223.89 e^{0.1979*15}= 4357.505$

$y=116.85(15) +111.15=1863.9$

As we can see the difference between the two models is higher.