Respuesta :
Answer:
a) For this case we have the linear model given:
[tex] P(t) = 223.89 e^{0.1979t}[/tex]
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.
[tex] t=0, P(0)= 223.89 e^{0.1979*0}= 223.89 , e= |216.8-223.89|=7.09[/tex]
[tex] t=1, P(1)= 223.89 e^{0.1979*1}= 272.89 , e= |256.6-272.89|=16.29[/tex]
[tex] t=2, P(2)= 223.89 e^{0.1979*2}=332.60 , e= |361.6-332.60|=29.00[/tex]
[tex] t=3, P(3)= 223.89 e^{0.1979*3}= 405.39 , e= |425.8-405.39|=20.41[/tex]
[tex] t=4, P(4)= 223.89 e^{0.1979*4}= 494.11 , e= |481.6-494.11|=12.51[/tex]
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) [tex]y=116.85 x +111.15[/tex]
c) [tex] t=15, P(15)= 223.89 e^{0.1979*15}= 4357.505[/tex]
[tex]y=116.85(15) +111.15=1863.9[/tex]
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:
[tex] P(t) = 223.89 e^{0.1979t}[/tex]
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.
[tex] t=0, P(0)= 223.89 e^{0.1979*0}= 223.89 , e= |216.8-223.89|=7.09[/tex]
[tex] t=1, P(1)= 223.89 e^{0.1979*1}= 272.89 , e= |256.6-272.89|=16.29[/tex]
[tex] t=2, P(2)= 223.89 e^{0.1979*2}=332.60 , e= |361.6-332.60|=29.00[/tex]
[tex] t=3, P(3)= 223.89 e^{0.1979*3}= 405.39 , e= |425.8-405.39|=20.41[/tex]
[tex] t=4, P(4)= 223.89 e^{0.1979*4}= 494.11 , e= |481.6-494.11|=12.51[/tex]
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:
[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]
Where:
[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]
[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]
So we can find the sums like this:
[tex]\sum_{i=1}^n x_i =45[/tex]
[tex]\sum_{i=1}^n y_i =6369.4[/tex]
[tex]\sum_{i=1}^n x^2_i =285[/tex]
[tex]\sum_{i=1}^n y^2_i =5239157[/tex]
[tex]\sum_{i=1}^n x_i y_i =38302.3[/tex]
With these we can find the sums:
[tex]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[/tex]
[tex]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-[/tex]
And the slope would be:
[tex]m=\frac{9640}{82.5}=116.85[/tex]
Nowe we can find the means for x and y like this:
[tex]\bar x= \frac{\sum x_i}{n}=\frac{45}{10}=4.5[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{6369.4}{10}=636.94[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x=636.94-(116.85*4.5)=111.15[/tex]
So the line would be given by:
[tex]y=116.85 x +111.15[/tex]
Part c
[tex] t=15, P(15)= 223.89 e^{0.1979*15}= 4357.505[/tex]
[tex]y=116.85(15) +111.15=1863.9[/tex]
As we can see the difference between the two models is higher.
