Respuesta :
Answer:
[tex]y=0.259 x +10.447[/tex]
Now we can find the residulls like this:
[tex] e_1 = 17.3 - 17.375 = -0.075[/tex]
[tex] e_2 = 17.1 - 17.052 = 0.049[/tex]
[tex] e_3 = 17.3 - 17.311 = -0.011[/tex]
[tex] e_4 = 17.5 - 17.440 = 0.06[/tex]
[tex] e_5 = 16.9 - 16.922 = -0.022[/tex]
So then we can see that the residuals are not with an specified pattern (alternating sign) so then we can conclude that are distributed normally
Step-by-step explanation:
We have the following data given:
Height (inches), x 26.75 25.5 26.5 27 25
Head Circumference (inches), y 17.3 17.1 17.3 17.5 16.9
We need to find a linear model [tex] y = mx +b[/tex]
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 =130.75[/tex]
[tex]\sum_{i=1}^n y_i =86.1[/tex]
[tex]\sum_{i=1}^n x^2_i =3422.06[/tex]
[tex]\sum_{i=1}^n y^2_i = 1482.85[/tex]
[tex]\sum_{i=1}^n x_i y_i =2252.28[/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}=3422.06-\frac{130.75^2}{5}=2.95[/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)}=2252.28-\frac{130.75*86.1}{5}=0.765[/tex]
And the slope would be:
[tex]m=\frac{0.765}{2.95}=0.259[/tex]
Nowe we can find the means for x and y like this:
[tex]\bar x= \frac{\sum x_i}{n}=\frac{130.75}{5}=26.15[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{86.1}{5}=17.22[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x=17.22-(0.259*26.15)=10.447[/tex]
So the line would be given by:
[tex]y=0.259 x +10.447[/tex]
Now we can find the residulls like this:
[tex] e_1 = 17.3 - 17.375 = -0.075[/tex]
[tex] e_2 = 17.1 - 17.052 = 0.049[/tex]
[tex] e_3 = 17.3 - 17.311 = -0.011[/tex]
[tex] e_4 = 17.5 - 17.440 = 0.06[/tex]
[tex] e_5 = 16.9 - 16.922 = -0.022[/tex]
So then we can see that the residuals are not with an specified pattern (alternating sign) so then we can conclude that are distributed normally
