Question

A pediatrician wants to determine the relation that may exist between a​ child's height and head circumference. She randomly selects 5 children and measures their height and head circumference. The data are summarized below. Find the residuals from the regression and verify that the residuals are approximately normally distributed. 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

Answer 1

Answer:

$y=0.259 x +10.447$

Now we can find the residulls like this:

$e_1 = 17.3 - 17.375 = -0.075$

$e_2 = 17.1 - 17.052 = 0.049$

$e_3 = 17.3 - 17.311 = -0.011$

$e_4 = 17.5 - 17.440 = 0.06$

$e_5 = 16.9 - 16.922 = -0.022$

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 $y = mx +b$

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:

$\sum_{i=1}^n x_i =130.75$

$\sum_{i=1}^n y_i =86.1$

$\sum_{i=1}^n x^2_i =3422.06$

$\sum_{i=1}^n y^2_i = 1482.85$

$\sum_{i=1}^n x_i y_i =2252.28$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=3422.06-\frac{130.75^2}{5}=2.95$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=2252.28-\frac{130.75*86.1}{5}=0.765$

And the slope would be:

$m=\frac{0.765}{2.95}=0.259$

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

$\bar x= \frac{\sum x_i}{n}=\frac{130.75}{5}=26.15$

$\bar y= \frac{\sum y_i}{n}=\frac{86.1}{5}=17.22$

And we can find the intercept using this:

$b=\bar y -m \bar x=17.22-(0.259*26.15)=10.447$

So the line would be given by:

$y=0.259 x +10.447$

Now we can find the residulls like this:

$e_1 = 17.3 - 17.375 = -0.075$

$e_2 = 17.1 - 17.052 = 0.049$

$e_3 = 17.3 - 17.311 = -0.011$

$e_4 = 17.5 - 17.440 = 0.06$

$e_5 = 16.9 - 16.922 = -0.022$

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