Question

"Select the equation of the least squares line for the data: (34.0, 1.30), (32.5, 3.25), (35.0, .65), (31.0, 6.50), (30.0, 5.85), (27.5, 8.45), (29.0, 6.50)."a. ŷ= 37.643-1.0543r b. ŷ= 1.0543x - 37.643 c. ŷ= 37.643 1.1597x d. ŷ= 41.407-1.1597x e. ŷ= -37.643 1.05433x f. None of the above

Answer 1

Answer:

$y=-1.055 x +37.643$

And the best option would be:

a. ŷ= 37.643-1.0543x

Step-by-step explanation:

We assume that the data is this one:

x: 34.0, 32.5, 35.0, 31.0, 30.0, 27.5, 29.0

y: 1.30, 3.25, 0.65, 6.50, 5.85, 8.45, 6.50

Find the least-squares line appropriate for this data.  

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:

$\sum_{i=1}^n x_i = 34.0+ 32.5+ 35.0+ 31.0+ 30.0+ 27.5+ 29.0 =219$

$\sum_{i=1}^n y_i =1.3+3.25+0.65+6.5+5.85+8.45+6.5=32.5$

$\sum_{i=1}^n x^2_i = 34.0^2+ 32.5^2+ 35.0^2+ 31.0^2+ 30.0^2+ 27.5^2+ 29.0^2 =6895.5$

$\sum_{i=1}^n y^2_i =1.3^2+3.25^2+0.65^2+6.5^2+5.85^2+8.45^2+6.5^2=202.8$

$\sum_{i=1}^n x_i y_i =34*1.3 + 32.5*3.25+ 35*0.65+ 31*6.5+30*5.585 +27.5*8.45 +29*6.5=970.45$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=6895.5-\frac{219^2}{7}=43.929$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=970.45-\frac{219*32.5}{7}=-46.336$

And the slope would be:

$m=-\frac{46.336}{43.929}=-1.055$

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

$\bar x= \frac{\sum x_i}{n}=\frac{219}{7}=31.286$

$\bar y= \frac{\sum y_i}{n}=\frac{32.5}{7}=4.643$

And we can find the intercept using this:

$b=\bar y -m \bar x=4.643-(-1.055*31.286)=37.643$

So the line would be given by:

$y=-1.055 x +37.643$

And the best option would be:

a. ŷ= 37.643-1.0543x