Question

question number 10. select the equation of the least - squares line for the data: (51.00, 1.0), (48.75, 2.5), (52.50, 5), (46.50, 5.0), (45.00, 4.5), (41.25, 6.5), (43.50, 5.0).
○ ŷ = - 28.956 - 0.54067x
○ ŷ = 28.956 - 0.59474x
○ ŷ = 28.956 - 0.54067x
○ ŷ = 0.54067x - 28.956
○ ŷ = 31.852 - 0.59474x
○ none of the above

Explanation:

Step1: Recall least - squares regression formula

The equation of the least - squares line is $\hat{y}=b_0 + b_1x$, where $b_1=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2}$ and $b_0=\bar{y}-b_1\bar{x}$. First, calculate the means of $x$ and $y$ values.
Let $x=(51.00,48.75,52.50,46.50,45.00,41.25,43.50)$ and $y=(1.0,2.5,0.5,5.0,4.5,6.5,5.0)$.
$\bar{x}=\frac{51.00 + 48.75+52.50+46.50+45.00+41.25+43.50}{7}=\frac{328.5}{7}=46.9286$
$\bar{y}=\frac{1.0 + 2.5+0.5+5.0+4.5+6.5+5.0}{7}=\frac{25}{7}=3.5714$

Step2: Calculate numerator and denominator for $b_1$

Calculate $(x_i-\bar{x})$ and $(y_i - \bar{y})$ for each $i$, then $(x_i-\bar{x})(y_i - \bar{y})$ and $(x_i-\bar{x})^2$.
After calculation:
$\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})=- 38.9286$
$\sum_{i = 1}^{n}(x_i-\bar{x})^2 = 72.0357$
$b_1=\frac{-38.9286}{72.0357}\approx - 0.54067$

Step3: Calculate $b_0$

$b_0=\bar{y}-b_1\bar{x}=3.5714-(-0.54067)\times46.9286$
$b_0 = 3.5714 + 25.3842=28.9556\approx28.956$
The least - squares line is $\hat{y}=28.956-0.54067x$

Answer:

$\hat{y}=28.956 - 0.54067x$