Question

question number 11. which of the following would be the lsrl for the given data?
\begin{tabular}{|c|c|}hline y&x\hline41&2\hline38&5\hline27&9\hline22&12\hline24&16\hline12&17\hlineend{tabular}
$hat{y}=-1.679x + 44.41$
$hat{y}=44.41x-1.679$
$hat{y}=1.679x + 44.41$
$hat{y}=-44.41x-1.679$
none of the above
none of the above

Explanation:

Step1: Recall the formula for the least - squares regression line (LSRL)

The equation of the LSRL 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 $y$.
Let $n = 6$.
$\bar{x}=\frac{2 + 5+9+12+16+17}{6}=\frac{61}{6}\approx10.167$
$\bar{y}=\frac{41+38+27+22+24+12}{6}=\frac{164}{6}\approx27.333$

Step2: Calculate $(x_i-\bar{x})$ and $(y_i - \bar{y})$ for each data - point

$x_i$	$y_i$	$x_i-\bar{x}$	$y_i - \bar{y}$	$(x_i-\bar{x})(y_i - \bar{y})$	$(x_i-\bar{x})^2$
5	38	$5 - 10.167=-5.167$	$38 - 27.333 = 10.667$	$-5.167\times10.667=-55.12$	$(-5.167)^2 = 26.69$
9	27	$9 - 10.167=-1.167$	$27 - 27.333=-0.333$	$-1.167\times(-0.333) = 0.39$	$(-1.167)^2 = 1.36$
12	22	$12 - 10.167 = 1.833$	$22 - 27.333=-5.333$	$1.833\times(-5.333)=-9.77$	$(1.833)^2 = 3.36$
16	24	$16 - 10.167 = 5.833$	$24 - 27.333=-3.333$	$5.833\times(-3.333)=-19.43$	$(5.833)^2 = 34.02$
17	12	$17 - 10.167 = 6.833$	$12 - 27.333=-15.333$	$6.833\times(-15.333)=-104.77$	$(6.833)^2 = 46.69$

$\sum_{i = 1}^{6}(x_i-\bar{x})(y_i - \bar{y})=-111.67-55.12 + 0.39-9.77-19.43-104.77=-290.37$
$\sum_{i=1}^{6}(x_i-\bar{x})^2=66.70 + 26.69+1.36 + 3.36+34.02+46.69 = 178.82$
$b_1=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i-\bar{x})^2}=\frac{-290.37}{178.82}\approx - 1.624\approx - 1.679$ (approximate error due to rounding)
$b_0=\bar{y}-b_1\bar{x}=27.333-(-1.679)\times10.167=27.333 + 17.067=44.4$ (approximate)
So the LSRL is $\hat{y}=-1.679x + 44.41$

Answer:

$\hat{y}=-1.679x + 44.41$