Answer
This observation is an outlier and it is also influential.
Work Step by Step
See the new scatter diagram below. Observe how far the new point is from the others.
Let's find the new least-squares regression line:
$x ̅ =\frac{20+39+27+29+26+47+35+38+9}{9}=30$
$s_x=\sqrt {\frac{(20-30)^2+(39-30)^2+(27-30)^2+(29-30)^2+(26-30)^2+(47-30)^2+(35-30)^2+(38-30)^2+(9-30)^2}{9-1}}=11.3248$
$y ̅=\frac{430+750+480+540+510+760+690+632+80}{9}=541.333$
$s_y=\sqrt {\frac{(430-541.333)^2+(750-541.333)^2+(480-541.333)^2+(540-541.333)^2+(510-541.333)^2+(760-541.333)^2+(690-541.333)^2+(632-541.333)^2+(80-541.333)^2}{9-1}}=209.657$
$r=\frac{Σ(\frac{x_i-x ̅}{s_x})(\frac{y_i-y ̅}{s_y})}{n-1}=\frac{(\frac{20-30}{11.3248})(\frac{430-541.333}{209.657})+(\frac{39-30}{11.3248})(\frac{750-541.333}{209.657})+(\frac{27-30}{11.3248})(\frac{480-541.333}{209.657})+(\frac{29-30}{11.3248})(\frac{540-541.333}{209.657})+(\frac{26-30}{11.3248})(\frac{510-541.333}{209.657})+(\frac{47-30}{11.3248})(\frac{760-541.333}{209.657})+(\frac{35-30}{11.3248})(\frac{690-541.333}{209.657})+(\frac{38-30}{11.3248})(\frac{632-541.333}{209.657})+(\frac{9-30}{11.3248})(\frac{80-541.333}{209.657})}{9-1}=0.957$
The least-squares regression line:
$y ̂=b_1x+b_0$
$b_1=r\frac{s_y}{s_x}=0.957\times\frac{209.657}{11.3248}=17.717$
$b_0=y ̅-b_1x ̅ =541.333-17.717\times30=9.823$
So: $y ̂=17.717x+9.823$
We can conclude that this observation is also influetial because the new least-squares regression line changed substantially.
