An issue of BARRON’S presented information on top wealth managers in the United States, based...

Given: ${\displaystyle n=\text{Sample size}=38}$ a) Managers is on the horizontal axis and Assets is on the vertical axis. b) It is reasonable to find a regression lien for the data if there is no strong curvature present in the scatterplot. We note that there is no strong curvature in the scatterplot of part (a) and thus it is reasonable to find a regression line for the data. c) Let us first determine the necessary sums: ${\displaystyle \sum x_i=31386}$
${\displaystyle \sum x_i^2=174931090}$
${\displaystyle \sum y_i=3103.9}$
${\displaystyle \sum x_i{y}_i=8095211}$ Next, we can determine ${\displaystyle S_{xx}\text{and}{S}_{xy}}$
${\displaystyle S_{xx}=\sum x_i^2-\frac{{(\sum x_i)}^2}{n}=174931090-\frac{{31386}^2}{38}=149007905.8947}$
${\displaystyle S_{xy}=\sum x_i{y}_i-\frac{(\sum x_i)(\sum y_i)}{n}=8095211-\frac{31386\cdot 3103.9}{38}=5531552.9632}$ The estimate b of the slope ${\displaystyle \beta \text{is the ratio of}{S}_{xy}\text{and}{S}_{xx}}$:
${\displaystyle b=\frac{S_{xy}}{S_{xx}}=\frac{5531552.9632}{149007905.8947}=0.0371}$ The mean is the sum of all values divided by the number of values: ${\displaystyle \bar{x}=\frac{\sum x_i}{n}=\frac{31386}{38}=825.9474}$
${\displaystyle \bar{y}=\frac{\sum y_i}{n}=\frac{3103.9}{38}=81.6816}$ The estimate a of the intercept ${\displaystyle \alpha }$ is the average of y decreased by the product of the estimate of the slope and the average of x. ${\displaystyle a=\bar{y}-b\bar{x}=81.6816-0.0371\cdot 825.9474=51.0203}$ General least-squares equation: ${\displaystyle \hat{y}=\alpha +\beta x\text{Replace}\alpha \text{by}a=51.0203\text{and}\beta \text{by}b=0.0371}$ in the general least-squares equation: ${\displaystyle y=a+bx=51.0203+0.0371x}$ d) The right most point appears to be an influential outlier, because it lies extremely close to the regression line and lies far from the other points in the data set. The point at the top of the scatterplot appears to be an outlier, because it lies far from the data as well. There appears to be a few other outliers as well, but we will only use these two points in parts (e) and (f). e) Let us first determine the necessary sums: ${\displaystyle \sum x_i=29498}$
${\displaystyle \sum x_i^2=171366546}$
${\displaystyle \sum y_i=2473.9}$
${\displaystyle \sum x_i{y}_i=6905771}$ Next, we can determine