Box Office Mojo collects and posts data on movie grosses. For a random sample of 50 movies, we obtained both the domestic (U.S.) and overseas grosses,

Given: ${\displaystyle n=\text{Sample size}=50}$ a) Domestic is on the horizontal axis and Overseas 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: ${\displaystyle \sum x_i=3588.9}$
${\displaystyle \sum x_i^2=712440.81}$
${\displaystyle \sum y_i=5233.5}$
${\displaystyle \sum x_i{y}_i=968209.51}$ Next, we can determine ${\displaystyle S_{xx}}$ and ${\displaystyle S_{xy}}$
${\displaystyle S_{xx}=\sum x_i^2-\frac{{(\sum x_i)}^2}{n}=712440.81-\frac{{3588.9}^2}{50}=4548367458}$
${\displaystyle S_{xy}=\sum x_i{y}_i-\frac{(\sum x_i)(\sum y_i)}{n}=968209.51-\frac{3588.9\cdot 5233.5}{50}=592559.347}$ The estimate b of the slope ${\displaystyle \beta }$ is the ratio of ${\displaystyle S_{xy}}$ and ${\displaystyle S_{xx}}$: ${\displaystyle b=\frac{S_{xy}}{S_{xx}}=\frac{592559.374}{454836.7458}=1.3028}$ The mean is the sum of all values divided by the number of values: ${\displaystyle \bar{x}=\frac{\sum x_i}{n}=\frac{3588.9}{50}=71.778}$
${\displaystyle \bar{y}=\frac{\sum y_i}{n}=\frac{5233.5}{50}=104.67}$ 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. ${\displaystyle a=\bar{y}-b\bar{x}=104.67-1.3028\cdot 71.778=11.1579}$ General least-squares equation: ${\displaystyle \hat{y}=\alpha +\beta x}$. Replace ${\displaystyle \alpha }$ by ${\displaystyle a=11.1579}$ and ${\displaystyle \beta }$ by ${\displaystyle b=1.3028}$ in the general least-squares equation: ${\displaystyle y=a+bx=11.1579+1.3028x}$ d) There appear to be two outliers, because the two rightmost points lie far from the group of other points. There appear to be an influential obsevation, because the point in the rightmost corner lies very close to the regression line while the point is a potential outlier. e) Let us first determine the necessary sums: ${\displaystyle \sum x_i=2732.8}$
${\displaystyle \sum x_i^2=345183.2}$
${\displaystyle \sum y_i=4314}$
${\displaystyle \sum x_i{y}_i=567537.88}$ Next, we can determine ${\displaystyle S_{xx}}$ and

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