Use the technology of your choice to do the following tasks. The National Oceanic and Atmospheric Administration publishes temperature and precipitati

Given: ${\displaystyle n=Sa\mp \le size=48}$ a) IMR is on the horizontal axis and LE is on the vertical axis. b) When there is no strong curvature presents in the scatterplot, then it is safe to assume that there is a linear relationship between the variables and thus it is then reasonable to find a regression line. We note that the scatterplot of part (a) does not contain strong curvature and thus is reasonable to find the regression line. c) We determine all necessary sums: ${\displaystyle \sum x_i=3749}$
${\displaystyle \sum y_i=123.3}$
${\displaystyle \sum x_i{y}_i=9681.9}$
${\displaystyle \sum x_i^2=300561}$
${\displaystyle \sum y_i^2=584.99}$ Next, we can determine ${\displaystyle S_{xx}\text{and}{S}_{xy}}$
${\displaystyle S_{xx}=\sum x_i^2=300561-\frac{{3749}^2}{48}=39602.1698}$
${\displaystyle S_{xy}=\sum x_i{y}_i-\frac{(\sum x_i)(\sum y_i)}{n}=9681.9-\frac{3749\cdot 123.3}{48}=-14009.0727}$ The estimate b of the slope ${\displaystyle \beta \text{is the ratio of}{S}_{xy}\text{and}{S}_{xx}:}$
${\displaystyle b=\frac{S_{xy}}{S_{xx}=\frac{-14009.0727}{39602.1698}=0.0067}}$ The mean is the sum of all values divided by the number of values: ${\displaystyle \bar{x}=\frac{\sum x_i}{n}=\frac{3749}{48}=78.1042}$
${\displaystyle \bar{y}=\sum \{y_i\}\left\{n\right\}=\frac{123.3}{48}=2.5688}$ The estimate a of the intercept ${\displaystyle \alpha \text{is the average of y decresed by the product of the estimate of the slope and the average of x}}$
${\displaystyle a=\bar{y}-b\bar{x}=2.5688-0.0067\cdot 78.1042=2.0481}$ General least-squares equation: ${\displaystyle \hat{y}=\alpha +\beta x.\text{Replace}\alpha bya=2.0481\text{and}\beta byb=0.0067\text{in the general least-squres equation}}$
${\displaystyle \hat{y}=a+bx=2.0481+0.0067x}$
d) Let us evalute the regression line in part (c) at ${\displaystyle x=83:}$

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