The following bivariate data set contains an outlier. Enter the data into StatCrunch. Using it'sgraphing ability, draw a scatter plot of the data. Sketch the plot on your paper in your notebook.У61-37465-1567-2967681763-98967-510121-6407What is the correlation coefficient with the outlier? Use StatCrunch.rwith{3 decimal places}%DWhat is the correlation coefficient without the outlier? Use StatCrunch.rwithout{3 decimal places}%3DWould inclusion of the outlier change the evidence for or against a significant linear correlation?O Yes. Including the outlier changes the evidence regarding a linear correlation.O No. Including the outlier does not change the evidence regarding a linear correlation.

Given data : X Y 61 -374 65 -15 67 -296 76 817 63 -989 67 -510 121 -6407 Outlier is a data point which lies far away from the acceptable / normal range of the data and differs significantly from other observations First we will draw a scatter plot to see for the outlier: The above scatter plot shows one outlier i.e. point X Y 121 -6407a) The below table will be used for correlation : X Y X⋅YX \cdot Y X2X^2 Y2Y^2 61 -374 -22814 3721 139876 65 -15 -975 4225 225 67 -296 -19832 4489 87616 76 817 62092 5776 667489 63 -989 -62307 3969 978121 67 -510 -34170 4489 260100 121 -6407 -775247 14641 41049649 Sum = 520 -7774 -853253 41310 43183076 Calculating Sum of squares : Xˉ=1n∑i=1nXi=5207=74.285 Yˉ=1n∑i=1nYi=−77747=−1110.571 SSXX=∑i=1nXi2−1n(∑i=1nXi)2=41310−5202/7=2681.428 SSYY=∑i=1nYi2−1n(∑i=1nYi)2=43183076−−77742/7=34549493.714 SSXY=∑i=1nXiYi−1n(∑i=1nXi)(∑i=1nYi)=−853253−520×−7774/7=−275755.857 The correlation coefficient : r=SSXYSSXX⋅SSYY=−275755.8572681.429⋅34549493.714=−0.906 \begin{array}{ccl} r & = & \displaystyle \frac{ SS_{XY}}{ \sqrt{SS_{XX} \cdot SS_{YY}}} \\\\ \\\\ & = & \displaystyle \frac{ -275755.857 }{ \sqrt{2681.429 \cdot 34549493.714}} \\\\ \\\\ & = & -0.906 \end{array} The value of Rwith is -0.906. This is a strong negative correlation, which means that high X variable scores go with low Y variable scores (and vice versa). Now , The below table will be used for correlation : X Y 61 -374 65 -15 67 -296 76 817 63 -989 67 -510 The below table will be used for correlation : X Y X.Y X2X^2 Y2Y^2 61 -374 -22814 3721 139876 65 -15 -975 4225 225 67 -296 -19832 4489 87616 76 817 62092 5776 667489 63 -989 -62307 3969 978121 67 -510 -34170 4489 260100 Sum = 399 -1367 -78006 26669 2133427 Calculating Sum of squares : Xˉ=1n∑i=1nXi=3996=66.5\bar X = \frac{1}{n} \sum_{i=1}^{n} X_i = \frac{ 399}{ 6} = 66.5 Yˉ=1n∑i=1nYi=−13676=−227.83333333333\bar Y = \frac{1}{n} \sum_{i=1}^{n} Y_i = \frac{ -1367}{ 6} = -227.83333333333 SSXX=∑i=1nXi2−1n(∑i=1nXi)2=26669−3992/6=135.5\large SS_{XX} = \sum_{i=1}^{n} X_i^2 - \displaystyle\frac{1}{n}\left(\sum_{i=1}^{n} X_i\right)^2 = 26669 - 399^2/6 = 135.5 SSYY=∑i=1nYi2−1n(∑i=1nYi)2=2133427−−13672/6=1821978.8333333\large SS_{YY} = \sum_{i=1}^{n} Y_i^2 - \displaystyle\frac{1}{n}\left(\sum_{i=1}^{n} Y_i\right)^2 = 2133427 - -1367^2/6 = 1821978.8333333 SSXY=∑i=1nXiYi−1n(∑i=1nXi)(∑i=1nYi)=−78006−399×−1367/6=12899.5\large SS_{XY} = \sum_{i=1}^{n} X_i Y_i - \displaystyle\frac{1}{n}\left(\sum_{i=1}^{n} X_i\right) \left(\sum_{i=1}^{n} Y_i\right) = -78006 - 399 \times -1367/6 = 12899.5 The correlation coefficient: r=SSXYSSXX⋅SSYY=12899.5135.5⋅1821978.833=0.821 \begin{array}{ccl} r & = & \displaystyle \frac{ SS_{XY}}{ \sqrt{SS_{XX} \cdot SS_{YY}}} \\\\ \\\\ & = & \displaystyle \frac{ 12899.5 }{ \sqrt{135.5 \cdot 1821978.833}} \\\\ \\\\ & = & 0.821 \end{array} The value of Rwithout is 0.821. This is a strong positive correlation, which means that high X variable scores go with high Y variable scores (and vice versa). Here, we can see that the correlation changes from negative to positive which tells us a complete change from a strong negative correlation to strong positive correlation. But in both the conditions , there exists a strong linear correlation. Hence, Correct option :