Question

In: Statistics and Probability

Using the data set from the previous question (scatterplot.xlsx). With the data point with the smallest...

Using the data set from the previous question (scatterplot.xlsx).

With the data point with the smallest x value removed, the scatter plot now suggests a ________   ["nonlinear" OR "linear"]         relationship between the dependent and independent variables.

Data:

x y
45 2358
56 4204
26 287
54 3849
24 925
23 3273
34 -678
45 3748
47 2898
43 1974
32 226

Solutions

Expert Solution

With the data point with the smallest x value removed, the scatter plot now suggests a linear  relationship between the dependent and independent variables.

Line of Regression Y on X i.e Y = bo + b1 X
X Y (Xi - Mean)^2 (Yi - Mean)^2 (Xi-Mean)*(Yi-Mean)
45 2358 36 68263.424 1567.636
56 4204 289 4440598.232 35823.636
26 287 169 3275112.9 23526.455
54 3849 225 3070459.615 26284.091
24 925 225 1372944.866 17575.91
23 3273 256 1383617.465 -18820.363
34 -678 25 7699111.589 13873.637
45 3748 36 2726701.53 9907.636
47 2898 64 642037.94 6410.182
43 1974 16 15061.99 -490.909
32 226 49 3499620.631 13095.091

calculation procedure for regression
mean of X = ∑ X / n = 39
mean of Y = ∑ Y / n = 2096.7273
∑ (Xi - Mean)^2 = 1390
∑ (Yi - Mean)^2 = 28193530.18
∑ (Xi-Mean)*(Yi-Mean) = 128753.002
b1 = ∑ (Xi-Mean)*(Yi-Mean) / ∑ (Xi - Mean)^2
= 128753.002 / 1390
= 92.628
bo = ∑ Y / n - b1 * ∑ X / n
bo = 2096.7273 - 92.628*39 = -1515.767
value of regression equation is, Y = bo + b1 X
Y'=-1515.767+92.628* X          


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