Question

# 1. Here is a bivariate data set. x y 83.7 47.7 73.3 55.5 68.5 42.3 59...

1. Here is a bivariate data set.

x y
83.7 47.7
73.3 55.5
68.5 42.3
59 43.5
67 41.8
61.6 40.8
64.3 39.2
99.7 71.7
60.8 34.7
69.8 44
86 59.4
47.3 30.8
37.6 20.8
58.6 37.4
57.3 41.1

Find the correlation coefficient and report it accurate to three decimal places.
r =

2.Run a regression analysis on the following bivariate set of data with y as the response variable.

x y
3.8 74.8
54.7 49.8
26.5 64.7
-7.8 86.8
22.6 56.6
60.7 36.1
39.3 54.3
-3.4 91.2
30.4 58.8
50.2 55.2
32 52.9

Verify that the correlation is significant at an α=0.05. If the correlation is indeed significant, predict what value (on average) for the explanatory variable will give you a value of 43.3 on the response variable.

What is the predicted explanatory value?
x =

(Report answer accurate to one decimal place.)

1:

Following table shows the calculations:

 X Y X^2 Y^2 XY 83.7 47.7 7005.69 2275.29 3992.49 73.3 55.5 5372.89 3080.25 4068.15 68.5 42.3 4692.25 1789.29 2897.55 59 43.5 3481 1892.25 2566.5 67 41.8 4489 1747.24 2800.6 61.6 40.8 3794.56 1664.64 2513.28 64.3 39.2 4134.49 1536.64 2520.56 99.7 71.7 9940.09 5140.89 7148.49 60.8 34.7 3696.64 1204.09 2109.76 69.8 44 4872.04 1936 3071.2 86 59.4 7396 3528.36 5108.4 47.3 30.8 2237.29 948.64 1456.84 37.6 20.8 1413.76 432.64 782.08 58.6 37.4 3433.96 1398.76 2191.64 57.3 41.1 3283.29 1689.21 2355.03 Total 994.5 650.7 69242.95 30264.19 45582.57

2:

Following is the output of regression analysis:

 SUMMARY OUTPUT Regression Statistics Multiple R 0.938477378 R Square 0.880739789 Adjusted R Square 0.867488654 Standard Error 5.966106502 Observations 11 ANOVA df SS MS F Significance F Regression 1 2365.791977 2365.791977 66.46524 1.90269E-05 Residual 9 320.3498411 35.59442679 Total 10 2686.141818 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 80.6792263 2.919996322 27.62990682 5.17E-10 74.07373572 87.28471688 x -0.667545273 0.081881059 -8.152621419 1.9E-05 -0.852773096 -0.48231745

The p-value of slope is 0.000. Since p-value is less than 0.05 so we can conclude that correlation coefficient between the variables is significant.

The regression equation is

y' = 80.679 - 0.668*x

The predicted value for x = 43.3 is

y' = 80.679 - 0.668*43.3 = 51.7546