Question

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.)

Answer 1

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

Answer: 51.8