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

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

x y
61.2 32.1
42.8 12.4
30.8 20.7
47.4 28.4
46.7 44.4
64.7 38.6
36.6 25.4
52.1 13.2
54.9 46.2
21.1 16
41.2 43.9
56.4 26.3
50.6 29.9
29.1 35.9

Verify that the correlation is significant at an α=0.05α=0.05. If the correlation is indeed significant, predict what value (on average) for the explanatory variable will give you a value of 15.1 on the response variable.

What is the predicted explanatory value?

math   Statistics-And-Probability   
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Answer #1

(Y - My)2 X - MX 15.800 -2.600 - 14.600 2.000 1.300 19.300 -8.800 6.700 9.500 -24.300 -4.200 11.000 5.200 -16.300 Y - My 2.57

X Values
∑ = 635.6
Mean = 45.4
∑(X - Mx)2 = SSx = 2082.18

Y Values
∑ = 413.4
Mean = 29.529
∑(Y - My)2 = SSy = 1684.989

X and Y Combined
N = 14
∑(X - Mx)(Y - My) = 622.47

R Calculation
r = ∑((X - My)(Y - Mx)) / √((SSx)(SSy))

r = 622.47 / √((2082.18)(1684.989)) = 0.3323

NULL HYPOTHESIS H0: \rho=0

ALTERNATIVE HYPOTHESIS Ha: \rho\ne0

LEVEL OF SIGNIFICANCE= 0.05

t= r*sqrt(n-2)/sqrt(1-r^2)

t= 0.3323*sqrt(14-2)/sqrt(1-0.1104)

t= 0.3323*sqrt(12)/sqrt(0.8896)

t=0.3323*3.46/0.9432

t= 1.219

degrees of freedom= 14-2=12

The p-value is .246265.The result is not significant at p > .05.

Sum of X = 635.6
Sum of Y = 413.4
Mean X = 45.4
Mean Y = 29.5286
Sum of squares (SSX) = 2082.18
Sum of products (SP) = 622.47

Regression Equation = ŷ = bX + a

b = SP/SSX = 622.47/2082.18 = 0.29895

a = MY - bMX = 29.53 - (0.3*45.4) = 15.95619

ŷ = 0.29895X + 15.95619

X=15.1

ŷ = 0.29895X + 15.95619*15.1

ŷ =241.237

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