Question

For the result of Performance rating and Aptitude test of the following 5 employees answer the following questions:

1. Draw Scatter Plot
2. Coefficient of Correlation and its interpretation
3. Coefficient of Determination and Non-Determination and their interpretation
4. Regression line and interpretation of that. Where does the line crosses the y-axis
5. At Aptitude test result of 59 what is the residual?
6. What is your predicted Performance rating at Aptitude test result of 85!?
7. At 0.05 Significance level is the association between the two variables in the population greater than zero?

Let X be the Aptitude test

Answer 1

ANSWER::

a) Scatter plot:

b) Correlation Coefficient :

X Values
∑ = 307
Mean = 61.4
∑(X - Mx)2 = SSx = 489.2

Y Values
∑ = 14
Mean = 2.8
∑(Y - My)2 = SSy = 2.8

X and Y Combined
N = 5
∑(X - Mx)(Y - My) = -4.6

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

r = -4.6 / √((489.2)(2.8)) = -0.1243

Interpretation: A negative correlation, the relationship between these two variables is only weak ( The nearer the value is to zero, the weaker the relationship).

c)

The value of R^2, the coefficient of determination, is 0.0155 and the value of the coefficient of non determination, is 1-0.0155= 0.9845

d)

Sum of X = 307
Sum of Y = 14
Mean X = 61.4
Mean Y = 2.8
Sum of squares (SSX) = 489.2
Sum of products (SP) = -4.6

Regression Equation = ŷ = bX + a

b = SP/SSX = -4.6/489.2 = -0.0094

a = MY - bMX = 2.8 - (-0.01*61.4) = 3.37735

ŷ = -0.0094X + 3.37735

Interpretation: For every unit change in aptitude test performance rating decrease by 0.0094. The line crosses in downward direction.

e)

X=59

ŷ = -0.0094*59 + 3.37735

ŷ = 2.82

residual= y-ŷ = 3-2.82= 0.18

f)

X= 85

ŷ = -0.0094*85 + 3.37735

ŷ =2.578

ŷ =3 (APPROX)

g)

NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS Ha:

Level of significance=0.05

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

t= -0.1243*sqrt(3)/sqrt(0.9845)

t= -0.1243*1.73/0.9922

t= -0.2167

Degrees of freedom= 3

The p-value is .419996. The result is not significant because p > .05.

Decision: DO NOT REJECT NULL HYPOTHESIS H0.

Conclusion: We don't have sufficient evidence to show that the association between these two variables are greater than zero.

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