Question

Correlation and Simple Linear Regression Analysis

Determine the following using data about "hours studying per weekend" and "grade point average" below. a) scatter diagram with hours the independent variable b) coefficient of correlation c) coefficient of determination d) coefficient of nondetermination e) average error for predicting grades f) .01 level of significance test for slope being 0 g) regression equation h) expected grades for people who study 5 hours Hours Grades 3 3.0 2 2.0 6 3.8 3 2.6 4 3.2 8 3.7 2 2.1 3 2.8

Answer 1

a)

b)

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
3	3	0.77	0.01	-0.09
2	2	3.52	0.81	1.69
6	3.8	4.52	0.81	1.91
3	2.6	0.77	0.09	0.26
4	3.2	0.02	0.09	0.04
8	3.7	17.02	0.64	3.30
2	2.1	3.52	0.64000	1.5000
3	2.8	0.77	0.01000	0.088

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	31.00	23.20	30.88	3.10	8.7
mean	3.88	2.90	SSxx	SSyy	SSxy

correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.8893

c)

R² =    (Sxy)²/(Sx.Sy) =    0.7908

d)

coefficient of nondetermination = 1 - 0.7908 = 0.2092

e)

SSE=   (SSxx * SSyy - SS²xy)/SSxx =    0.6485
      
std error ,Se =    √(SSE/(n-2)) =    0.3288

f)

Ho:   ß1=   0          
H1:   ß1╪   0          
n=   8              
alpha =   0.01              
estimated std error of slope =Se(ß1) = Se/√Sxx =    0.329   /√   30.88   =   0.0592
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    0.2818   /   0.0592   =   4.7625
                  
t-critical value=    3.7074   [excel function: =T.INV.2T(α,df) ]          
Degree of freedom ,df = n-2=   6              
p-value =    0.003118              
decison :    p-value<α , reject Ho              
Conclusion:   Reject Ho and conclude that slope is significantly different from zero              

g)

sample size ,   n =   8          
here, x̅ = Σx / n=   3.875   ,     ȳ = Σy/n =   2.900  
                  
SSxx =    Σ(x-x̅)² =    30.8750          
SSxy=   Σ(x-x̅)(y-ȳ) =   8.7          
                  
estimated slope , ß1 = SSxy/SSxx =   8.7   /   30.875   =   0.28178
                  
intercept,   ß0 = y̅-ß1* x̄ =   1.80810          
                  
so, regression line is   Ŷ =   1.808   +   0.282   *x

h)

Predicted Y at X=   5   is          
Ŷ =   1.8081   +   0.2818   *5=   3.22