Question

A researcher wants to investigate the relationship between self-regulation and GPA. She randomly selects 200 undergraduate students and asks them to complete a scale measuring self-regulation (Self-control Schedule - SCS) and obtains their GPA from university administrative records (with their consent). The mean GPA is 2.9 with a standard deviation of 1.15. The mean for SCS is 22 with a standard deviation of 24. The correlation between the SCS and GPA is 0.40.

r = 0.40   

MEAN Self-regulation (SCS) = 22

SD = 24

MEAN GPA = 2.9

SD = 1.15

Is the correlation significant? Meaningfully describe the finding

What is the coefficient of determination? Explain the meaning of this value.

What is the 95% CI for r? Interpret this confidence interval

If one student has a score of 36 on the SCS, what do you predict their GPA will be?

What is the standard error of the estimate of the above prediction? What does this value represent?

Answer 1

a)

correlation hypothesis test                          
Ho:   ρ = 0                      
Ha:   ρ ╪ 0                      
                          
n=   200                      
alpha,α =    0.05                      
correlation , r =   0.4000                      
                          
Df =    n-2 =   198                  
t-test statistic = r*√(n-2)/√(1-r²) =        0.4000   * √    198   / √ ( 1 -    0.4000   ² ) =
                          
p-value =    0.0000   [excel function: =t.dist.2t(t-stat,df) ]                   
decision:   p value < α , so, reject the null hypothesis                      
correlation is significant.

so there is relationship between two

............

b)

coefficient of determination= (0.4^2)

= 0.16

16% of variation is explained by X

...............

c)

Confidence interval for correlation coefficient-                              
Transformed r into z = 1/2*ln((1+r)/(1-r)) =   0.4236                           -1.649898073
the std error = 1/√(N-3)=   1/√(200- 3 ) =    0.07124705                      
                              
Z critical value=   1.9600   [excel function: "=NORMSINV(α/2)" ]                      
                              
95%   confidence interval = transformed r ± critical value*std error = 0.4236±1.96*0.0712                          
=   (   0.2840   ,   0.5633   )          
                              
now, need to transform these limits(t) back to correlation coefficient scale ( r) by taking (e^2t - 1)/(e^2t + 1)                              
So, lower limit =    0.2766                          
Upper limit =   0.5104          

r will lie betwen these limits

..............

              

d)

slope =0.4*1.15/24

=0.0192

intercept =ß0 = y̅-ß1* x̄

=2.9-(0.0192*22)

=2.4776
Y =2.4776+0.0192*X

x= 36

Y=2.4776+0.0192*36

=3.1688

................

e)

S(errors) = (SQRT(1 minus R-squared)) x STDEV of y

=0.9165*1.15

=1.054

..................

THANKS

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