Question

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The following data is exa score and hours looking at social media in that week of the exa from a sample of 10 students.

Student	1	2	3	4	5	6	7	8	9	10
Exa Score	50	67	72	73	79	83	85	86	89	92
Hours	9.5	9.5	8.2	7.8	6.7	5.9	5.5	4.2	1.8	0.1

a)  (3 points) Using the IQR rule, are there any outliers for the exa scores? If so write down the number(s).  Give the interval that defines the outliers.

b)  (4 points) Determine the correlation coefficient, r, between exa score and hours studied. Interpret what this number means.

c)  (3 points) Determine the least-square regression equation to predict exa scores based on hours studied.  

d)  (3 points) If a student studies for 5 hours what is the predicted exa score?

e) (3 points) Give the residual for student 2.

f)  (4 points) Give the coefficient of determination, , for this least-squares equation. Give the interpretation of this number.

Answer 1

a)

quartile , Q1 = 0.25(n+1)th value=   2.75th value of sorted data=   70.75
      
Quartile , Q3 = 0.75(n+1)th value=   8.25th value of sorted data=   86.75
      
IQR = Q3-Q1 =    86.75-70.75=   16.0
      
1.5IQR =    24.00  
      
lower bound=Q1-1.5IQR=   46.75  
upper bound=Q3+1.5IQR=   110.75  

no outlier

(46.75 , 110.75)

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

b)

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	59.20	776.00	88.96	1420.40	-305.52
mean	5.92	77.60	SSxx	SSyy	SSxy

correlation coefficient ,    r = SSxy/√(SSx.SSy) =   -0.8595
...

c)

sample size ,   n =   10          
here, x̅ = Σx / n=   5.920   ,     ȳ = Σy/n =   77.600  
                  
SSxx =    Σ(x-x̅)² =    88.9560          
SSxy=   Σ(x-x̅)(y-ȳ) =   -305.5          
                  
estimated slope , ß1 = SSxy/SSxx =   -305.5   /   88.956   =   -3.4345
                  
intercept,   ß0 = y̅-ß1* x̄ =   97.93228          
                  
so, regression line is   Ŷ =   97.932   +   -3.435   *x

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

d)

Predicted Y at X=   5   is          
Ŷ =   97.9323   +   -3.4345   *5=   80.760
..........

e)

residual for student 2 =1.6955

..........

f)

R² =    (SSxy)²/(SSx.SSy) =    0.7387
73.87%   of variation in observation of variable Y, is explained by variable x  
.................

thnaks

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