Question

A psychologist conducted an experiment analysing the relationship between student scores in an exam and the amount of attention they paid in class. The latter was measured using a type of brain monitor. The Psychologist believed that scores would increase by 1 for every two unit increase in attention. The data are listed in the excel spreadsheet.

Estimate a linear regression between the score (Y) and the measure of attention(X).

(a) Write out the equation for Y in the form , but with coefficients. Show the estimated standard errors in parenthesis below the coefficients. What is the R2 of the regression? Calculate a 99 percent confidence interval for β.                                                               [5 pts]

(b) What are the mean and the estimated standard deviation of the estimated residuals? [2 pts]

Hint: the first answer is definitional and the second answer is easily seen from the output.

(c )Test the hypothesis that there is no relationship between the variables at the 90 percent significance level.                                                                                                    [3 pts]

(d) Test the hypothesis that the coefficient β=0.5 at the 99% significance level.               [3 pts]

(e) The Psychologist concluded from the experiment that test scores increase significantly if students pay attention in class. In one word, how would you describe the results of this experiment based on the data you have?                                                                                                       [2 pts]

DATA:

Regression data for Psychology Experiment
Attention	Score
18	80
35	90
86	80
22	50
72	76
102	74
86	75
30	80
35	85
94	82
16	80
42	41
50	50
96	96
60	80
106	70
80	65
14	14
11	14
80	85
12	14
37	43
26	80
86	70
5	20
17	20
35	80
76	68
50	70
15	16
90	86
96	80
7	16
10	14
35	65
88	88
20	32
22	70
50	65
22	62
35	50
64	92
68	84
13	15
102	102
86	85
18	24
78	64
98	78
70	80
60	70
98	98
9	14
50	90
104	72
35	45
60	60
74	72
88	88
80	95
22	58
8	14
86	110
60	75
92	84
60	100
80	75
86	95
16	18
86	90
35	75
35	60
80	60
80	70
104	104
80	100
60	90
86	100
62	96
60	65
39	41
50	80
50	75
6	18
60	95
22	54
21	40
100	100
94	94
80	90
48	41
106	106
50	43
46	41
90	90
60	85
92	92
22	80
35	70
66	88
80	60
50	60
80	80
100	76
50	45
86	65
19	28
50	85
22	75
86	105

Answer 1

a)

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	6262	7445	100985.4182	74369.9	63083.45
mean	56.93	67.68	SSxx	SSyy	SSxy

sample size ,   n =   110          
here, x̅ = Σx / n=   56.93   ,     ȳ = Σy/n =   67.68  
                  
SSxx =    Σ(x-x̅)² =    100985.4182          
SSxy=   Σ(x-x̅)(y-ȳ) =   63083.5          
                  
estimated slope , ß1 = SSxy/SSxx =   63083.5   /   100985.418   =   0.6247
                  
intercept,   ß0 = y̅-ß1* x̄ =   32.1206          
                  
so, regression line is   Ŷ =   32.1206   +   0.6247   *x
(3.65) (0.057)
SSE=   (SSxx * SSyy - SS²xy)/SSxx =    34962.964          
                  
std error ,Se =    √(SSE/(n-2)) =    17.993          
                  
correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.7279          
                  
R² =    (Sxy)²/(Sx.Sy) =    0.5299   

α=   0.01              
t critical value=   t α/2 =    1.982   [excel function: =t.inv.2t(α/2,df) ]      
estimated std error of slope = Se/√Sxx =    17.99253   /√   100985.42   =   0.057
                  
margin of error ,E= t*std error =    1.982   *   0.057   =   0.112
estimated slope , ß^ =    0.6247              
                  
                  
lower confidence limit = estimated slope - margin of error =   0.6247   -   0.112   =   0.5124
upper confidence limit=estimated slope + margin of error =   0.6247   +   0.112   =   0.7369

B)

ANOVA
	df	SS	MS	F	Significance F
Regression	1	39406.9	39406.9	121.7272	2.08E-19
Residual	108	34962.96	323.7311
Total	109	74369.86

c)

slope hypothesis test               tail=   2
Ho:   ß1=   0          
H1:   ß1╪   0          
n=   110              
alpha =   0.1              
estimated std error of slope =Se(ß1) = Se/√Sxx =    17.993   /√   100985.42   =   0.0566
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    0.6247   /   0.0566   =   11.0330
                  
t-critical value=    1.6591   [excel function: =T.INV.2T(α,df) ]          
Degree of freedom ,df = n-2=   108              
p-value =    0.0000              
decison :    p-value<α , reject Ho              
Conclusion:   Reject Ho and conclude that slope is significantly different from zero              

e)

Relationship is moderate positive relatioship. It means although score are increasing but not at the rate expected.

for each 0.6247 increase in attention, score is increaseed by 1.

Please revert back in case of any doubt.

Please upvote. Thanks in advance.