Question

1. A psychologist is interested in the relationship between intelligence and academic motivation. She recruits eight participants for a study. Each participant completes an intelligence test and a measure of academic motivation. Scores for the eight participants are in the table below. Use intelligence scores to predict motivation scores with regression analysis by hand.

Participant	Intelligence Score	Motivation Score
1	61	100
2	56	90
3	56	117
4	29	66
5	43	100
6	41	92
7	45	86
8	31	78
Mean	45.25	91.13
S.D.	11.77	15.39

1. Calculate the slope coefficient.

2. Calculate the intercept.

3. Write out the prediction equation and write one sentence interpreting the slope.

4. Calculate the standard error of the estimate.

5. Use the regression equation to predict a motivation score and then calculate its 95% confidence interval when intelligence is 50. Is this an appropriate prediction to make? Briefly explain why or why not.

6. Use the regression equation to predict a motivation score and then calculate its 95% confidence interval when intelligence is 65. Is this an appropriate prediction to make? Briefly explain why or why not.

7. Does the regression model significantly predict motivation scores? (H₀: β = 0; H₁: β ≠ 0, α = .05) Provide the test statistic, appropriate degrees of freedom, and whether p is greater than or less than α. (Note: there are two ways that you can use to test model significance.)

Answer 1

X	Y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
61	100	248.06	78.8	139.8
56	90	115.56	1.3	-12.1
56	117	115.56	669.5	278.2
29	66	264.06	631.3	408.3
43	100	5.06	78.8	-20.0
41	92	18.06	0.8	-3.7
45	86	0.06	26.3	1.3
31	78	203.06	172.3	187.0

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	362	729	969.5	1658.9	978.75
mean	45.250	91.13	SSxx	SSyy	SSxy

sample size ,   n =   8          
here, x̅ =   45.250   ,   ȳ =   91.1250  
                  
SSxx =    Σ(x-x̅)² =    969.50          
SSxy=   Σ(x-x̅)(y-ȳ) =   978.8          
                  
a)

slope ,    ß1 = SSxy/SSxx =   1.0095

b)   
                  
intercept,   ß0 = y̅-ß1* x̄ =   45.4433     

c)

so, regression line is   Ŷ =   45.4433   +   1.00954   *x
for every unit increase in intelligence score, motivation score get increase by 1.00954

d)

SSE=   (Sx*Sy - S²xy)/Sx =    670.79
      
std error ,Se =    √(SSE/(n-2)) =    10.5734

e)

X Value=   50
Confidence Level=   95%
  
  
Sample Size , n=   8
Degrees of Freedom,df=n-2 =   6
critical t Value=tα/2 =   2.447

margin of error,E=t*Std error=t* S(ŷ) =t*Se*√ (1/n+(X-X̅)²/Sxx) =  9.9624

Confidence Lower Limit=Ŷ -E =   85.958
Confidence Upper Limit=Ŷ +E =   105.883

f)

X Value=   65
Confidence Level=   95%
  
  
Sample Size , n=   8
Degrees of Freedom,df=n-2 =   6
critical t Value=tα/2 =   2.447

margin of error,E=t*Std error=t* S(ŷ) = t*Se*√ (1/n+(X-X̅)²/Sxx) =   18.7879

Confidence Lower Limit=Ŷ -E =   92.2756
Confidence Upper Limit=Ŷ +E =   129.8513

g)

slope hypothesis test      
Ho:   ß1=   0
H1:   ß1╪   0

n=   8              
alpha=   0.05              
estimated std error of slope =Se(ß1) =                s/√Sxx =    0.3396
                  
t stat =    ß1 /Se(ß1) =        2.973
                  
  
p-value =    0.0249              
decision :    p-value<α , reject Ho , so slope is significant

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second way is

Anova table
variation	SS	df	MS	F-stat	p-value
regression	988.088	1	988	8.838	0.0249
error,	670.787	6	111.798
total	1658.875	7

F-stat = 8.838

p-value = 0.0249

p-value<α , reject Ho , so model is significant at α=0.05