Question

3. The following data represents data collected by a social researcher to examine the relationship

between “HAPPINESS” and “INCOME PER YEAR for workers 30 to 40 years old living on Long

Island, NY. Each worker completed a personality inventory scale that yields an HAPPINESS LEVEL

Scale Value(1 is LOW and 5 HIGH) and reported their respective yearly INCOME.

HAPPINESS

LEVEL

3

1

3

4

3

2

5

1

3

4

2

4

5

3

4

5

4

4

5

2

YEARLY

INCOME (1000$)

46 24 36 74 75 19 137 23 54 85 25 55 88 56 21 91 99 105 79 82

Using the data:

a. Plot a scatter diagram of YEARLY INCOME (x) against HAPPINESS LEVEL (y). (4pts)

b. Compute a Pearson correlation coefficient for part a. (2pts)

c. Would you consider the correlation coefficient weak, moderately weak, moderately strong

or strong? Why? (2pts)

d. Based on the data, do you believe an HAPPINESS LEVEL is related to YEARLY INCOME ?

Explain using your answer to part c (4pt)

e. Calculate the coefficient of determination for the correlation coefficient computed in part b.

Explain the meaning of the coefficient of determination in the context of the variables. Be

sure to identify the independent variable and dependent variable(4pts)

f. Determine the LINEAR regression Equation for the variables in this problem. Be sure to be

consistent in your use of the independent and dependent variables (4pts)

g. Using the regression formula from part f predict HAPPINESS LEVEL for a YEARLY INCOME

value of $150,000. (2pts)

h. Is the prediction value in part g an example of Interpolation or Extrapolation? Explain.

(2pts)

Answer 1

X	Y	XY	X²	Y²
46	3	138	2116	9
24	1	24	576	1
36	3	108	1296	9
74	4	296	5476	16
75	3	225	5625	9
19	2	38	361	4
137	5	685	18769	25
23	1	23	529	1
54	3	162	2916	9
85	4	340	7225	16
25	2	50	625	4
55	4	220	3025	16
88	5	440	7744	25
56	3	168	3136	9
21	4	84	441	16
91	5	455	8281	25
99	4	396	9801	16
105	4	420	11025	16
79	5	395	6241	25
82	2	164	6724	4

Ʃx =	1274
Ʃy =	67
Ʃxy =	4831
Ʃx² =	101932
Ʃy² =	255
Sample size, n =	20
x̅ = Ʃx/n = 1274/20 =	63.7
y̅ = Ʃy/n = 67/20 =	3.35
SSxx = Ʃx² - (Ʃx)²/n = 101932 - (1274)²/20 =	20778.2
SSyy = Ʃy² - (Ʃy)²/n = 255 - (67)²/20 =	30.55
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 4831 - (1274)(67)/20 =	563.1

a) Scatter pot:

b) Correlation coefficient, r = SSxy/√(SSxx*SSyy) = 563.1/√(20778.2*30.55) = 0.7068

c) the correlation coefficient moderately strong.

d)

Null and alternative hypothesis:

Ho: ρ = 0 ; Ha: ρ ≠ 0

Test statistic :  

t = r*√(n-2)/√(1-r²) = 0.7068 *√(20 - 2)/√(1 - 0.7068²) = 4.2386

df = n-2 = 18

p-value = T.DIST.2T(ABS(4.2386), 18) = 0.0005

Conclusion:

p-value < α Reject the null hypothesis. There is a correlation between x and y.

e)

Coefficient of determination, r² = (SSxy)²/(SSxx*SSyy) = (563.1)²/(20778.2*30.55) = 0.4995

49.95% variation in y is explained by the least squares model.

f)

Slope, b = SSxy/SSxx = 563.1/20778.2 = 0.0271005

y-intercept, a = y̅ -b* x̅ = 3.35 - (0.0271)*63.7 = 1.623697

Regression equation :

ŷ = 1.6237 + (0.0271) x

g)

Predicted value of y at x = 150

ŷ = 1.6237 + (0.0271) * 150 = 5.6888

h) Extrapolation