Answer To: The submission should adhere to APA conventions including spacing and the formatting of tables. The...
Pooja answered on Feb 24 2021
1)
The correlation matrix to know the strength of direction of linear relationship between divorce, education, and well-being is given below.
divorce
education
well_being
divorce
1
education
0.649979
1
well_being
0.153103
-0.14929
1
With correlation coefficient r = 0.65, I can say that there is a moderate positive linear relationship between education and divorce. That is as the value of education increases, the value of divorce also increases moderately.
With correlation coefficient r = 0.15, I can say that there is a very weak positive linear relationship or almost no linear relationship between well-being and divorce. That is as the value of well-being increases, the value of divorce is either increases very slightly or not affected at all.
With correlation coefficient r = -0.15, I can say that there is a very weak negative linear relationship or almost no linear relationship between well-being and education. That is as the value of well-being increases, the value of education is either decreases very slightly or not affected at all.
2)
The dependent variable is well-being. The independent variables are Divorce, Education, Social support and Freedom. The regression analysis is applied to predict well-being on the basis of Divorce, Education, Social support and Freedom. The regression output is given below.
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.755779
R Square
0.571202
Adjusted R Square
0.544814
Standard Error
0.728178
Observations
70
ANOVA
df
SS
MS
F
Significance F
Regression
4
45.91187
11.47797
21.6466
2.19E-11
Residual
65
34.46582
0.530243
Total
69
80.37769
Coefficients
Standard Error
t Stat
P-value
Lower 95%
Upper 95%
Lower 95.0%
Upper 95.0%
Intercept
1.815897
0.466008
3.896705
0.000233
0.885214
2.74658
0.885214
2.74658
divorce
0.016121
0.012586
1.280813
0.204811
-0.00902
0.041257
-0.00902
0.041257
education
-0.1
0.043797
-2.28333
0.025688
-0.18747
-0.01253
-0.18747
-0.01253
social_support
0.039589
0.007931
4.991855
4.75E-06
0.02375
0.055428
0.02375
0.055428
Freedom
0.018544
0.009907
1.87174
0.065743
-0.00124
0.03833
-0.00124
0.03833
a)
Consider the null hypothesis, ho: model is not significant. This is tested against an alternative hypothesis, h1: model is significant. With F(4, 65) = 21.6466, P<5%, the null hypothesis is rejected at 5% level of significance and conclude that model is significant. Thus, there is sufficient evidence to support the claim that model is significant.
The coefficient of determination, R2 is 57%. There is 57% variation in the well-being which is explained by Divorce, Education, Social support and Freedom. This percentage Is less and fitted model is said to be a moderate fit to the data.
b)
Null hypothesis, ho1: the coefficient of Divorce Is not significant. Beta1 = 0. Alternative hypothesis, h11: the coefficient of Divorce Is significant. Beta1 =/= 0. With t = 1.28, p>5%, I fail to reject the null hypothesis and conclude that the coefficient of Divorce Is not significant. Beta1 = 0.
Null hypothesis, ho1: the coefficient of Education Is not significant. Beta2 = 0. Alternative hypothesis, h11: the coefficient of Education Is significant. Beta2 =/= 0. With t = -2.28, p<5%, I reject the null hypothesis and conclude that the coefficient of education Is significant. Beta2 =/= 0.
Null hypothesis, ho1: the coefficient of social support Is not significant. Beta3 = 0. Alternative hypothesis, h11: the coefficient of social support Is significant. Beta3 =/= 0. With t = 4.99, p<5%, I reject the null hypothesis and conclude that the coefficient of social support Is significant. Beta3 =/= 0.
Null hypothesis, ho1: the coefficient of Freedom Is not significant. Beta4 = 0. Alternative hypothesis, h11: the coefficient of Freedom Is significant. Beta4 =/= 0. With t = 1.87, p>5%, I fail to reject the null hypothesis and conclude that the coefficient of Freedom Is not significant. Beta4 = 0.
c)
With 1 unit increase in divorce, the well-being is increased by 0.016121 units only. But this value is not significant at 5% level of significance.
With 1 unit increase in education, the well-being is decreased by 0.1 units only. But this value is significant at 5% level of significance.
With 1 unit increase in social support, the well-being is increased by 0.039589 units only. But this value is significant at 5% level of significance.
With 1 unit increase in Freedom, the well-being is increased by 0.018544 units only. But this value is not significant at 5% level of significance.
d)
Assumption 1: normality of residuals
Since the normal probability plot is S shaped, I can say that the assumption of normality of residuals is followed.
Assumption 2: homogeneity of variance of residuals
All points in the residual plot are randomly distributed. Hence the variance is constant. Thus the assumption of homogeneity of error variances is also satisfied.
e)
The limitations of the regression analysis lie in its assumptions. Regression model should not have multi-co-linearity. multi-co-linearity is a problem where independent variables are correlated to each other. There should be no auto-correlation, that is error terms in prediction must not be correlated with each other. The variance of error terms must be constant, that there should not be hetero-scedasticity. Residuals should be normally distributed.
References:
Gunst, R. F. (2018). Regression analysis and its application: a data-oriented approach. Routledge.
Schroeder, L. D., Sjoquist, D. L., & Stephan, P. E. (2016). Understanding regression analysis: An introductory guide (Vol. 57). Sage Publications.
Brook, R. J. (2018). Applied...