SPSS Bivariate and Multivariate Regression Analysis for Desklib

Statistics
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SPSS Bivariate Regression Assignment
1. As with all data, it is important to take a look at the data to examine potential
univariate and bivariate outliers. Use procedures described in earlier assignments
to examine outliers and to get to know your data (if you see potential outliers you
can transform the data, delete the outliers, or leave them in the data; justify your
reasoning). For this assignment focus on only two variables: timedrs, stress.
Supply a box plot and/or histogram of the two variables.
Answer
From table 1 below, it can be seen that the average number of visits to health
professionals was 7.80 with a standard deviation of 10.95. The skewness value is
3.25 implying the variable is highly skewed.
Table 1: Statistics
Visits to health
professionals
Stressful life events
N Valid 465 465
Missing 1 1
Mean 7.90 204.22
Median 4.00 178.00
Mode 2.00 0.00
Std. Deviation 10.95 135.79
Variance 119.87 18439.66
Skewness 3.25 1.04
Std. Error of Skewness 0.11 0.11
Kurtosis 13.10 1.80
Std. Error of Kurtosis 0.23 0.23
Range 81.00 920.00
Minimum 0.00 0.00
Maximum 81.00 920.00
Percentiles 25 2.00 98.00
50 4.00 178.00
75 10.00 278.00

Histogram and Boxplot for the stressful life events

Figure 2: Boxplot for visits to the health professional
As can be seen from the histogram, it is evident that the variable (timedrs) is skewed. The
data is skewed to the right (longer tail to the right). The boxplot clearly shows that there
are several outliers in the dataset.
Histogram and Boxplot for the number of
visits to the health professionals
Figure 1: Histogram for visits to health professional

Figure 4: Boxplot for stressful life events
Again, just like for the case of the number of visits to the health professional, it is evident
that the variable (stress) is skewed. The data is skewed to the right (longer tail to the
right). The boxplot also clearly shows that there are several outliers in the dataset.
2. Run a correlation between stress and timedrs (ANALYZE-CORRELATE-
BIVARIATE). What is the correlation between these two variables?
The correlation can run between -1 and +1. The closer to the extremes (-1 and
+1) the stronger the correlation. A correlation of 0 means that there is no
relationship. How strong is this relationship?
Answer
Table 2: Correlations
Visits to health
professionals
Stressful
life events
Visits to health
professionals
Pearson
Correlation
1 .287**
Sig. (2-tailed) .000
N 465 465
Stressful life events Pearson
Correlation
.287** 1

Sig. (2-tailed) .000
N 465 465
**. Correlation is significant at the 0.01 level (2-tailed).
Results shows that the there is a weak positive relationship between stress and
timedrs (r = 0.287, p = 0.000).
3. Run a bivariate regression (ANALYZE-REGRESSION-LINEAR) with stress as a
INDEPENDENT variable and timedrs as an DEPENDENT variable. That is, you
are trying to predict the number of visits a person takes to the doctor with how
much stress they have in their life.
a. The second table in the output gives you a few important values, R and R-Square.
What are these values and what do they mean?
Answer
Table 3: Model Summary
Model R R Square Adjusted R
Square
Std. Error of
the Estimate
1 .287a .082 .080 10.501
a. Predictors: (Constant), Stressful life events
The value of R is 0.287; this means that a weak positive correlation exists
between the variables.
The value of R-Square is 0.082; this implies that 8.2% of the variation in the
dependent variable (timedrs) is explained by the one independent variable (stress)
in the model.
b. The third table is also important. It is labeled ANOVA (hey, wait, I thought we
were doing regression). This table tells you whether your regression equation is

significant (thus does the regression line significantly predict the DV). Is your
regression line significant?
Answer
Table 4: ANOVA
Model Sum of
Squares
df Mean
Square
F Sig.
1 Regressi
on
4568.402 1 4568.402 41.432 .000b
Residual 51051.047 463 110.261
Total 55619.449 464
a. Dependent Variable: Visits to health professionals
b. Predictors: (Constant), Stressful life events
Looking at the ANOVA table, we can see that the p-value for the F-Statistics is
0.000 (a value less than 5% level of significance), we therefore reject the null
hypothesis and conclude that the regression line significant at 5% level of
significance.
c. The Coefficients table gives you the values you need for your regression equation.
The ‘Constant’ variable is your intercept or a. The Stressful Life Events is you
slope or b coefficient. Each one of those values has a significance value telling
you whether each value is significant (note: the above ANOVA table tells you
whether the whole equation is significant, these values are for individual parts of
the equation). Please give me the regression equation for this data (Y = a + bX)
and explain what that equation means.
Answer
Table 5: Regression coefficients
Model Unstandardized
Coefficients
Standardiz
ed
t Sig.

Coefficient
s
B Std. Error Beta
1 (Constant) 3.182 .880 3.616 .000
Stressful life
events
.023 .004 .287 6.437 .000
a. Dependent Variable: Visits to health professionals
The regression equation is given as follows;
Timedrs=3.182+0.023 (Stress)
The constant variable (intercept) is given as 3.182; this means that holding stress
constant (zero value for stress0 we would expect the average number of visits to
the health professional to be 3.182.
The slope coefficient is 0.023; this implies that a unit increase in the stress would
result to an increase in the number of visits to the professional (timedrs).
Similarly, a unit decrease in the stress would result to a decrease in the number of
visits to the professional (timedrs).
d. Write up a simple results section for this analysis. Feel free to use a bivariate
scatterplot as a figure. Make sure you describe the regression equation and R-
square in your results.
Answer
This part sought to find out how best stress predicts timedrs.
In regard to the descriptive statistics, it was found that the average number of
visits to health professionals was M = 7.90 with SD = 10.95, N = 365. Also, the
average number of stressful events was M = 204.22, SD = 135.79, N = 465.