# spss

One-way anova

Name

Institution affiliation
date

Recoding marital (marital status) variable.
The variable has been recorded successfully as follows; 1—> “Never Married”,2—> “Married”,3—-> “Living alone”
Running frequencies and data cleaning.
The reason for data cleaning is to take into consideration the present of outliers or extreme values. These values ought to be deleted to avoid invalidating the results and conclusion
After running frequencies, we noted that all values are valid.
Check the assumptions of ANOVA.
a. Normality assumption
We are instructed to use skewness of the dependent variable (total IPA score) to check this assumption.
Skewness = (mean-median)/SD
Statistics
total IPA score
N Valid 279
Missing 22
Mean 153.5197
Median 156.0000
Std. Deviation 29.86912
Skewness = (153.5197-156)/29.86912
Skewness=-0.083
According to the Hildebrand guideline:
A skewness less than 1 indicates minor skewness and so this is our conclusion
b. Assumption of homogeneity of variance
For this assumption to hold p-values must be greater than the level of significance (0.05)
Test of Homogeneity of Variances
total IPA score
Levene Statistic df1 df2 Sig.
11.080 2 264 .000

As you can see, dependent variables have their p-values less than 0.05. this implies that both variable have equal variances and therefore the assumption holds.
c. Assumption of independence.
For this assumption hold, dependent variables should be measured in ratio or interval scale.in our case, dependent variables are age and total IPA score. Independent variables should be categorical (gender).in all instances, this assumption hold.

Running one-way anova
The aim of the analysis was to determine whether person’s marital status differ significantly in their positive attitudes toward health outcomes.
We formulate both null and alternate hypothesis as follows;
Null hypothesis: person’s marital status does not differ significantly in their positive attitudes toward health outcomes.
Alternate hypothesis: person’s marital status differs significantly in their positive attitudes toward health outcomes.
ANOVA
total IPA score
Sum of Squares df Mean Square F Sig.
Between Groups 4283.593 2 2141.796 2.651 .072
Within Groups 213264.707 264 807.821
Total 217548.300 266

The results of the ANOVA are not significant (F (between group degree of freedom, within group degree of freedom) = 2.651, p < .05). Therefore, we reject the null hypothesis and therefore person’s marital status differs significantly in their positive attitudes toward health outcomes.
Post-hoc analysis is irrelevant in this case because ANOVA results indicate that three groups are not significantly different.