The whole point of organizing data using SPSS software is so that we can manipulate the data to answer our descriptive and inferential questions.  This next section will walk you through how to run specific tests within SPSS.  Note the level of measurement for variables that are used for specific tests.

1.      Let's start with a descriptive question:

• For the variable #45 age for patients, how much does the sample vary and what is its most appropriate measure of central tendency?
• Age is at least interval level, so you can use SPSS to calculate mean (central tendency) and range and standard deviation (variability) with a few clicks of the cursor.
• Go to Analyze at the top of the sheet: Descriptive statistics: Descriptives, and click on patient's age in the left hand window, then click on the arrow to the right to move the variable into the variable window on the right, then click OK.
• This will produce an output box that tells you there were n = 40 participants.
• The most appropriate measure of central tendency for a variable that is at least interval level is usually the mean, and mean age was 59.06 years.  Variability is indicated by the age range which was from 42.48-84.33 years and, in particular, the standard deviation which was 9.75.  (Beats having to learn the formulas and do the calculations by hand!)

2.      Let's say you wanted to know if there is a difference in marital status between male and female partners and if the difference is statistically significant.  Choose variables #59 fsex and #60 fmarst:

• First, determine the level of your variables: the variable 'partner gender' is nominal with 2 levels (male and female); the variable 'partner marital status' is nominal with 5 levels (married, separated, single, widowed, and divorced; though only 4 groups have subjects).

• Go to Analyze at the top of the sheet, scroll down to Descriptive statistics: Crosstabs; because both variables are nominal and partner marital status has 4 groups, a Chi-Square test is appropriate; click on partner's sex in the left hand window, then click on the arrow to the right to move the variable into the variable window 'Row' on the right; click on partner's marital status in the left hand window, then click on the arrow to the right to move the variable into the variable window 'Column' on the right; click on Statistics at the right of the box and check Chi-Square; click continue and then OK.

• This will produce an output box that tells you there were 33 female and 7 male partners; 31 women and 6 men were married, 1 female was separated; 1 male was single; and 1 female was widowed; the Pearson Chi-Square statistic = 5.181 and the p-value = .159 ('Asymp. Sig. 2-tailed')
• The null hypothesis is supported. There is no statistically significant difference in marital status between male and female partners; Chi-Square = 5.181, p = .159, so p > alpha or p > .05. OR the alternate/research hypothesis that there is a relationship between marital status and gender is not supported (probably because almost all of the partners are married to the "patients" in the study given patients and their partners were both subjects).

3.      Let's say you wanted to compare demographic data between patients in the control and experimental groups to see if they are different.  Choose variable #2 group and another variable such as variable #49 educyr (years of formal education):

• First, determine the level of your variables: the IV is 'patient group' and it is nominal with 2 levels (control and experimental); the DV is patient years of formal education which is at least interval level, so you can calculate a mean score.
• Go to Analyze, scroll down to compare means; because there are 2 independent groups (experimental and control) at the nominal level and the DV is at the interval level, you could use an independent samples t-test (a 1-way ANOVA could also work, but t-test is more common); click on the variable you have chosen from the left window to move it to the test variable window on the right (in this case you would choose years of formal education) by clicking on the upper arrow (and you will then see years of formal education in the upper box on right.  Then click study group (group) and the lower arrow so group (??) appears in the lower box on right.  Under group click on define groups and beside group 1 type in 1 and beside group 2 type in 2; click continue, and then click OK.
• An output box will appear that will tell you that there were 23 participants in the intervention group and 17 in the control group.
• At 13.59, the mean education score for patients in the control group was higher compared to the intervention group whose mean was 11.  Looking at the other output data: which group had more variability in education? _________________________* (answer below)
• You also would want to know if the means for the two groups were statistically significant.  As you can see, the significance ('Sig.') value for the Levene test is .089, so > .05 therefore can assume equal variances.  The equal variances t-test statistic is -2.306 with p = .027 ('Sig. 2-tailed'). Given the standard of p = .05, are the results statistically significant?
• The null hypothesis is refuted (or the statistical/research hypothesis is supported).  There is a statistically significantly difference in years of formal education between patients in the control group and those in the intervention group with intervention group patients having fewer years of education; t = -2.306, p = .027, so p < alpha or p < .05.

*The control group had more variability because the standard deviation was 4.3 compared to 2.8 for the experimental group.

4.      Let's say you wanted to do some testing of relationships.  For example, you might want to see if there is a relationship between individual quality of life and uncertainty ratings for patients at the end of the study.  Choose variables #22 pqoli5 and #12 MUIS5:

• First, determine the level of your variables: quality of life is an interval level variable, and so is uncertainty.
• Because both variables are at the interval level a Pearson's r correlation is an appropriate test.  Click on Analyze, scroll down to Correlate and then Bivariate.
• This takes you to a two-side window box where you can click on QOL Total - Patient end highlight it and then click the right arrow key to move the variable to the right side window; then click on the MUIS Patient Time 5 variable, click on the arrow and also move this variable over to the right side window.
• Make sure that there are check marks for Pearson, two tailed, and flag significant correlations then click OK.
• An Output viewer will come up that shows you the results of the analysis.  You can see that at the end of 8 weeks the correlation between patients' quality of life and uncertainty was r = -.431.  The p-value was reported as .006, the ** means p < .01.
• The null hypothesis is refuted.  There is a statistically significant moderate, negative correlation between patients' quality of life and uncertainty at the end of the study, so patients with higher quality of life have less uncertainty; r = -.431, p = .006, so p < alpha or p < .05 (in this case, there's even stronger evidence as p < .01).

5.      Let's say you wanted to compare data between partners with various marital status to see if they are different.  Choose variable #60 fmarst and another variable such as #25 fbai1 (anxiety at time 1):

• First, determine the level of your variables: 'partner marital status' is nominal with 5 levels (married, separated, single, widowed, and divorced; though only 4 actually have subjects in them); 'anxiety' is interval.

• More than 2 groups, so cannot use a t-test; there is a single IV (marital status) and a single DV (anxiety), so test of choice is a 1-way ANOVA.
• Go to Analyze, scroll down to compare means; then One-Way ANOVA; click on the variable you have chosen from the left window to move it to the Dependent List window on the right (in this case you would choose fbai1) by clicking on the upper arrow (and you will then see anxiety in the upper box on right. Then click partner's marital status and the lower arrow so the variable moves to the lower box on right (Factor). Under options click on Descriptive; click continue, and then click OK.
• An output box will appear that tells you the mean scores on anxiety at baseline for partners who are married (7.1351); separated (15.00); single (0.00); and widowed (7.00). The F-statistic for the ANOVA is .701 and the p-value = .558.
• The null hypothesis is supported. For partners, there is no statistically significant difference in baseline anxiety scores across marital status; F = .701, p = .558, so p > alpha or p > .05.