# SPSS Tutorial – t tests in SPSS by ozhan

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```									SPSS Tutorial – t tests in SPSS
January 11, 2001

To demonstrate SPSS’s one sample t-test command, we’ll use the anagram data that were presented last week. Remember that the dependent variable is the amount of money allocated to the “least helpful” partner in each of 10 5person groups (the person who had the P10 anagrams). To review, our null and alternative hypotheses were: Ho: µ = 2 (they allocate the \$10 equally, so each person gets \$2) H1: µ ≠ 2 The data to the left are from the file anagram.sav (available on the class website on the “data sets” link). We want to subject these data to a one sample t-test of the null hypothesis above. Use a two-tailed test, α = .05. Click here for the data set (you must save it to disk; save it in the c:/temp directory).

Once you have the data, it is fairly simple to compute a one-sample t-test in SPSS. Choose, from the menu, ANALYZE / COMPARE MEANS / ONE SAMPLE T TEST You will get the dialog box that appears below:

Your test variable is amount; click this over into the test variable column. What is your test value? It is the value of the mean under the null, in this case, 2. Enter 2 into the test value box. You may want to check out the “options” box to see what is available to you. The box is below.

Note that will automatically get the SPSS default 95% confidence interval for the data. The missing values command tells SPSS how to deal with missing data; the SPSS default is fine for most purposes (this will become more critical when you are dealing with more variables and have missing data in your set). Click OK once you have set up your dialog boxes correctly, and SPSS will run the test.

The output that you will receive is below:
One-Sample Statistics Std. Error Mean .3004

N AMOUNT 10

Mean 1.1770

Std. Deviation .9499

One-Sample Test Test Value = 2 95% Confidence Interval of the Difference Lower Upper -1.5026 -.1434

AMOUNT

t -2.740

df 9

Sig. (2-tailed) .023

Mean Difference -.8230

Take a moment to see what you have here. Of course, your critical information is the t value (-2.74), your degrees of freedom (9), and the significance of your test (.023; notice that you can get exact p-values in SPSS). You also get your sample mean, the standard deviation (this is the estimated standard deviation), and the estimated standard error of the mean. You could compute the t-test yourself with this information: t = 1.177 – 2 = -2.74 .9499 / √10 Also note that your 95% confidence interval of the difference ranges from –1.50 to -.14. It’s strange that SPSS computes the confidence intervals this way, but you can easily get the confidence intervals around the mean as follows: If you were to compute the confidence intervals by hand, you would calculate Mean +/– tcrit (α=.05, 2 tailed) * σ hat / √N Or 1.177 +/– 2.262 * .9499 / √10, the confidence interval is .4975 to 1.8565 Because you use +/- .6795 to give the bounds of your interval. Using these values around the mean difference (-.8230) you get the SPSS values; using these values around the mean (1.177) you get .4975 to 1.8565. These are the values we would use.

What would a confidence interval look like if you were using a lower alpha level (i.e., having a greater % confidence)? Let’s try using a 99% confidence interval. Run the t-test as before, but select in the options that you want a 99% c.i..

One-Sample Test Test Value = 2 99% Confidence Interval of the Difference Lower Upper -1.7993 .1533

AMOUNT

t -2.740

df 9

Sig. (2-tailed) .023

Mean Difference -.8230

You will get the table above. Notice that your confidence interval is now larger (this makes intuitive sense; to have a greater degree of confidence, you would need a larger range of possible values). This time, the confidence interval of the difference contains 0 (this is essentially the same as stating that the confidence interval around the mean contains 2 (c.i. around the mean would be .201 to 2.153). In other words, if you use an α level of .01, you will fail to reject the null hypothesis, and your confidence interval would contain the null value. You can see that your exact p-value is .023, too high to reject the null if α is .01. Now let’s do a paired-groups t-test. The example for this problem comes from the Implicit Assumptions Test (IAT). If you are interested in this test (developed by Tony Greenwald at UW), you can go to the following link: http://buster.cs.yale.edu/implicit/index.html . The IAT method is used to assess automatic evaluations (i.e., evaluations that occur outside of conscious awareness and individuals’ control). The purpose of this study was to see if the IAT method could be used to assess automatic associations toward specific people who play significant roles in our lives (e.g., mother, romantic partner). The hypothesis was that if the IAT is able to assess automatic attitudes, it should be able to discriminate between a “positive” significant person and a “negative” significant person. Subjects in the study were asked to nominate a positive person from their lives and a negative person, and then two IATs were performed to assess their automatic associations toward each person. The two variables represent the strength of

positive automatic associations toward each person (larger numbers indicating more liking). Use a two-tailed test, α = .05. The data set is below (click here for the data):

Why is this a paired-samples t test (also called a repeated measures t test)? Because each participant is involved in both independent variable manipulations (in this case, both positive and negative significant person associations). The key to this type of test is that a line of data represents each participant’s performance. Thus, the two dependent variables are paired in that they apply to each participant.

To run this test, choose the commands ANALYZE / COMPARE MEANS / PAIRED-SAMPLES T TEST from the menu system. You will get the following dialog box:

Here, you need to click on both variables (the positive and negative dependent variables). Click one, hold down the SHIFT key, and click the other. Then select them to move to the test variable box. Your options are the same as they were for the one-sample t test. The output follows:
Paired Samples Statistics Std. Error Mean 43.6309 39.7637

Mean Pair 1 IAT effect for positive significant person IAT effect for negative significant person 371.1356 114.8433

N 20 20

Std. Deviation 195.1234 177.8288

Paired Samples Correlations N Pair 1 IAT effect for positive significant person & IAT effect for negative significant person 20 Correlation .418 Sig. .067

Paired Samples Test Paired Differences 95% Confidence Interval of the Difference Lower Upper 161.8703 350.7142

Mean Pair 1 IAT effect for positive significant person IAT effect for negative significant person 256.2923

Std. Deviation 201.7501

Std. Error Mean 45.1127

t 5.681

df 19

Sig. (2-tailed) .000

For this study, the null hypothesis is that there is NO DIFFERENCE between conditions (positive or negative), so Ho: µpositive = µnegative. The alternative hypothesis is H1: µpositive ≠ µnegative. The first table of your output gives you the mean, standard deviation, and standard error of the mean of the two conditions. Don’t worry about the second table for now, other than knowing that it shows how the two dependent variables correlate with one another. The third table is the t test itself. Notice that you have a t of 5.681, with 19 DFs, and that the significance is listed as .000. What kind of p-value is that? Well, since SPSS only displays three decimal places, you know that p is at least less than .005. Your 95% confidence interval is constructed around the difference between the means. If the null hypothesis is that there is no difference, this value is 0. Since 0 is not in your 95% confidence interval, you can reject the null at the alpha level of .05 (actually even less than that, as you know from the exact p-value). SO, a paired samples t test is essentially the same as a one-sample t test of the difference scores. Let’s prove that you would get the same result if you were to run the test that way: First, compute a variable of the difference scores. Select: TRANSFORM / COMPUTE You will get the following dialog box:

Select that your target variable (diff) is equal to pmsiat – nmsiat. Then click OK. Now, run the ANALYZE / COMPARE MEANS / ONE SAMPLE T TEST command again. This time, your test variable is “diff.” You will get the following:
One-Sample Test Test Value = 0 95% Confidence Interval of the Difference Lower Upper 161.8703 350.7142

DIFF

t 5.681

df 19

Sig. (2-tailed) .000

Mean Difference 256.2923

Notice that this is the same t, with the same dfs, significance level, and confidence intervals as when you ran the paired-samples t test.

Independent-samples t test: Now, let’s get a new data set that applies to independent groups. In this study, the IAT is again used, this time to assess automatic associations with “mother.” The variable of interest is gender; that is, do males and females differ in their automatic associations with their mothers? Ho: µmales = µfemales. The alternative hypothesis is H1: µmales ≠ µfemales. The data set is as follows (click here for the data set):

Notice that you still have two variables, but here the first variable is the independent variable (sex) and the second is the dependent variable (momiatms, or a measure of automatic associations to “mother.” This should look quite different from the paired-samples t test; put simply, nothing is paired in this study. To run this t test, select: ANALYZE / COMPARE MEANS / INDEPENDENT-SAMPLES T TEST The dialog box that you will get follows on the next page.

Your test variable (the DV) is momiatms; the grouping variable is sex. You need to define the groups, which are 0 and 1 (what the experimenter has used, but you can use anything; here 1 is male and 0 is female). Once you click OK, you will get the following:
Group Statistics Std. Error Mean 43.9549 42.0613

mom IAT effect (in ms)

subjects' sex females males

N 13 13

Mean 246.3593 373.1423

Std. Deviation 158.4815 151.6543

Independent Samples Test Levene's Test for Equality of Variances

t-test for Equality of Means 95% Confidence Interval of the Difference Lower Upper -252.3452 -252.3581 -1.2209 -1.2080

F mom IAT effect (in ms) Equal variances assumed Equal variances not assumed .001

Sig. .975

t -2.084 -2.084

df 24 23.954

Sig. (2-tailed) .048 .048

Mean Difference -126.7830 -126.7830

Std. Error Difference 60.8374 60.8374

Not much is new here; Levene’s test is a measure that tests an important assumption of the independent-samples t-test; that is, the two groups have equal variances. Here we’ll assume that the variances are equal (use the first line in the second table for your t).

What can you say about this study? It looks like your t, with 24 degrees of freedom (N1 + N2 – 2), is –2.084. The p-value is .048. You can reject the null hypothesis at the alpha level of .05, but what do you conclude? It appears that males have more positive automatic associations with “mother” than females.

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