One Sample T-Test
Dr. Scott Marley
EdPsy 511
Spring 2007
SPSS Tutorial
Please note that SPSS only gives results with p-values based upon two-tailed statistical tests. All examples presented here are two-tailed tests. In order to verify the hand calculations for one-tailed tests on your homework, Dr. Marley recommends the following methodologies: 1) After computing the t-test value by hand, compare this value to the t-test value seen in the SPSS output for two-tailed tests. & 2) Divide the p-value in the SPSS output in half and apply it to the alpha level assigned in the homework problem.
One Sample T-Test
The One-Sample T Test procedure tests whether the mean of a single variable differs from a specified constant.
Examples. A researcher might want to test whether the average IQ score for a group of students differs from 100. Or, a cereal manufacturer can take a sample of boxes from the production line and check whether the mean weight of the samples differs from 1.3 pounds at the 95% confidence level. The following example is designed for the purposes of humor and instruction.
We have volunteered to collect data for the people of Mozambique on the circumferences of mythical giraffe necks. We were able to measure 10 adult mythical giraffes whose neck sizes in feet had the following dimensions: 37, 68, 45, 30, 40, 50, 70, 35, 26, 47. We have approximately 500 feet of gauze material that we are hoping is sufficient to create neck bows for each giraffe. Thus, we are testing the null hypothesis that the average neck size of our sample of mythical giraffes is 50 ft. We are using an alpha level of .05.
H0: µ = 50
HA: µ ≠ 50
Enter the following neck sizes of the giraffes 37, 68, 45, 30, 40, 50, 70, 35, 26, 47. Then choose One-Sample T-Test
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Set Test Value at 50.
Click OK.
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Conclusion Based Upon SPSS Output for a One-Sample T-Test.
The mean size of our giraffe necks was 44.8 ft. with a standard deviation of 14.75 ft. The standard error of the mean was 4.66 ft.. We find a p-value of .294, which is greater than our alpha level of .05. Thus, the mean difference between neck sizes in our sample of giraffes is not statistically significantly different from 50 ft and we fail to reject the null hypothesis. The people of Mozambique are sure to cast admiring gazes at the beautifully mythical and gauzed giraffes.
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Independent-Samples T Test
The Independent-Samples T Test procedure compares means for two groups of cases. Ideally, for this test, the subjects should be randomly assigned to two groups, so that any difference in response is due to the treatment (or lack of treatment) and not to other factors. This is not the case if you compare average income for males and females. A person is not randomly assigned to be a male or female. In such situations, you should ensure that differences in other factors are not masking or enhancing a significant difference in means. Differences in average income may be influenced by factors such as education (and not by sex alone).
Example. Patients with high blood pressure are randomly assigned to a placebo group and a treatment group. The placebo subjects receive an inactive pill, and the treatment subjects receive a new drug that is expected to lower blood pressure. After the subjects are treated for two months, the two-sample t test is used to compare the average blood pressures for the placebo group and the treatment group. Each patient is measured once and belongs to one group.
1. Select Independent Samples T –Test as follows: Please find the Data Set entitled, ‘New Drug’ on Dr. Marley’s website to follow along with this exercise as practice for your homework.
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2. Enter the drug condition as the grouping variable and the blood pressure results as the test variable.
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Using Option Subcommand enter values (in this case 1 & 2) specified for your control and treatment drug groups. Then hit the continue button followed by the ok button on the original screen.
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3. Analyze SPSS Output for the Independent Samples T-Test. Find the Mean, Std. Deviation and Std. Error of the Mean for each of the drug conditions.
Group Statistics
| |Drug |N |Mean |Std. Deviation |Std. Error Mean|
|Blood Pressure |New Drug |6 |149.000 |18.6976 |7.6333 |
| |Placebo |6 |129.833 |16.4732 |6.7252 |
I apologize for the blurriness of the following output, which is due to size restrictions. The t-test for Equality of Means yields a p-value of .089, which informs us that the mean difference of 19.1667in blood pressure between the two groups was not statistically significant at the two-tailed alpha = .05 level. We note that a larger sample size or a one-tailed test might/would give different results.
Paired Samples T-Test
The Paired-Samples T Test procedure compares the means of two variables for a single group. The procedure computes the differences between values of the two variables for each case and tests whether the average differs from 0.
Example. In a study on coronary artery disease, the time a patient can spend on a treadmill is measured while still smoking and measured again after six months of having quit smoking. Thus, each subject has two measures, often called before and after measures.
1. Select Paired Samples T–Test as follows: Please find the Data Set entitled, ‘Coronary Artery Data’ on Dr. Marley’s website to follow along with this exercise as practice for your homework.
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2. Select ‘Treadmill Time Before’ and ‘Treadmill Time After’ in the left window and notice how Variable 1 and Variable 2 are filled in as you do so. Then press the arrow button.
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3. Notice how the two variables are paired into a relationship that indicates that the ‘time before’ variable precedes the ‘time after’ variable. Verify that the correct sequence is established. Then hit the okay button.
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4. Analyze SPSS output. Notice the mean, sample size, standard deviation and standard error of the mean for each of the paired variables.
Paired Samples Statistics
| |Mean |N |Std. Deviation |Std. Error Mean|
|Pair 1 |Treadmill time in seconds before |837.44 |18 |197.653 |46.587 |
| |Treatment | | | | |
| |Treadmill time in seconds after |617.9444 |18 |204.98335 |48.31504 |
| |Treatment | | | | |
The correlation between the paired variables is .810, which is statistically significant at the alpha = .05 level with a p-value ................
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