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Radford PSYC 201 - Study References

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PSYC 201 -- T-test labEntering the data into the SPSS spreadsheetIndependent-samples t-test using SPSSUsing the SPSS output window for an Independent-Samples T TestPSYC 201 -- T-test labAn investigator does an experiment to test the idea that caffeine has an adverse effect on memory function. She assigns 10 participants at random toa group that will get the equivalent of four cups of coffee before they take a test of memory function. She assigns 10 additional participants to a group that gets a placebo before taking the same test of memory function. In the test of memory function, participants are given a list of twenty words to remember and then asked to recall as many words from the list as they can. The researcher predicts that participants in the caffeine condition will remember significantly fewer words than participants in the placebo condition.The null hypothesis for the test is that participants in the caffeine condition do not remember significantly fewer words than participants in the placebo condition.The alternative hypothesis for the test is that participants in the caffeine condition remember significantly fewer words than participants in the placebocondition.The researcher decides to test this null hypothesis using an alpha level of .05.The data for the participants in the two groups are presented below. Caffeine Group Placebo Group ------------------- ------------------- 8 12 10 9 7 14 9 10 11 11 9 13 8 14 10 10 9 9 11 11You can use an independent-samples t-test to address this question using SPSS in the following way.Entering the data into the SPSS spreadsheetThere are two things that you know about each participant. First, you know which level of the independent variable they’ve been assigned to. In other words, each participant is in the caffeine condition (group 1) or the placebo condition (group 1). That means for a variable that we might name caffgrp each persons will get a score of “1” if they’re in the caffeine group and a score of “2” if they’re in the placebo group. The variable name for this column would be caffgrp. In this example, there will be 20 rows in the spreadsheet because there are 20 participants. The only other thing weknow about each participant is their score on the dependent variable, memory function. These scores go in a second column that is labeled with thevariable name memory.So that it. The data set has 20 rows (for the 20 participants) and it has two columns of numbers (for the two variables).Independent-samples t-test using SPSS- Click Analyze- Click on the Compare Means pull-down menu- Click on Independent-Samples T Test. You should now see the Independent-Samples T Test window.- Move the variable memory over to the Test Variable(s) box. The “test variable” is the dependent variable.- Move the variable caffgrp over to the Grouping Variable box. The “grouping variable” is the independent variable.- Click on the Define Groups button. There is where you tell SPSS which groups you want to compare. Enter a “1” for the Group 1 box and a “2” for the Group 2 box. Then click Continue.- Click OK back at the Independent-Samples T Test window.Using the SPSS output window for an Independent-Samples T Test- In the Group Statistics box you get the means and standard deviations for both groups. - In the Independent-Samples T Test box, you get the results for the t-test. The row to look in is labeled “Equal variances assumed”. - Look in the section of the box labeled t-test for Equality of Means. Thevalue for t is –2.885. The test has 18 degrees of freedom. The significance level is .010. Because the alpha level has been set at .05, all you have to do to see is the test is significant is to see is the significance level reported in the out is .05 or less. .010 is less than .05, so we can saythat the test is significant. - Because the test is significant, our decision is to reject the null hypothesis and accept the alternative hypothesis.- The sentence that we allowed to write as our conclusion is the following: Participants in the caffeine condition remember significantly fewer words than participants in the placebo condition, t (18) = -2.89, p = .010.- Notice that at the end of the sentence the researcher has given her statistical justification for why she made that statement. She has reportedthe value for t, the number degrees of freedom for the test, and the significance level of the test. Also notice that the symbols “t” and “p” are in italics. This is the format we want you to use in reporting the result for a t-test. Use the same format whether or not the test is significant.Problems1. An investigator predicts that students taking an achievement test inthe morning will do better than students taking the same test in the afternoon. Eight students are assigned to take the test in the morning and 8 students are assigned to take the test in the afternoon. Morning Group: 25, 31, 44, 36, 29, 42, 39, 41Afternoon Group: 22, 25, 24, 27, 33, 28, 31, 34a. State the null and alternative hypotheses.b. Write the decision rule for rejecting this null hypothesis.c. Using an alpha level of .05, test the null hypothesis, and state your decision about whether or not the researcher should reject the null hypothesis.d. In APA format, write a conclusion sentence that states what the researcher has learned from doing the test.e. What is independent variable in this study. What is the dependent variable in this study.f. Which is the experimental condition. Which is the control condition.2. An investigator theorizes that people who participate in a regular program of exercise will have lower levels of anxiety than people who do not participate in a regular program of exercise. To test this idea the investigator randomly assigns 12 subjects to an exercise program for 10 weeks and 12 subjects to a non-exercise comparisongroup. Lower anxiety scores reflect lower levels of anxiety. Please test the investigator's theory using an alpha level of .05. Exercise Group: 18, 23, 21, 19,


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Radford PSYC 201 - Study References

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