201 t-test lab

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PSYC 201 -- T-test lab

An investigator does an experiment to test the idea that caffeine has anadverse 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 atest 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 ofmemory function, participants are given a list of twenty words to rememberand then asked to recall as many words from the list as they can. Theresearcher predicts that participants in the caffeine condition will remembersignificantly fewer words than participants in the placebo condition.

The data for the participants in the two groups are presented below.

Caffeine Group Placebo Group------------------- -------------------

8 1210 97 149 10

11 119 138 14

10 109 9

11 11

Please address each of the following questions.

a. Please write both the null and alternative hypotheses.b. Please write the decision rule for rejecting the null hypothesis.c. Using SPSS, please determine the observed value for t.d. State your decision as to whether to reject the null hypothesis or not.e. Please write a conclusion sentence for this test in APA format.

Below are instructions for using SPSS to get the observed value for t

Entering the data into the SPSS spreadsheet

There are two things that you know about each participant. First, you knowwhich level of the independent variable theyve been assigned to. In otherwords, each participant is in the caffeine condition (group 1) or the placebocondition (group 1). That means for a variable that we might name caffgrpeach persons will get a score of 1 if theyre in the caffeine group and a scoreof 2 if theyre in the placebo group. The variable name for this column wouldbe caffgrp. In this example, there will be 20 rows in the spreadsheet becausethere are 20 participants. The only other thing we know about each participantis their score on the dependent variable, memory function. These scores go ina second column that is labeled with the variable name memory.

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So that it. The data set has 20 rows (for the 20 participants) and it has twocolumns 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 theIndependent-Samples T Test window.

Move the variable memory over to the Test Variable(s) box. The testvariable is the dependent variable.

Move the variable caffgrp over to the Grouping Variable box. The groupingvariable is the independent variable.

Click on the Define Groups button. There is where you tell SPSS whichgroups you want to compare. Enter a 1 for the Group 1 box and a 2for the Group 2 box. Then click Continue.

Click OKback 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 forboth 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. Thesignificance level is .010. Because the alpha level has been set at .05, allyou have to do to see is the test is significant is to see is the significancelevel 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 are allowed to write as our conclusion is the following:Participants in the caffeine condition remember significantly fewer wordsthan participants in the placebo condition, t(18) = -2.89,p < .05.

Notice that at the end of the sentence the researcher has given herstatistical justification for why she made that statement. She has reportedthe value for t, the number degrees of freedom for the test, and the

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significance level of the test. Also notice that the symbols t and p arein italics. This is the format we want you to use in reporting the result for at-test. Use the same format whether or not the test is significant.

Answers to the sample problem

a. Null hypothesis: Participants in the caffeine condition do not remembersignificantly fewer words than participants in the placebo condition.

Alternative hypothesis: Participants in the caffeine condition remembersignificantly fewer words than participants in the placebo condition.

b. Decision rule: If t -1.73, reject HO.

c. t = -2.885

d. Decision: reject HO

e. Participants in the caffeine condition remember significantly fewer wordsthan participants in the placebo condition, t(18) = -2.89,p < .05.

Problems

1. An investigator predicts that students taking an achievement test in

the morning will perform differently than students taking the sametest in the afternoon. Eight students are assigned to take the test inthe morning and 8 students are assigned to take the test in theafternoon.

Morning Group: 25, 31, 44, 36, 29, 42, 39, 41Afternoon Group: 22, 25, 24, 27, 33, 28, 31, 34

a. 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 rejectthe 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.

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2. An investigator theorizes that people who participate in a regularprogram of exercise will have lower levels of anxiety than people whodo not participate in a regular program of exercise. To test this ideathe investigator randomly assigns 12 subjects to an exercise program

for 10 weeks and 12 subjects to a non-exercise comparison group.Lower anxiety scores reflect lower levels of anxiety. Please test theinvestigator's theory using an alpha level of .05.

Exercise Group: 18, 23, 21, 19, 17, 16, 20, 25, 18, 19, 23, 21No-Exercise Group: 26, 28, 25, 24, 31, 29, 27, 28, 27, 25, 23, 28


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.

3. An investigator thinks that people under the age of forty have bettervocabularies than people over sixty years of age. The investigatoradministers a vocabulary test to a group of 12 younger subjects and

to a group of 31 older subjects. Higher scores reflect betterperformance.

Under 40 Group: 31, 36, 33, 41, 37, 40, 36, 29, 31, 34, 41, 38Over 40 Group: 43, 41, 46, 44, 48, 39, 37, 46, 44, 50, 48, 45


your decision about whether or not the researcher should rejectthe null hypothesis.

d. In APA format, write a conclusion sentence that states what theresearcher has learned from doing the test.

e. What is independent variable in this study. What is thedependent variable in this study.

f. Which is the experimental condition. Which is the controlcondition.

Print out and hand in the output window for these tests.

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Answers Independent t-test problems

1.

HO: Students taking an achievement test in the morning do not performdifferently than students taking the same test in the afternoon.

H1: Students taking an achievement test in the morning performdifferently than students taking the same test in the afternoon.

Decision Rule: If t -2.14 or if t +2.145 , reject HO.

Decision: Reject HO

Conclusion sentence: Students taking an achievement test in the

morning perform differently than students taking the same test in theafternoon, t(14) = 2.75,p < .05.

2.HO: People who participate in a regular program of exercise do not havelower levels of anxiety than people who do not participate in a regularprogram of exercise.

H1: People who participate in a regular program of exercise have lowerlevels of anxiety than people who do not participate in a regularprogram of exercise.

Decision Rule: If t -1.72 , reject HO.

Decision: Reject HO

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Conclusion sentence: People who participate in a regular program ofexercise have lower levels of anxiety than people who do not participatein a regular program of exercise, t(22) = -5.39,p < .05.

3.

HO: People under the age of forty do not have better vocabularies thanpeople over sixty years of age.

H1: People under the age of forty have better vocabularies than peopleover sixty years of age.

Decision Rule: If t -1.72, reject HO.

Decision: Fail to reject HO

Conclusion sentence: People under the age of forty do not have better

vocabularies than people over sixty years of age,p > .05. [Note: This isbecause the results are in the opposite direction from the prediction ofthe researcher!]

201 t-test lab

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