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Aron chpt 8 ed

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Introduction to Introduction to the the t t Test Test Chapter 8 Copyright © 2011 by Pearson Education, Inc. All rights reserved
Transcript
Page 1: Aron chpt 8 ed

Introduction to the Introduction to the tt TestTest

Chapter 8

Copyright © 2011 by Pearson Education, Inc. All rights reserved

Page 2: Aron chpt 8 ed

t t TestsTests

Hypothesis-testing procedure in which the population variance is unknown◦ compares t scores from a sample to a

comparison distribution called a t distributiont Test for a single sample

◦ hypothesis-testing procedure in which a sample mean is being compared to a known population mean but the population variance is unknown

◦ Works basically the same way as a Z test, but: because the population variance is unknown, with a

t test you have to estimate the population variance With an estimated variance, the shape of the

distribution is not normal, so a special table is used to find cutoff scores.

Copyright © 2011 by Pearson Education, Inc. All rights reserved

Page 3: Aron chpt 8 ed

Basic Principle of the Basic Principle of the tt Test: Test: Estimating the Population Variance Estimating the Population Variance from the Sample Scoresfrom the Sample Scores

You can estimate the variance of the population of individuals from the scores of people in your sample.◦ The variance of the scores from your sample will be slightly

smaller than the variance of scores from the population. Using the variance of the sample to estimate the variance of the

population produces a biased estimate. Unbiased Estimate

◦ estimate of the population variance based on sample scores, which has been corrected so that it is equally likely to overestimate or underestimate the true population variance The bias is corrected by dividing the sum of squared deviation by

the sample size minus 1 S2 = ∑(X – M)2

N – 1

Copyright © 2011 by Pearson Education, Inc. All rights reserved

Page 4: Aron chpt 8 ed

The The tt Distribution Distribution When the population variance is estimated, you

have less true information and more room for error.◦ The shape of the comparison distribution will not be a

normal curve; it will be a t distribution. t distributions look like the normal curve—they are bell

shaped, unimodal, and symmetrical—but there are more extreme scores in t distributions. Their tails are higher.

There are many t distributions, the shapes of which vary according to the degrees of freedom used to calculate the distribution. There is only one t distribution for any particular degrees of

freedom.

Page 5: Aron chpt 8 ed

The The tt Score Score• The sample’s mean score on the comparison distribution• It is calculated in the same way as a Z score, but it is

used when the variance of the comparison distribution is estimated.

• It is the sample’s mean minus the population mean divided by the standard deviation of the distribution of means.

• If your sample’s mean was 35, the population mean was

46, and the estimated standard deviation was 5, then the t score for this example would be -2.2.

• This sample’s mean is 2.2 standard deviations below the mean.

Copyright © 2011 by Pearson Education, Inc. All rights reserved

Page 6: Aron chpt 8 ed

Hypothesis Testing When the Hypothesis Testing When the Population Variance Is UnknownPopulation Variance Is Unknown

1. Restate the question about the research hypothesis and a null hypothesis about the populations.

2. Determine the characteristics of the comparison distribution.

3. Determine the significance cutoff.4. Determine your sample’s score on the

comparison distribution.

5. Decide whether to reject the null hypothesis.Compare the t score of your sample and the cutoff score from the t table.

• Population mean• Population variance

• Standard deviation of the distribution of means • Shape of the comparison distribution

Page 7: Aron chpt 8 ed

EXAMPLE– Aron, Coups, Aron (pg 264)EXAMPLE– Aron, Coups, Aron (pg 264)

Eight participants are tested after being given an experimental procedure. Their scores are 14, 8, 6, 5, 13, 10, 10, 6. The population (of people not given this procedure) is normally distributed with a mean of 6. Using the .05 level, does the experimental procedure make a difference? (a) Use the five steps of hypothesis testing and (b) sketch the distinctions involved.

Page 8: Aron chpt 8 ed

Step 1: Restate the question about Step 1: Restate the question about the research hypothesis and a null the research hypothesis and a null hypothesis about the populations.hypothesis about the populations.

Research Question – “Does the experimental procedure make a difference for the participants?”

Page 9: Aron chpt 8 ed

Step 1: Restate the question about Step 1: Restate the question about the research hypothesis and a null the research hypothesis and a null hypothesis about the populations.hypothesis about the populations.Research Question – “Does the experimental procedure make a difference for the participants?”

Population 1: Participants given the experimental procedure.

Population 2: The general population

Ha: “Participants given the experimental procedure will score differently than the general population”

Ho: “Participants given the experimental procedure will not score differently than the general population”

Page 10: Aron chpt 8 ed

Step 2: Determine the Step 2: Determine the characteristics of the comparison characteristics of the comparison distributiondistribution

◦ population mean This is the same as the known population mean.

◦ population variance Figure the estimated population variance.

(df = N-1) (N = #

scores in sample)

Figure the variance of the distribution of means. (N = # scores in sample)

◦ standard deviation of the distribution of means Figure the standard deviation of the distribution of means.

◦ shape of the comparison distribution

t distribution with N – 1 degrees of freedom

Page 11: Aron chpt 8 ed

Step 2: Determine the Step 2: Determine the characteristics of the comparison characteristics of the comparison distribution (cont.)distribution (cont.)

◦ population mean = 6

◦ population variance

X M (X-M) (X-M)2

14 - 9 "= 5 258 - 9 "= "-1 16 - 9 "= "-3 95 - 9 "= "-4 16

13 - 9 "= 4 1610 - 9 "= 1 110 - 9 "= 1 1

6 - 9 "= "-3 972 78

8Sample M = 9

df = 8-1 (7)

Page 12: Aron chpt 8 ed

Step 2: Determine the Step 2: Determine the characteristics of the comparison characteristics of the comparison distribution (cont.)distribution (cont.)

◦ Figure the variance of the distribution of means.

◦ standard deviation of the distribution of means Figure the standard deviation of the distribution of

means.

S2 = 11.14

N = 8 (# scores in sample)

Page 13: Aron chpt 8 ed

Step 2: Determine the Step 2: Determine the characteristics of the comparison characteristics of the comparison distribution (cont.)distribution (cont.)

◦shape of the comparison distribution t distribution with N – 1 df t distribution with 7 df

Page 14: Aron chpt 8 ed

Step 3: Determine the significance cutoffStep 3: Determine the significance cutoff

Decide the significance level and whether to use a one- or a two-tailed test.p< .05 (two-tailed)

◦Look up the appropriate cutoff in a t table.◦Cutoffs = -2.365 & 2.365

Ha = “Participants given the experimental procedure will score differently than the general population”

Page 15: Aron chpt 8 ed

Step 4: Determine your sample’s score on the Step 4: Determine your sample’s score on the comparison distribution.comparison distribution.

M = 9Pop. M = 6SM = 1.18

Page 16: Aron chpt 8 ed

Step 5: Decide whether to reject the Step 5: Decide whether to reject the null hypothesis.null hypothesis.

Compare the t score of your sample and the cutoff score from the t table.

◦t of 2.54 is more extreme than the t cutoff of 2.365.◦Reject Ho

◦Found support for Ha

◦The experimental procedure makes a difference.

Page 17: Aron chpt 8 ed

The The tt Test for Dependent Test for Dependent MeansMeansIt is common when conducting research to have two sets of scores and not to know the mean of the population.Repeated Measures Design (Within Subjects Design)

◦ research design in which each person is tested more than once

◦ For this type of design, a t test for dependent means is used. The means for each group of scores are from the

same people and are dependent on each other. A t test for dependent means is calculated the same

way as a t test for a single sample; however: Difference scores are used . You assume that the population mean is 0.

Page 18: Aron chpt 8 ed

Difference ScoresDifference ScoresFor each person, you subtract one score

from the other.If the difference compares before

versus after, difference scores are also called change scores.

Once you have the difference score for each person in the study, you do the rest of the hypothesis testing with difference scores.◦ You treat the study as if there were a single

sample of scores.

Copyright © 2011 by Pearson Education, Inc. All rights reserved

Page 19: Aron chpt 8 ed

Population of Difference Population of Difference Scores with a Mean of 0Scores with a Mean of 0

Null hypothesis in a repeated measured design◦ On average, there is no difference

between the two groups of scores. When working with difference scores, you

compare the population of difference scores from which your sample of difference scores comes (Population 1) to a population of difference scores (Population 2) with a mean of 0.

Copyright © 2011 by Pearson Education, Inc. All rights reserved

Page 20: Aron chpt 8 ed

Steps for a Steps for a tt Test for Test for Dependent MeansDependent Means1. Restate the question as a research

hypothesis and a null hypothesis about the populations.

2. Determine the characteristics of the comparison distribution.

3. Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected.

4. Determine the sample’s score on the comparison distribution.

5. Decide whether to reject the null hypothesis.

Page 21: Aron chpt 8 ed

EXAMPLE– Aron, Coups, Aron (pg 265)EXAMPLE– Aron, Coups, Aron (pg 265)

A researcher tests 10 individuals before and after an experimental procedure.

Test the hypothesis that there is an increase in scores, using the .05 significance level. (a) Use the five steps of hypothesis testing and (b) sketch the distribution involved.

Page 22: Aron chpt 8 ed

Step 1: Restate the question about Step 1: Restate the question about the research hypothesis and a null the research hypothesis and a null hypothesis about the populations.hypothesis about the populations.Population 1: Participants given the experimental procedure.

Population 2: Those who show no change from before or after the procedure

Ha: “Participants given the experimental procedure will have an increase in scores following the procedure”

Ho: “Participants given the experimental procedure will not have an increase in scores following the procedure”

Page 23: Aron chpt 8 ed

Step 2: Determine the characteristics of Step 2: Determine the characteristics of the comparison distributionthe comparison distribution◦ Make each person’s two scores into a difference score.

Do all of the remaining steps using these difference scores.◦ Figure the mean of the difference scores.◦ Assume the mean of the distribution of means of difference scores =

0.◦ Find the standard deviation of the distribution of means of difference

scores. Figure the estimated population variance of difference scores.

df = N-1

Figure the variance of the distribution of means of difference scores.

Figure the standard deviation of the distribution of means of difference scores.

◦ The shape is a t distribution with N – 1 degrees of freedom.

Page 24: Aron chpt 8 ed

Step 2: Determine the characteristics of Step 2: Determine the characteristics of the comparison distribution (cont.)the comparison distribution (cont.)

◦ Make each person’s two scores into a difference score.

◦ Figure the mean of the difference scores.

Participant Before After Difference (After-Before)

1 10.4 10.8 0.42 12.6 12.1 -0.53 11.2 12.1 0.94 10.9 11.4 0.55 14.3 13.9 -0.46 13.2 13.5 0.37 9.7 10.9 1.28 11.5 11.5 09 10.8 10.4 -0.4

10 13.1 12.5 -0.6117.7 119.1 1.4

Page 25: Aron chpt 8 ed

Step 2: (cont.)Step 2: (cont.)◦ Find the standard deviation of the distribution of means

of difference scores. Figure the estimated population variance of difference scores.

Participant Before After Difference (After-Before)

Deviation (Difference - M)

Squared Deviation

1 10.4 10.8 0.4 0.26 0.0682 12.6 12.1 -0.5 -0.64 0.4103 11.2 12.1 0.9 0.76 0.5784 10.9 11.4 0.5 0.36 0.1305 14.3 13.9 -0.4 -0.54 0.2926 13.2 13.5 0.3 0.16 0.0267 9.7 10.9 1.2 1.06 1.1248 11.5 11.5 0 -0.14 0.0209 10.8 10.4 -0.4 -0.54 0.292

10 13.1 12.5 -0.6 -0.74 0.548117.7 119.1 1.4 3.484

M = 0.14

(df = N-1)

Page 26: Aron chpt 8 ed

Step 2: (cont.)Step 2: (cont.)

Figure the variance of the distribution of means of difference scores.

Figure the standard deviation of the distribution of means of difference scores.

◦ The shape is a t distribution with N – 1 degrees of freedom.◦ t distribution with 9 df

Page 27: Aron chpt 8 ed

Step 3: Determine the cutoff sample score on the Step 3: Determine the cutoff sample score on the comparison distribution at which the null comparison distribution at which the null hypothesis should be rejected.hypothesis should be rejected.◦ Decide the significance level and whether to

use a one- or a two-tailed test.

◦ p<.05 (one-tailed)

◦ Look up the appropriate cutoff in a t table.

“Participants given the experimental procedure will have an increase in scores following the procedure”

Page 28: Aron chpt 8 ed

Step 4: Step 4: Determine the sample’s score on Determine the sample’s score on the comparison distribution.the comparison distribution.

◦ Assume the mean of the distribution of means of difference scores = 0.

Page 29: Aron chpt 8 ed

Step 5: Decide whether to reject the null hypothesis.

Compare the t score for your sample to the cutoff score found using the t tables

◦ t of .71 is less extreme than the t cutoff of 1.833.

◦ Fail to reject Ho

◦ Findings inconclusive for Ha

◦ Cannot say the experimental procedure made a difference.

Page 30: Aron chpt 8 ed

Review of the Z test, Review of the Z test, tt Test for a Single Test for a Single Sample, and Sample, and t t Test for Dependent MeansTest for Dependent Means

Z Test◦ Population variance is known.◦ Population mean is known.◦ There is 1 score for each participant.◦ The comparison distribution is a Z distribution.◦ Formula Z = (M – Population M) / Population SDM

◦ The best estimate of the population mean is the sample mean. t Test for a Single Sample

◦ Population variance is not known.◦ Population mean is known.◦ There is 1 score for each participant.◦ The comparison distribution is a t distribution.◦ df = N – 1◦ Formula t = (M – Population M) / Population SM

t Test for Dependent Means◦ Population variance is not known.◦ Population mean is not known.◦ There are 2 scores for each participant.◦ The comparison distribution is a t distribution.◦ df = N – 1◦ Formula t = (M – Population M) / Population SM

Copyright © 2011 by Pearson Education, Inc. All rights reserved

Page 31: Aron chpt 8 ed

Assumptions of the Assumptions of the tt Test for a Test for a Single Sample and Single Sample and tt Test for Test for Dependent MeansDependent Means

Assumption◦ a condition required for carrying out a particular

hypothesis-testing procedure◦ It is part of the mathematical foundation for the

accuracy of the tables used in determining cutoff values.

A normal population distribution is an assumption of the t test.◦ It is a requirement within the logic and mathematics

for a t test.◦ It is a requirement that must be met for the t test to

be accurate.

Copyright © 2011 by Pearson Education, Inc. All rights reserved

Page 32: Aron chpt 8 ed

Effect Size for the Effect Size for the tt Test for Test for Dependent MeansDependent Means

Mean of the difference scores divided by the estimated standard deviation of the population of difference scores

estimated effect size =

M = mean of the difference scores S = estimated standard deviation of the

population of individual difference scores

Copyright © 2011 by Pearson Education, Inc. All rights reserved

Page 33: Aron chpt 8 ed

Step 2: (cont.)Step 2: (cont.)◦ Find the standard deviation of the distribution of means

of difference scores. Figure the estimated population variance of difference scores.

Participant Before After Difference (After-Before)

Deviation (Difference - M)

Squared Deviation

1 10.4 10.8 0.4 0.26 0.0682 12.6 12.1 -0.5 -0.64 0.4103 11.2 12.1 0.9 0.76 0.5784 10.9 11.4 0.5 0.36 0.1305 14.3 13.9 -0.4 -0.54 0.2926 13.2 13.5 0.3 0.16 0.0267 9.7 10.9 1.2 1.06 1.1248 11.5 11.5 0 -0.14 0.0209 10.8 10.4 -0.4 -0.54 0.292

10 13.1 12.5 -0.6 -0.74 0.548117.7 119.1 1.4 3.484

M = 0.14

(df = N-1)

Page 34: Aron chpt 8 ed

Effect Size for the Effect Size for the tt Test for Test for Dependent MeansDependent Means

Mean of the difference scores divided by the estimated standard deviation of the population of difference scores

estimated effect size =

Copyright © 2011 by Pearson Education, Inc. All rights reserved


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