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Psych 5500/6500

t Test for Two Independent Means

Fall, 2008

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Cause and Effect Relationships

Does Variable X (cause/influence/affect) Variable Y?

Variable X would be the independent variable (the cause).

Variable Y would be the dependent variable (the effect).

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t Test for 2 Independent Means

Used to compare two independent samples, specifically we are comparing the means of the samples to see whether or not we can infer that the means of the two populations they represent are different.

Usually we are testing a theory which proposes that an independent variable has made the population means different.

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Example 1

Question: Does gender influence respiration rate?

Design: randomly sample 20 males and 20 females from some population and measure their respiration rate while at rest.

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Example 1

I.V.=

D.V.=

H0: μfemale= μmale (or) μfemale-μmale =0

HA: μfemale μmale (or) μfemale-μmale 0

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HypothesesNote that the hypotheses are always about

populations.

H0: μfemale= μmale HA: μfemale μmale

NOT about samples:

We want to be able to generalize the results to the populations from which we sampled, not just to the specific participants who are in our sample.

malesfemalesAmalesfemales0 Y Y :H Y Y :H

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Example 2Question: Is maze running ability affected by

the presence or absence of some specific drug?

Design: Select 11 rats and randomly divide them into two groups. The rats in Group 1 are given the drug, the rats in Group 2 are not. The number of wrong turns each rat makes before reaching the end of the maze is recorded.

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Example 2

I.V.=

D.V.=

H0: μdrug= μno_drug (or) μdrug-μno_drug = 0

HA: μdrug μno_drug (or) μdrug-μno_drug 0

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Statistic

The statistic upon which we will base our decision about H0 is:

Challenge: if H0 is true then the two population means are equal, and we expect the two sample means to be similar, but even if H0 is true the two sample means will not exactly equal each other due to chance (i.e. the samples have random bias). So if the two sample means differ is that due to differences in the population means or due to chance?

21 YY

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The Solution

Set up the ‘Sampling Distribution of

assuming H0 is true’ and label it

as such. Then:

• The mean of that sampling distribution is based upon H0.

• The standard deviation of that sampling distribution is estimated from the data from the two samples

21 YY

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Sampling Distribution of the statistic if H0 is true

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Sampling Distribution (cont.)

021 YY

2121

21YY N

1

N

1

2NN

SSSSσ est.

21

2NN1N1Nd.f. 2121

(based upon H0 being true)

Computational formula for the standard error of the difference:

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Standard Error of the Difference

21

2YY N

1

N

1est.σσ est.

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Let’s take a conceptual look at the standard deviation of the teststatistic This t test assumes that both populations have the samevariance, and uses this to ‘pool’ the two estimates of this populationvariance into one good estimate using the following formula, whichweights the estimate based upon the size of each sample was.:

)Y-Y( 21

2NN

1)est.σ(N1)est.σ(Nσ est.

21

222

2112

Conceptual formula for the standard error of the difference:

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Drug Example Data

Group 1 (Drug) Group 2 (No Drug)

7 4

8 6

6 5

10 3

6 2

5

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Group Statistics

166

42310SS

76

42Y

310Y

42Y

6N

2

1

1

2

1

1

1

105

0209SS

45

20Y

09Y

02Y

5N

2

2

2

2

2

2

2

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t Computations

03.107.137.89.22.17.9

26

5

1

6

1

256

0161

N

1

N

1

2NNσ est.

2121

21YY 21

SSSS

17

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Setting up the Rejection Regions

2.262t

92562NNd.f.

c

21

19

20

tobt

21

21

YY

YY21

obt est.σ

μYYt

91.2

03.1

047tobt

21

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Decision

We ‘reject H0’. We can conclude that the two samples represent populations that have different means.

We would like to then go on and state that the independent variable (drug vs. no drug) must have made the two population means different from each other, but at this point we can’t, first we have to show that there is no other reason for why the two populations might have different means. This takes us to the topic of ‘confounding variables’, which we will cover in the next lecture.

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Confidence Intervals

Confidence interval for the true difference between μ1 and μ2 :

21 YYtail-df,2,:c21 est.σtYY

5.33 0.67,2.3331.032.26247

95% confidence interval: 0.67 (μ1 - μ2) 5.33

If a hypothesis states a value for μ1 - μ2 that is outside of thatconfidence interval (e.g. H0: μ1 - μ2 = 0) then you can reject it

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Reporting ResultsA standard way of reporting the results of t tests is to

use the following format:

t(df)=tobt, p=?

From last example:t(9)=2.91, p=...

We can use the ‘t distribution tool’ in ‘Oakley Stat Tools’ to get the exact value of p, or we can use SPSS to do the t test for us. t(9) = 2.91, p=.017

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One-Tailed TestsSo far the examples have involved non-directional

theories (which predict the two population means will be different but don’t predict one which will be greater than the other). This is done with a 2-tail test.

It is also possible to test theories which are directional (predict specifically which population mean should be greater). This is done with a 1-tail test, which will influence how we write our hypotheses, and will lead to just one rejection region in our sampling distribution.

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One-tailed Tests

Again, express the prediction made by the theory when you write HA, H0 is then everything else. For testing a theory which predicts that the mean of population one (drugged rats) should be greater than the mean of population two (undrugged rats):

H0: μ1 μ2 (or equivalently) μ1-μ2 0

HA: μ1 > μ2 (or equivalently) μ1-μ2 > 0

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Rejection Region

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one-tailed tests

For testing a theory which predicts that the mean of population one (drugged rats) should be less than the mean of population two (undrugged rats):

H0: μ1 μ2 (or equivalently) μ1-μ2 0

HA: μ1 < μ2 (or equivalently) μ1-μ2 < 0

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Rejection Region

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Null Hypothesis

I have given two ways of expressing H0.

H0: μdrug= μno_drug (or) μdrug - μno_drug = 0

The first way is somewhat conceptually easier, but the second has an advantage as well. I have mentioned before that the null hypothesis is usually, but not always, the hypothesis of ‘no difference’. Let me now give an example where that wouldn’t be true…

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Let us say that the difference in population means between rats on the drug and those not on the drug has been previously established as equaling 1.5. In this experiment, however, we will have the rats swim through the maze and we are testing a theory which predicts that the difference between the two groups should be influenced by swimming (i.e. it will no longer equal 1.5) The null hypothesis is that performance won’t be influenced by swimming. Now our hypotheses would look like this:

H0: μdrug-μno_drug = 1.5

HA: μdrug-μno_drug 1.5

Note that conceptually H0 can still be thought of as ‘no difference’, in this case we are saying that the difference between the drug and no drug groups will not be different when the rats are swimming.

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Sampling Distribution if H0 is true

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tobtained

The tobtained score is still simply a standard score on the curve: i.e. the test statistic minus the mean of the curve divided by the standard deviation (standard error) of the curve.

5.1μ case in thisbut

est.

μ)YY(t

)YY(

)YY(

)YY(21

obtained

21

21

21

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Assumptions underlying this t test

1. Independence of scores (within and between groups)

2. Both populations normally distributed3. The two populations have identical

variances

We will examine the assumption of identical variances in a latter lecture.

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Effect Size

Again the measures of effect size include:

1. Simply reporting the ‘raw’ effect size.

2. Reporting a standardized effect size.

3. Reporting the strength of association (which we will cover next semester).

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‘Raw’ Effect Size

This would be simply reporting the mean of the two groups and the difference between those means. In our experiment the mean number of wrong turns made by the ‘drug’ group was 7, the mean number for the ‘no drug’ group was 4, which is a difference of 3 between the two means.

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Standardized Effect Size

We will take a look at three standardized measures of effect size:

1. Cohen’s d

2. Hedges’s g

3. Glass’s Δ

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Cohen’s dFor the effect size in the population (exact, not estimated):

Y

21

σ

μμδ

Remember that an assumption underlying this t test is that bothpopulations have the same variance (σY).

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Cohen’s dFor the effect size in the sample:

If you examine the formula for Spooled you can see that it is simply the standard deviation of all the scores lumped into one group.

NN

SSSS

NN

SNSNS where

S

YYd

21

21

21

222

211

pooled

pooled

21

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Hedges’s gHedges’s g gives us an estimate of the effect size in the populationform which we sampled.

est. σpooled is a ‘pooled’ estimate of the standard deviation of Y, the formula combines the estimates of σY from the two groups, weighting each estimate based upon how many scores were in the group.

2NN

1)est.σ-(N1)est.σ-(Nest.σ where

est.σ

YYg

21

222

211

pooled

pooled

21

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Glass’s ΔGlass’s delta is similar to Hedges’s g, but instead of using thescores from both groups to estimate the standard deviation of Y, delta just uses the data from the control group to estimate it.

est.σest.σ where

est.σ

YYΔ

2ControlControl

Control

21

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Glass’s ΔIn an experiment designed to test the effect of some

treatment it is common to include a group that is handled exactly like the group that gets the treatment except without the treatment, this is called the ‘Control Group’ and the group that gets the treatment is called the ‘Treatment Group’. The reason the control group is named as it is will be covered in the lecture on confounding variables. In our example the rats given the drug constitute the ‘treatment group’ and those not given the drug are the ‘control group’.

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Glass’s ΔGlass’s delta is used when there is reason to

believe that the treatment applied to the treatment group might have affected not only the mean of the group but the variance of that group as well. If this is the case then the assumption of equal variances is not met (we will see how to analyze the data anyway later) and it makes no sense to use the variance of both groups to estimate the variance of Y, as they would actually be measuring two different variances, not one.

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Using these Formulas

3.205

16

1N

SSest.σ

67.26

16

N

SS S

166

42310SS

76

42Y

6N

1

121

1

121

2

1

1

1

50.24

10

1N

SSest.σ

00.25

10

N

SS S

105

0209SS

45

20Y

5N

2

222

2

222

2

2

2

2

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Using these Formulas

1.95 1.54

47d

54.111

0161

NN

SSSS

NN

SNSNS where

S

YYd

21

21

21

222

211

pooled

pooled

21

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Using these Formulas

76.1 1.7

47g

70.1256

(4)(2.50)(5)(3.2)

2NN

1)est.σ-(N1)est.σ-(Nest.σ where

est.σ

YYg

21

222

211

pooled

pooled

21

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Using these Formulas

90.11.58

47Δ

58.12.50 est.σest.σ where

est.σ

YYΔ

2ControlControl

Control

21

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Useful Conversions

2NN

NNgd

21

21

2 groupin prop. p 1, groupin scores of prop.p where

2)N(Npp

td

21

2121

obt

‘prop’ means ‘proportion’, so if 14 of the 20 scores were in group 1, then p1 wouldbe 14/20 while p2 would be 6/20.

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Useful Conversions

21

21

NN

2NNdg

2 groupin prop. p 1, groupin scores of prop.p where

)N(Npp

tg

21

2121

obt

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Useful Conversions

2 groupin prop. p 1, groupin scores of prop.p where

2)N(Nppdt

21

2121obt

2 groupin prop. p 1, groupin scores of prop.p where

)N(Nppgt

21

2121obt