statistics for the social sciences psychology 340 spring 2005 using t-tests

Statistics for the Social Sciences

Psychology 340Spring 2005

Using t-tests


Outline (for 2 weeks)

• Review t-tests– One sample, related samples, independent samples

– Additional assumptions• Levene’s test


Statistical analysis follows design

• The one-sample z-test can be used when:

– 1 sample

€

zX

=X − μ

X

σX

– One score per subject

– Population mean () and standard deviation ()are known



• The one-sample t-test can be used when:

– 1 sample– One score per subject

– Population mean () is known

– but standard deviation () is NOT known

€

t =X − μ

X

sX


Testing Hypotheses

– Step 1: State your hypotheses– Step 2: Set your decision criteria– Step 3: Collect your data – Step 4: Compute your test statistics

• Compute your estimated standard error• Compute your t-statistic• Compute your degrees of freedom

– Step 5: Make a decision about your null hypothesis

• Hypothesis testing: a five step program


Performing your statistical test

• What are we doing when we test the hypotheses?– Consider a variation of our memory experiment example

Population of memory patients

MemoryTest is known

Memory treatment

Memory patients

MemoryTestX

Compare these two means

Conclusions: • the memory treatment sample are the same as those in the population of memory patients.• they aren’t the same as those in the population of memory patients

H0

:HA:



• What are we doing when we test the hypotheses?

Real world (‘truth’)

H0: is false (is a treatment effect)

Two populations

XA

they aren’t the same as those in the population of memory patients

H0: is true (no treatment effect)

One population

XA

the memory treatment sample are the same as those in the population of memory patients.



• What are we doing when we test the hypotheses?– Computing a test statistic: Generic test

€

test statistic =observed difference

difference expected by chance

Could be difference between a sample and a population, or between different samples

Based on standard error or an estimate

of the standard error



€



€

t =X − μ

X

sX

€

zX

=X − μ

X

σX

Test statistic

One sample z One sample tidentical



€



€

t =X − μ

X

sX

€

zX

=X − μ

X

σX

Test statistic

Diff. Expected by chance

€

X

=σ

nStandard error

One sample z One sample t

different

don’t know this,so need to estimate it



€



€

t =X − μ

X

sX

€

zX

=X − μ

X

σX

€

sX

=s

n

Test statistic


€

X

=σ

nStandard error

Estimated standard error

One sample z One sample t

different

don’t know this,so need to estimate it

€

df = n −1Degrees of freedom


Proportion in one tail0.10 0.05 0.025 0.01 0.005

Proportion in two tailsdf 0.20 0.10 0.05 0.02 0.011 3.078 6.314 12.706 31.821 63.6572 1.886 2.920 4.303 6.965 9.9253 1.638 2.353 3.182 4.541 5.8414 1.533 2.132 2.776 3.747 4.6045 1,476 2.015 2.571 3.365 4.0326 1.440 1.943 2.447 3.143 3.707:

15:

:1.341

:

:1.753

:

:2.131

:

:2.602

:

:2.947

:

One sample t-test

• The t-statistic distribution (a transformation of the distribution of sample means transformed)– Varies in shape according to the degrees of freedom

• New table: the t-table




15:

:1.341

:

:1.753

:

:2.131

:

:2.602

:

:2.947

:

One sample t-test

If test statistic is here Fail to reject H0

Distribution of the t-statistic If test

statistic is here Reject H0

– To reject the H0, you want a computed test statistics that is large• The alpha level gives us the decision criterion• New table: the t-table

• The t-statistic distribution (a transformation of the distribution of sample means transformed)




15:

:1.341

:

:1.753

:

:2.131

:

:2.602

:

:2.947

:

One sample t-test

• New table: the t-table

One tailed - or -

Two-tailed

levels

Critical values of ttcrit

Degrees of freedomdf




15:

:1.341

:

:1.753

:

:2.131

:

:2.602

:

:2.947

:

One sample t-test

• What is the tcrit for a two-tailed hypothesis test with a sample size of n = 6 and an -level of 0.05?

Distribution of the t-statistic= 0.05

Two-tailedn = 6

df = n - 1 = 5

tcrit = +2.571




15:

:1.341

:

:1.753

:

:2.131

:

:2.602

:

:2.947

:

One sample t-test

Distribution of the t-statistic= 0.05

One-tailedn = 6

df = n - 1 = 5

tcrit = +2.015

• What is the tcrit for a one-tailed hypothesis test with a sample size of n = 6 and an -level of 0.05?


One sample t-test

An example: One sample t-test

Memory experiment example:• We give a n = 16 memory patients a memory improvement treatment.

• How do they compare to the general population of memory patients who have a distribution of memory errors that is Normal, = 60?

• After the treatment they have an average score of = 55, s = 8 memory errors.

€

X

• Step 1: State your hypotheses

H0

:the memory treatment sample are the same as those in the population of memory patients.HA: they aren’t the same as those in the population of memory patients

Treatment > pop = 60

Treatment < pop = 60


One sample t-test


• Step 2: Set your decision criteria

H0: Treatment > pop = 60 HA: Treatment < pop = 60

= 0.05One -tailed



€

X



One sample t-test



• Step 2: Set your decision criteria


= 0.05One -tailed



€

X


One sample t-test



= 0.05One -tailed

• Step 3: Collect your data




€

X


One sample t-test



= 0.05One -tailed

• Step 4: Compute your test statistics

€

t =X − μ

X

sX

€

=55 − 60

816

⎛ ⎝ ⎜

⎞ ⎠ ⎟

= -2.5




€

X


One sample t-test



= 0.05One -tailed

• Step 4: Compute your test statistics

€

df = n −1

t = -2.5




€

X

€

=16 −1=15


One sample t-test



= 0.05One -tailed

€

t(15) = −2.5

• Step 5: Make a decision about your null hypothesis




€

X

€

df =15


Proportion in two tailsdf 0.20 0.10 0.05 0.02 0.01:

15:

:1.341

:

:1.753

:

:2.131

:

:2.602

:

:2.947

:

tcrit = -1.753


One sample t-test



= 0.05One -tailed

€

t(15) = −2.5

• Step 5: Make a decision about your null hypothesis




€

X

€

df =15

-1.753 = tcrit

tobs=-2.5

- Reject H0


t Test for Dependent Means

• Unknown population mean and variance• Two situations

– One sample, two scores for each person• Repeated measures design

– Two samples, but individuals in the samples are related

• Same procedure as t test for single sample, except– Use difference scores– Assume that the population mean is 0



• The related-samples t-test can be used when:

– 1 sample

t =D−D

sD

– Two scores per subject



• The related-samples t-test can be used when:

– 1 sample– Two scores per subject

t =D−D

sD

– 2 samples

– Scores are related

- OR -



€



€

sD

=sD

nD


Estimated standard error of the differences

€

df = nD −1

Test statistic

€

t =D − μ

D

sD

What are all of these “D’s” referring to?

Mean of the differences

Number of difference scores

• Difference scores– For each person, subtract one score from the other

– Carry out hypothesis testing with the difference scores

• Population of difference scores with a mean of 0– Population 2 has a mean of 0



€




Differencescores

2

65

9

22 €

t =D − μ

D

sD

PersonPre-testPost-test

1

23

4

45

5540

60

43

4935

51

(Pre-test) - (Post-test)

H0

:There is no difference between pre-test and post-test

HA: There is a difference between pre-test and post-test

D = 0

D ≠ 0



€




Differencescores

2

65

9

22= 5.5

€

t =D − μ

D

sD

PersonPre-testPost-test

1

23

4

45

5540

60

43

4935

51

€

nD = 4

(Pre-test) - (Post-test)

D∑nD

=D



€




PersonPre-testPost-testDifferencescores

1

23

4

45

5540

60

43

4935

51

2

65

9

22D = 5.5

€

t =D − μ

D

sD D

D

s

−=

5.5

€

nD = 4



€




PersonDifferencescores

1

23

4

Pre-testPost-test

45

5540

60

43

4935

51

2

65

9

22D = 5.5

€

sD

=sD

nD

€

t =D − μ

D

sD

-3.5- 5.5 =

0.5- 5.5 =-0.5- 5.5 =

3.5- 5.5 =

12.25

0.250.25

12.25

25 = SSD

D - D(D - D)2

€

sD =SSD

nD −19.2

14

25=

−=

D

D

s

−=

5.5

€

nD = 4



€




Person

1

23

4

Pre-testPost-test

45

5540

60

43

4935

51

2

65

9

22D = 5.5

€

sD

=sD

nD

€

t =D − μ

D

sD

25 = SSD

(D - D)2

€

sD =SSD

nD −19.2

14

25=

−=

D

D

s

−=

5.5 -3.5

0.5-0.5

3.5

D - DDifferencescores

12.25

0.250.25

12.25

€

nD = 4



€




Person

1

23

4

Pre-testPost-test

45

5540

60

43

4935

51

2

65

9

22D = 5.5

€

t =D − μ

D

sD

25 = SSD

(D - D)2

€

sD

=sD

nD

45.14

9.2==

2.9 = sD

D

D

s

−=

5.5 -3.5

0.5-0.5

3.5


12.25

0.250.25

12.25

€

nD = 4



€




Person

1

23

4

Pre-testPost-test

45

5540

60

43

4935

51

2

65

9

22D = 5.5

€

t =D − μ

D

sD

25 = SSD

(D - D)2

2.9 = sD

45.1

5.5 D−=

1.45 = sD

?Think back to the null hypotheses

-3.5

0.5-0.5

3.5


12.25

0.250.25

12.25

€

nD = 4



€




Person

1

23

4

Pre-testPost-test

45

5540

60

43

4935

51

2

65

9

22D = 5.5

€

t =D − μ

D

sD

25 = SSD

(D - D)2

2.9 = sD

45.1

5.5 D−=

1.45 = sD

H0: Memory performance at the post-test are equal to memory performance at the pre-test.

€

D

= 0

-3.5

0.5-0.5

3.5


12.25

0.250.25

12.25

€

nD = 4



€




Person

1

23

4

Pre-testPost-test

45

5540

60

43

4935

51

2

65

9

22D = 5.5

€

t =D − μ

D

sD

25 = SSD

(D - D)2

2.9 = sD

45.1

05.5 −=

1.45 = sD

8.3=This is our tobs

-3.5

0.5-0.5

3.5


12.25

0.250.25

12.25

€

nD = 4



€




Person

1

23

4

Pre-testPost-test

45

5540

60

43

4935

51

2

65

9

22D = 5.5

€

t =D − μ

D

sD

25 = SSD

(D - D)2

2.9 = sD

45.1

05.5 −=

1.45 = sD

8.3=tobs


Proportion in two tailsdf 0.20 0.10 0.05 0.02 0.011 3.078 6.314 12.706 31.821 63.6572 1.886 2.920 4.303 6.965 9.9253 1.638 2.353 3.182 4.541 5.8414 1.533 2.132 2.776 3.747 4.604

= 0.05Two-tailedtcrit

€

df = nD −1

=±3.18

-3.5

0.5-0.5

3.5


12.25

0.250.25

12.25

€

nD = 4


+3.18 = tcrit

- Reject H0


€




Person

1

23

4

Pre-testPost-test

45

5540

60

43

4935

51

2

65

9

22D = 5.5

€

t =D − μ

D

sD

25 = SSD

(D - D)2

2.9 = sD

45.1

05.5 −=

1.45 = sD

8.3=tobs= 0.05Two-tailedtcrit

€

df = nD −1

=±3.18

-3.5

0.5-0.5

3.5


12.25

0.250.25

12.25

€

nD = 4 tobs=3.8



€




Person

1

23

4

Pre-testPost-test

45

5540

60

43

4935

51

2

65

9

22D = 5.5

€

t =D − μ

D

sD

25 = SSD

(D - D)2

2.9 = sD

45.1

05.5 −=

1.45 = sD

8.3=tobs= 0.05Two-tailedtcrit

€

df = nD −1

=±3.18

Tobs > tcrit

so we reject the H0

-3.5

0.5-0.5

3.5


12.25

0.250.25

12.25

€

nD = 4


Next time

• Finishing up related samples t-tests

• Independent samples t-tests• Using SPSS with t-tests (maybe next Monday)

statistics for the social sciences psychology 340 spring 2005 using t-tests

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