Statistics for the Social Sciences Psychology 340 Fall 2006 Using t-tests.

Statistics for the Social Sciences

Psychology 340Fall 2006

Using t-tests


Outline (for 2 weeks)

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

– Additional assumptions• Levene’s test


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


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


Statistical analysis follows design

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

– 1 sample

t =D−μD

sD

– Two scores per subject


Statistical analysis follows design

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

– 1 sample– Two scores per subject

t =D−μD

sD

– 2 samples

– Scores are related

- OR -


Performing your statistical test

€

test statistic =observed difference

difference expected by chance

€

sD

=sD

nD

Diff. Expected by chance

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

μD = 0

μD ≠ 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

μ−=

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

μ−=

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

μ−=

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

μ−=

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 Dμ−=

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 Dμ−=

1.45 = sD

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

€

μD

= 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 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 4.604

= 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


• Using SPSSPersonPre-testPost-test

1

23

4

45

5540

60

43

4935

51


Next time

• Independent samples t-tests

Statistics for the Social Sciences Psychology 340 Fall 2006 Using t-tests.

Documents

Transcript of Statistics for the Social Sciences Psychology 340 Fall 2006 Using t-tests.