Using t-tests (independent samples) Statistics for the Social Sciences Psychology 340 Spring 2010.

Using t-tests (independent samples)

Statistics for the Social SciencesPsychology 340

Spring 2010

PSY 340Statistics for the

Social Sciences

Statistical analysis follows design

• The one-sample t-test can be used when:– 1 sample

– One score per subject

– Population mean (μ) is known

– but Population standard deviation (σ) is NOT known

€

t =X − μ

X

sX


Social Sciences Independent samples

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

Memory treatment

Memory patients Memory

Test

• 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:

Memory placebo

MemoryTest

Compare these two means

XA

XB


Social Sciences

Statistical analysis follows design

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

– 2 samples

– Samples are independent

€

t =(X A − X B ) − (μA − μB )

sX A −X B


Social Sciences Performing your statistical test

€

test statistic =observed difference

difference expected by chance

Test statistic

€

t =(X A − X B ) − (μA − μB )

sX A −X B

€

t =X − μ

X

sX

One-sample tIndependent-samples t

Observed (sample) means



€



Test statistic

€

t =(X A − X B ) − (μA − μB )

sX A −X B

€

t =X − μ

X

sX


Hypothesized population means• from the Null hypothesis



€



Test statistic

€

t =(X A − X B ) − (μA − μB )

sX A −X B

€

t =X − μ

X

sX


Hypothesized population means• from the Null hypothesis

H0: Memory performance by the treatment group is equal to memory performance by the no treatment group.

So: (μA −μB) =0



€



Test statistic

€

t =X − μ

X

sX

One-sample t

€

t =(X A − X B ) − (μA − μB )

sX A −X B

Estimated standard error(difference expected by chance)

estimate is based on one sample

We have two samples, so the estimate is based

on two samples

The Estimate of the Standard Error is based on the variability of both

samples



€

sX A −X B

=sp

2

nA

+sp

2

nB

“pooled variance”We combine the

variance from the two samples

Number of subjects in group A

Number of subjects in group B


Social Sciences

s2 =SS

n−1variance

Performing your statistical test

€

sX A −X B

=sp

2

nA

+sp

2

nB

“pooled variance”We combine the

variance from the two samples

Recall “weighted means,” need to use

“weighted variances” here

€

sp2 =

SSA + SSB

dfA + dfB

sp2 =

sA2dfA( ) + sB

2dfB( )dfA +dfB

dfA =(nA −1)dfB =(nB −1)

Variance (s2) * degrees of freedom (df)

s2 (n −1) =SS

s2df =SS



€

sX A −X B

=sp

2

nA

+sp

2

nB

€

df = nA + nB − 2

€

t =(X A − X B ) − (μA − μB )

sX A −X B

€

=(nA −1) + (nB −1)

€

sp2 =

SSA + SSB

dfA + dfB

Independent-samples t• Compute your estimated standard error

sp2 =

sA2dfA( ) + sB

2dfB( )dfA +dfB

• Compute your t-statistic

• Compute your degrees of freedom

dfA =(nA −1)dfB =(nB −1)This is the one you use

to look up your tcrit



Exp.

groupControl group

45

5540

60

43

4935

51

Need to compute the mean and variability for each sample

Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples. He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use α = 0.05.



Exp.

groupControl group

45

5540

60

43

4935

51


Control group

= 50

(45-50)2 + (55-50)2 + (40-50)2 + (60-50)2

= 250

SS =A


XA =45 + 55 + 40 + 60

4

XA =50SSA =250



Exp. group

(43-44.5)2 + (49- 44.5)2 + (35- 44.5)2 + (51- 44.5)2

= 155

SS =B

Exp.

groupControl group

45

5540

60

43

4935

51



XB =43+ 49 + 35 + 51

4

XA =50SSA =250

XB =44.5SSB =155

= 44.5



€

sX A −X B

=sp

2

nA

+sp

2

nB

€

t =(X A − X B ) − (μA − μB )

sX A −X B

€

sp2 =

SSA + SSB

dfA + dfB

€

=250 +155

3+ 3= 67.5€

=67.5

4+

67.5

4= 5.81€

=(50 − 44.5) − (0)

5.81

€

dfA = (nA −1)

€

dfB = (nB −1)

Exp.

groupControl group

45

5540

60

43

4935

51XA =50

SSA =250XB =44.5

SSB =155


= 0.95



Tobs= 0.95Tcrit= ±2.447

€

sX A −X B

= 5.81

€

sp2 = 67.5

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 0.01: : : : : :5 1,476 2.015 2.571 3.365 4.0326 1.440 1.943 2.447 3.143 3.707: : : : : :

€

df = nA + nB − 2 = 6

α = 0.05 Two-tailed

Exp.

groupControl group

45

5540

60

43

4935

51


€

t =(X A − X B ) − (μA − μB )

sX A −X B

€

=(50 − 44.5) − (0)

5.81

€

dfA = (nA −1)

€

dfB = (nB −1)

XA =50SSA =250

XB =44.5SSB =155

= 0.95



Tobs= 0.95α = 0.05 Two-tailed Tcrit= ±2.447

Exp.

groupControl group

45

5540

60

43

4935

51

€

sX A −X B

= 5.81

€

sp2 = 67.5

€

df = nA + nB − 2 = 6


€

t =(X A − X B ) − (μA − μB )

sX A −X B

€

=(50 − 44.5) − (0)

5.81

€

dfA = (nA −1)

€

dfB = (nB −1)

XA =50SSA =250

XB =44.5SSB =155

+2.45 = tcrit

- Fail to Reject H0

tobs=0.95

= 0.95


Social Sciences Assumptions: Independent samples t

• Each of the population distributions follows a normal curve (this is an assumption of all t-tests)– T-tests are fairly ‘robust’ against this assumption

• This means that the results generally still hold even if this assumption is violated

• Homogeneity of variance: The two populations have the same variance– SPSS tests this using Levene’s Test

• Two rows in the SPSS output– Us the top row if the p-value for the Levene’s test is greater than 0.05– Use the bottom row if the p-value for the Levene’s test is less than 0.05

• Tests the Null hypothesis that the two groups have equal variances


Social SciencesEffect Size for the t Test for Independent Means

• Estimated effect size after a completed study

Estimated d =X1 −X2

sPooled

“pooled standard deviation”

not “pooled variance,” so take the square root of sP2


Social Sciences

Power for the t Test for Independent Means (.05 significance level)

8-4


Social Sciences

Approximate Sample Size Needed for 80% Power (.05 significance level)


Social Sciences Statistical Tests Summary

Design Statistical test (Estimated) Standard error

t =(XA −XB)−(μA −μB)

sXA −XB

sXA −XB=

sP2

nA

+sP2

nB

sX =sn

t =X−μX

sX

zX =X−μX

σ X

σ X =σ

nOne sample, σ known

One sample, σ unknown

Two related samples, σ unknown

Two independent samples, σ unknown €

sD

=sD

nD

t =D−μD

sD

Next time


Social Sciences Using SPSS: Independent samples t

• Entering the data– Different groups of observations go into

SAME column• e.g., Exp grp and control grp in a single

column

Person Cntrl-grp Exp-grp

1

23

4

45

5540

60

43

4935

51

• Performing the analysis– Analyze -> Compare means -> independent samples t-test

– Identify which columns have the observations (test variable) and which column has the group membership defined (grouping variable)

– Define groups: what numbers correspond to the two groups?

• Reading the output– Means of the different groups, the mean difference, the computed-t,

degrees of freedom, p-value (Sig.), Levene’s test

– Separate column defines the group membership for each observation– e.g., exp grp = 0, control grp = 1

Using t-tests (independent samples) Statistics for the Social Sciences Psychology 340 Spring 2010.

Documents

sample ttest

samples samples

independent samples

memory placebo memory

memory treatment sample

treatment group

statistical test t obs

sample t independentsamples