hypothesis tests: two independent samples cal state northridge 320 andrew ainsworth phd

Hypothesis Tests: Two Independent Samples

Cal State Northridge320

Andrew Ainsworth PhD

2

Major Points

•What are independent samples?

•Distribution of differences between means

•An example •Heterogeneity of Variance•Effect size•Confidence limits

Psy 320 - Cal State Northridge


3

Independent Samples

•Samples are independent if the participants in each sample are not related in any way

•Samples can be considered independent for 2 basic reasons▫First, samples randomly selected from 2 separate populations


4

Independent Samples

Ran

dom

S

elec

tionPopulation #1

Sample #1R

ando

m

Sel

ectio

n

Population #2

Sample #2

•Getting Independent Samples▫Method #1


5

Independent Samples

•Samples can be considered independent for 2 basic reasons▫Second, participants are randomly selected from a single population

▫Subjects are then randomly assigned to one of 2 samples


6

Independent Samples

Sample #1 Sample #2

Population

Random Selection and Random Assignment

•Getting Independent Samples▫Method #2


7

Sampling Distributions Again•If we draw all possible pairs of independent samples of size n1 and n2, respectively, from two populations.

•Calculate a mean and in each one.

•Record the difference between these values.

•What have we created?•The Sampling Distribution of the Difference Between Means

1X 2X


8

Sampling Distribution of the Difference Between Means

•Mean of sampling distribution is 1

•Mean of sampling distribution is 2

•So, the mean of a sampling distribution is 1- 2

Sample 1

Population 1

Sample 2

Sample 1 Sample 2

Sample 1 Sample 2

Sample 1 Sample 2

1

2

3

∞

... ...

Calculate Statistic

12 22X X

13 23X X

11 21X X

1 2X X

Population 2

1X

2X

1 2X X


9


•Shape▫approximately normal if

both populations are approximately normal▫or

N1 and N2 are reasonably large.


10

Sampling Distribution of the Difference Between Means•Variability (variance sum rule)

▫The variance of the sum or difference of 2 independent samples is equal to the sum of their variances

1 2

1 2 1 2

1 2 1

1 2 1

2 22 2 2

1 2

2 2

1 2 1

2 2 2

; Remembering that

X X

X X X X

X X X

X X X

n n

n n n

Think Pathagoras c a b


11


•Sampling distribution of

1 - 2

2 21 2

1 2n n

1 2X X


12

Converting Mean Differences into Z-Scores•In any normal distribution with a known mean and standard deviation, we can calculate z-scores to estimate probabilities.

•What if we don’t know the ’s? •You got it, we guess with s!

1 2

1 2 1 2( ) ( )

X X

X Xz


13

Converting Mean Differences into t-scores

•We can use to estimate

•If,▫The variances of the samples are roughly

equal (homogeneity of variance)▫We estimate the common variance by

pooling (averaging)

1 2

1 2 1 2( ) ( )

X X

X Xt

s

1 2X Xs 1 2X X


14

Homogeneity of Variance

•We assume that the variances are equal because:▫If they’re from the same population they

should be equal (approximately)▫If they’re from different populations they

need to be similar enough for us to justify comparing them (“apples to oranges” and all that)


15

Pooling Variances•So far, we have taken 1 sample estimate

(e.g. mean, SD, variance) and used it as our best estimate of the corresponding population parameter

•Now we have 2 samples, the average of the 2 sample statistics is a better estimate of the parameter than either alone


16

Pooling Variances

•Also, if there are 2 groups and one groups is larger than it should “give” more to the average estimate (because it is a better estimate itself)

•Assuming equal variances and pooling allows us to use an estimate of the population that is smaller than if we did not assume they were equal


17

Calculating Pooled Variance

•Remembering that

sp2 is the pooled estimate: we pool SS and df to get

2

1

SS SSs

df n

2 1 2 1 2 1 2

1 2 1 2 1 2( 1) ( 1) 2pooled

ppooled

SS SS SS SS SS SS SSs

df df df n n n n


18

Calculating Pooled Variance

2 1 2

1 2 2p

SS SSs

n n

21 1 1

22 2 2

Method 1

( )

( )

i

i

SS X X

SS X X

21 1 1

22 2 2

Method 2

( 1)

( 1)

SS n s

SS n s


19

Calculating Pooled Variance•When sample sizes are equal (n1 = n2) it is

simply the average of the 2

•When unequal n we need a weighted average

1 2

2 22

2X X

p

s ss

2 22 1 1 2 2

1 2

( 1) ( 1)

2p

n s n ss

n n


20

Calculating SE for the Differences Between Means

•Once we calculate the pooled variance we just substitute it in for the sigmas earlier

•If n1 = n2 then

1 2

2 2

1 2

p p

X X

s ss

n n

1 2

2 2 2 21 2

1 2 1 2

p p

X X

s s s ss

n n n n


21

Example: Violent Videos Games•Does playing violent video games increase

aggressive behavior•Two independent randomly

selected/assigned groups▫Grand Theft Auto: San Andreas (violent: 8

subjects) VS. NBA 2K7 (non-violent: 10 subjects)

▫We want to compare mean number of aggressive behaviors following game play


22

Hypotheses (Steps 1 and 2)•2-tailed

▫h0: 1 - 2 = 0 or Type of video game has no impact on aggressive behaviors

▫h1: 1 - 2 0 or Type of video game has an impact on aggressive behaviors

•1-tailed▫h0: 1 - 2 ≤ 0 or GTA leads to the same or

less aggressive behaviors as NBA▫h1: 1 - 2 > 0 or GTA leads to more

aggressive behaviors than NBA


23

Distribution, Test and Alpha(Steps 3 and 4)•We don’t know for either group, so we

cannot perform a Z-test•We have to estimate with s so it is a t-

test (t distribution)•There are 2 independent groups•So, an independent samples t-test•Assume alpha = .05


24

Decision Rule (Step 5)•dftotal = (n1 – 1) + (n2 – 1) = n1 + n2 – 2

▫Group 1 has 8 subjects – df = 8 – 1= 7▫Group 2 has 10 subjects – df = 10 – 1 = 9

•dftotal = (n1 – 1) + (n2 – 1) = 7 + 9 = 16 OR dftotal = n1 + n2 – 2 = 8 + 10 – 2 = 16

•1-tailed t.05(16) = ____; If to > ____ reject 1-tailed

•2-tailed t.05(16) = ____; If to > ____ reject 2-tailed

t Distribution

df1-tailed 0.1 0.05 0.025 0.01 0.005 0.00052-tailed 0.2 0.1 0.05 0.02 0.01 0.001

1 3.078 6.314 12.706 31.821 63.657 636.6192 1.886 2.92 4.303 6.965 9.925 31.5983 1.683 2.353 3.182 4.5415 5.841 12.9414 1.533 2.132 2.776 3.747 4.604 8.615 1.476 2.015 2.571 3.365 4.032 6.8596 1.44 1.943 2.447 3.143 3.707 5.9597 1.415 1.895 2.365 2.998 3.499 5.4058 1.397 1.86 2.306 2.896 3.355 5.0419 1.383 1.833 2.262 2.821 3.25 4.78110 1.372 1.812 2.228 2.764 3.169 4.58711 1.363 1.796 2.201 2.718 3.106 4.43712 1.356 1.782 2.179 2.681 3.055 4.31813 1.35 1.771 2.16 2.65 3.012 4.22114 1.345 1.761 2.145 2.624 2.977 4.1415 1.341 1.753 2.131 2.602 2.947 4.07316 1.337 1.746 2.12 2.583 2.921 4.01517 1.333 1.74 2.11 2.567 2.898 3.96518 1.33 1.734 2.101 2.552 2.878 3.92219 1.328 1.729 2.093 2.539 2.861 4.88320 1.325 1.725 2.086 2.528 2.845 3.8521 1.323 1.721 2.08 2.518 2.831 3.81922 1.321 1.717 2.074 2.508 2.819 3.79223 1.319 1.714 2.069 2.5 2.807 3.76724 1.318 1.711 2.064 2.492 2.797 3.74525 1.316 1.708 2.06 2.485 2.787 3.725

Critical Values of Student's t 25


26

Calculate to (Step 6)Subject GTA NBA

1 12 92 9 93 11 84 13 65 8 116 10 77 9 88 10 119 8

10 7

Mean 10.25 8.4s 1.669 1.647

s2 2.786 2.713

Data


27

Calculate to (Step 6)

•We have means of 10.25 (GTA) and 8.4 (NBA), but both of these are sample means.

•We want to test differences between sample means.▫Not between a sample and a population

mean


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•The t for 2 groups is:

•But under the null 1 - 2 = 0, so:1 2

1 2 1 2( ) ( )

X X

X Xt

s

1 2

1 2( )

X X

X Xt

s


29


•We need to calculate and to do that we need to first calculate

2 22 1 1 2 2

1 2

( 1) 1 __(____) __ ____

2 __ __ 2

_____ _____ _________

__ __

p

n s n ss

n n

1 2X Xs 2ps


30

Calculate to (Step 6)•Since is _____ we can insert this into

the equation

1 2

1 2

2 2

1 2

1 2

____ ____

__ __

___ ___ ____

,

___ ___ ______

___ ___

p pX X

o

X X

s ss

n n

Therefore

X Xt

s

1 2X Xs

2ps


31

Make your decision(Step 7)•Since ____ > ____ we would reject the null

hypothesis under a 2-tailed test•Since ____ > ____ we would reject the null

hypothesis under a 1-tailed test•There is evidence that violent video

games increase aggressive behaviors


32

Heterogeneous Variances•Refers to case of unequal population

variances•We don’t pool the sample variances just

add them•We adjust df and look t up in tables for

adjusted df•For adjustment use Minimum

df = smaller n - 1


33

Effect Size for Two Groups•Extension of what we already know.•We can often simply express the effect as

the difference between means (e.g. 1.85)•We can scale the difference by the size of

the standard deviation.▫Gives d (aka Cohen’s d)▫Which standard deviation?


34

Effect Size•Use either standard deviation or their

pooled average.▫We will pool because neither has any claim

to priority.▫Pooled st. dev. = 2.745 1.657


35

Effect Size, cont.10.25 8.4

1.657

1.851.12

1.657

viol nonviol

pooled

X Xd

s

•This difference is approximately 1.12 standard deviations

•This is a very large effect


36

Confidence Limits

1 21 2.95 .05,2 *

(____ ____) ____ ____

____ ____

____ ____

tailed X XCI X X t s


37

Confidence Limits•p = .95 that based on the sample values the

interval formed in this way includes the true value of 1 - 2

•The probability that interval includes 1 - 2

given the value of•Does 0 fall in the interval? What does that

mean?

1 2X X


38

Computer Example

•SPSS reproduces the t value and the confidence limits.

•All other statistics are the same as well.•Note different df depending on

homogeneity of variance. •Remember: Don’t pool if heterogeneity

and modify degrees of freedom


39

SPSS outputGro u p Sta tis tic s

8 1 0 .2 5 0 0 1 .6 6 9 0 5 .5 9 0 1 0

1 0 8 .4 0 0 0 1 .6 4 6 5 5 .5 2 0 6 8

GAME1 .0 0 g ta

2 .0 0 n b a

AGGRESSN Me a n Std . De v i a t i o n

Std . Erro rMe a n

Ind e p e n de nt Sa mple s Te s t

.0 0 5 .9 4 2 2 .3 5 5 1 6 .0 3 2 1 .8 5 0 0 .7 8 5 7 1 .1 8 4 3 6 3 .5 1 5 6 4

2 .3 5 1 1 5 .0 4 8 .0 3 3 1 .8 5 0 0 .7 8 6 9 7 .1 7 3 0 8 3 .5 2 6 9 2

Eq u a l v a ria n c e sa s s u me d

Eq u a l v a ria n c e sn o t a s s u me d

AGGRESSF Sig .

L e v e n e 's T e s t fo rEq u a lity o f Va ria n c e s

t d f Sig . (2 -ta ile d )Me a n

Diffe re n c eStd . Erro r

Diffe re n c e L o we r Up p e r

9 5 % Co n fid e n c eIn te rv a l o f th eDiffe re n c e

t-te s t fo r Eq u a lity o f Me a n s

hypothesis tests: two independent samples cal state northridge 320 andrew ainsworth phd

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