What if.... The two samples have different sample sizes (n)

What if. . . .

• The two samples have different sample sizes (n)

Results

Psychology

110

150

140

135

Sociology

90

95

80

98

Results

Psychology

110

150

140

135

Sociology

90

95

80

If samples have unequal n

• All the steps are the same!

• Only difference is in calculating the Standard Error of a Difference

Standard Error of a Difference

When the N of both samples is equal

If N1 = N2:

Sx1 - x2 =

Standard Error of a Difference

When the N of both samples is not equal

If N1 = N2:

N1 + N2 - 2

Results

Psychology

110

150

140

135

Sociology

90

95

80

X1= 535

X12=

72425

N1 = 4

X2= 265

X22=

23525

N2 = 3

N1 + N2 - 2

X1= 535

X12=

72425

N1 = 4

X2= 265

X22=

23525

N2 = 3

N1 + N2 - 2

X1= 535

X12=

72425

N1 = 4

X2= 265

X22=

23525

N2 = 3

535 265

N1 + N2 - 2

X1= 535

X12=

72425

N1 = 4

X2= 265

X22=

23525

N2 = 3

535 26572425 23525

4 + 3 - 2

X1= 535

X12=

72425

N1 = 4

X2= 265

X22=

23525

N2 = 3

535 26572425 23525

4 34 3

5

X1= 535

X12=

72425

N1 = 4

X2= 265

X22=

23525

N2 = 3

535 26572425 23525

4 34 3

71556.25 23408.33

.25+.33

5

X1= 535

X12=

72425

N1 = 4

X2= 265

X22=

23525

N2 = 3

535 26572425 23525

4 34 3

71556.25 23408.33

.25+.33197.08 (.58)

5

X1= 535

X12=

72425

N1 = 4

X2= 265

X22=

23525

N2 = 3

535 26572425 23525

4 34 3

71556.25 23408.33

.25+.33114.31

= 10.69

Practice• I think it is colder in Philadelphia than in

Anaheim ( = .10).

• To test this, I got temperatures from these two places on the Internet.

Results

Philadelphia

52

53

54

61

55

Anaheim

77

75

67

Hypotheses

• Alternative hypothesis– H1: Philadelphia < Anaheim

• Null hypothesis– H0: Philadelphia = or > Anaheim

Step 2: Calculate the Critical t

• df = N1 + N2 - 2

• df = 5 + 3 - 2 = 6 = .10

• One-tailed

• t critical = - 1.44

Step 3: Draw Critical Region

tcrit = -1.44

NowStep 4: Calculate t observed

tobs = (X1 - X2) / Sx1 - x2

6

X1= 275

X12=

15175

N1 = 5

X1 = 55

X2= 219

X22=

16043

N2 = 3

X2 = 73

275 21915175 16043

5 35 3

15125 15987.2 + .33

= 3.05

Step 4: Calculate t observed

-5.90 = (55 - 73) / 3.05

Sx1 - x2 = 3.05X1 = 55

X2 = 73

Step 5: See if tobs falls in the critical region

tcrit = -1.44

tobs = -5.90

Step 6: Decision

• If tobs falls in the critical region:

– Reject H0, and accept H1

• If tobs does not fall in the critical region:

– Fail to reject H0

Step 7: Put answer into words

• We Reject H0, and accept H1

• Philadelphia is significantly ( = .10) colder than Anaheim.

SPSS

5 55.0000 3.5355 1.5811

3 73.0000 5.2915 3.0551

PHILLY1.00

.00

TEMPN Mean

Std.Deviation

Std. ErrorMean

Group Statistics

.986 .359 -5.864 6 .001 -18.0000 3.0696 -25.5110 -10.4890

-5.233 3.104 .012 -18.0000 3.4400 -28.7437 -7.2563

Equalvariancesassumed

Equalvariancesnotassumed

TEMPF Sig.

Levene's Test forEquality of Variances

t dfSig.

(2-tailed)Mean

DifferenceStd. ErrorDifference Lower Upper

95% ConfidenceInterval of the Mean

t-test for Equality of Means

Independent Samples Test

So far. . . .

• We have been doing independent samples designs

• The observations in one group were not linked to the observations in the other group

Example

Philadelphia

52

53

54

61

55

Anaheim

77

75

67

Matched Samples Design

• This can happen with:– Natural pairs– Matched pairs– Repeated measures

Natural Pairs

The pairing of two subjects occurs naturally (e.g., twins)

Psychology (X) Sociology (Y)

Joe Smith 100 Bob Smith 90

Al Wells 110 Bill Wells 89

Jay Jones 105 Mike Jones 86

Matched Pairs

When people are matched on some variable (e.g., age)


Joe (20) 100 Bob (20) 90

Al (25) 110 Bill (25) 89

Jay (30) 105 Mike (30) 86

Repeated Measures

The same participant is in both conditions


Joe 100 Joe 90

Al 110 Al 89

Jay 105 Jay 86


• In this type of design you label one level of the variable X and the other Y

• There is a logical reason for paring the X value and the Y value


• The logic and testing of this type of design is VERY similar to what you have already done!

Example• You just invented a “magic math pill” that

will increase test scores.

• On the day of the first test you give the pill to 4 subjects. When these same subjects take the second test they do not get a pill

• Did the pill increase their test scores?

HypothesisOne-tailed

• Alternative hypothesis– H1: pill > nopill

– In other words, when the subjects got the pill they had higher math scores than when they did not get the pill

• Null hypothesis– H0: pill < or = nopill

– In other words, when the subjects got the pill their math scores were lower or equal to the scores they got when they did not take the pill

Results

Test 1 w/ Pill (X)

Mel 3

Alice 5

Vera 4

Flo 3

Test 2 w/o Pill (Y)

1

3

2

2


• N = Number of pairs

• df = N - 1

• 4 - 1 = 3 = .05

• t critical = 2.353

Step 3: Draw Critical Region

tcrit = 2.353


tobs = (X - Y) / SD


tobs = (X - Y) / SD


tobs = (X - Y) / SD

X = 3.75

Y = 2.00


tobs = (X - Y) / SD

Standard error of a difference


tobs = (X - Y) / SD

SD = SD / N

N = number of pairs

S =

S =

Test 1 w/ Pill (X)

Mel 3

Alice 5

Vera 4

Flo 3

Test 2 w/o Pill (Y)

1

3

2

2

S =

Test 1 w/ Pill (X)

Mel 3

Alice 5

Vera 4

Flo 3

Test 2 w/o Pill (Y)

1

3

2

2

Difference (D)

2

2

2

1

S =

Test 1 w/ Pill (X)

Mel 3

Alice 5

Vera 4

Flo 3

Test 2 w/o Pill (Y)

1

3

2

2

Difference (D)

2

2

2

1

D = 7

D2 =13

N = 4

S =

Test 1 w/ Pill (X)

Mel 3

Alice 5

Vera 4

Flo 3

Test 2 w/o Pill (Y)

1

3

2

2

Difference (D)

2

2

2

1

D = 7

D2 =13

N = 4

7

S =

Test 1 w/ Pill (X)

Mel 3

Alice 5

Vera 4

Flo 3

Test 2 w/o Pill (Y)

1

3

2

2

Difference (D)

2

2

2

1

D = 7

D2 =13

N = 4

713

S =

Test 1 w/ Pill (X)

Mel 3

Alice 5

Vera 4

Flo 3

Test 2 w/o Pill (Y)

1

3

2

2

Difference (D)

2

2

2

1

D = 7

D2 =13

N = 4

713

4

4 - 1

S =

Test 1 w/ Pill (X)

Mel 3

Alice 5

Vera 4

Flo 3

Test 2 w/o Pill (Y)

1

3

2

2

Difference (D)

2

2

2

1

D = 7

D2 =13

N = 4

713

4

3

12.25

.5 =

Test 1 w/ Pill (X)

Mel 3

Alice 5

Vera 4

Flo 3

Test 2 w/o Pill (Y)

1

3

2

2

Difference (D)

2

2

2

1

D = 7

D2 =13

N = 4

7

4

3

.75


tobs = (X - Y) / SD

SD = SD / N

N = number of pairs


tobs = (X - Y) / SD

.25=.5 / 4

N = number of pairs


7.0 = (3.75 - 2.00) / .25


tcrit = 2.353


tcrit = 2.353tobs = 7.0

Step 6: Decision






• Reject H0, and accept H1

• When the subjects took the “magic pill” they received statistically ( = .05) higher math scores than when they did not get the pill

SPSS

3.7500 4 .9574 .4787

2.0000 4 .8165 .4082

TIME1

TIME2

Pair 1Mean N

Std.Deviation

Std. ErrorMean

Paired Samples Statistics

4 .853 .147TIME1 &TIME2

Pair 1N Correlation Sig.

Paired Samples Correlations

1.7500 .5000 .2500 .9544 2.5456 7.000 3 .006TIME1 -TIME2

Pair 1Mean

Std.Deviation

Std. ErrorMean Lower Upper

95% ConfidenceInterval of the Difference

Paired Differences

t dfSig.

(2-tailed)

Paired Samples Test

Practice• You just created a new program that is suppose

to lower the number of aggressive behaviors a child performs.

• You watched 6 children on a playground and recorded their aggressive behaviors. You gave your program to them. You then watched the same children and recorded this aggressive behaviors again.

Practice

• Did your program significantly lower ( = .05) the number of aggressive behaviors a child performed?

Results

Time 1 (X)

Child1 18

Child2 11

Child3 19

Child4 6

Child5 10

Child6 14

Time 2 (Y)

16

10

17

4

11

12

HypothesisOne-tailed

• Alternative hypothesis– H1: time1 > time2

• Null hypothesis– H0: time1 < or = time2


• N = Number of pairs

• df = N - 1

• 6 - 1 = 5 = .05

• t critical = 2.015


tobs = (X - Y) / SD

1.21 =

(D)

2

1

2

2

-1

2

D = 8

D2 =18

N = 6

818

6

6 - 1

Time 1 (X)

Child1 18

Child2 11

Child3 19

Child4 6

Child5 10

Child6 14

Test 2 (Y)

16

10

17

4

11

12


tobs = (X - Y) / SD

.49=1.21 / 6

N = number of pairs


2.73 = (13 - 11.66) / .49

X = 13

Y = 11.66

SD = .49


tcrit = 2.015tobs = 2.73

Step 6: Decision






• Reject H0, and accept H1

• The program significantly ( = .05) lowered the number of aggressive behaviors a child performed.

SPSS

13.0000 6 4.9800 2.0331

11.6667 6 4.6762 1.9090

CTIME1

CTIME2

Pair 1Mean N

Std.Deviation

Std. ErrorMean

Paired Samples Statistics

6 .970 .001CTIME1& CTIME2

Pair 1N Correlation Sig.

Paired Samples Correlations

1.3333 1.2111 .4944 6.240E-02 2.6043 2.697 5 .043CTIME1 -CTIME2

Pair 1Mean

Std.Deviation

Std. ErrorMean Lower Upper

95% ConfidenceInterval of the Difference

Paired Differences

t dfSig.

(2-tailed)

Paired Samples Test

New Step

• Should add a new page

• Determine if – One-sample t-test– Two-sample t-test

• If it is a matched samples design

• If it is a independent samples with equal N

• If it is a independent samples with unequal N

Thus, there are 4 different kinds of designs

• Each design uses slightly different formulas

• You should probably make up ONE cook book page (with all 7 steps) for each type of design– Will help keep you from getting confused on a

test

Practice

• A research study was conducted to examine whether or not there were differences between older and younger adults on perceived life satisfaction. A pilot study was conducted to examine this hypothesis. Ten older adults (over the age of 70) and ten younger adults (between 20 and 30) were give a life satisfaction test (known to have high reliability and validity). Scores on the measure range from 0 to 60 with high scores indicative of high life satisfaction; low scores indicative of low life satisfaction.

• Older Adults= 44.5; S = 8.68; n = 10

• Younger Adults = 28.1; S = 8.54; n = 10

Practice

• Dr. Willard is studying the effects of a new drug (Drug-Y) on learning improvement following traumatic brain injury. To study this, Dr. Willard takes six rats and lesions a part of the brain responsible for learning. He puts each rat in a maze and counts the number of times it takes each rat to navigate through the maze without making a mistake. Dr. Willard then puts each rat on a regimen of Drug-Y for one week. After one week, he places each rat in a similar maze and counts the number of times it takes each rat to navigate through the maze without a mistake. Examine if Drug-Y had a positive impact on rats performance.

t Time 1 Time 2

Ben 28 24

Splinter 29 26

George 30 22

Jerry 33 30

Fievel 34 29

Patches 32 28

Practice

• A sample of ten 9th grades at James Woods High School can do an average of 11.5 pull-ups (chin-ups) in 30 seconds, with a sample standard deviation of s = 3 The US Department of Health and Human Services suggests that 9th grades be able to do a minimum of 9 pull-ups in 30 seconds, if not, they're watching too much Family Guy. Is this sample of 9th grades able to do significantly (alpha = .01) more pull-ups than the number recommended by the US Department of Health and Human Services?

• tobs = 4.257

• tcrit = 2.101

• There age is related to life satisfaction.

• t = (31 - 26.5)/0.764 = 4.500/0.764 = 5.890

• Because the obtained t-Value is larger that the critical t-Value, the mean difference between the number of maze navigation's at Time 1 and Time 2 is statistically significant. Thus, we can conclude that Drug-Y lead to a statistically significant decrease in the number of times it took rats to navigate through a maze without making a mistake.

• The obtained t-Value is (11.5 - 9)/1 = 2.5/1 = 2.500

•

• 6) Because the obtained t-Value (2.500) is less than the critical t-Value (2.689), the difference between the mean number of pull-ups that 9th grades from James Woods High School can do is not significantly greater than the number of pull-ups recommended by the US Department of Health and Human Services.

Cookbook

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What if.... The two samples have different sample sizes (n)

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