Example You give 100 random students a questionnaire designed to measure attitudes toward living in...

Example• You give 100 random students a questionnaire

designed to measure attitudes toward living in dormitories

• Scores range from 1 to 7 – (1 = unfavorable; 4 = neutral; 7 = favorable)

• You wonder if the mean score of the population is different then 4

Hypothesis

• Alternative hypothesis– H1: sample = 4

– In other words, the population mean will be different than 4

Hypothesis


• Null hypothesis– H0: sample = 4

– In other words, the population mean will not be different than 4

Results

• N = 100

• X = 4.51

• s = 1.94

• Notice, your sample mean is consistent with H1, but you must determine if this difference is simply due to chance

Results

• N = 100

• X = 4.51

• s = 1.94

• To determine if this difference is due to chance you must calculate an observed t value

Observed t-value

tobs = (X - ) / Sx

Observed t-value

tobs = (X - ) / Sx

This will test if the null hypothesis H0: sample = 4 is true

The bigger the tobs the more likely that H1: sample = 4 is true

Observed t-value

tobs = (X - ) / Sx

Sx = S / N

Observed t-value

tobs = (X - ) / .194

.194 = 1.94/ 100

Observed t-value

tobs = (4.51 – 4.0) / .194

Observed t-value

2.63 = (4.51 – 4.0) / .194

t distribution

t distribution

tobs = 2.63

t distribution

tobs = 2.63

Next, must determine if this t value happened due to chance or if represent a real difference in means. Usually, we want to be 95% certain.

t critical

• To find out how big the tobs must be to be significantly different than 0 you find a tcrit value.

• Calculate df = N - 1

• Page 747– First Column are df

– Look at an alpha of .05 with two-tails

t distribution

tobs = 2.63

t distribution

tobs = 2.63

tcrit = 1.98tcrit = -1.98

t distribution

tobs = 2.63


If tobs fall in critical area reject the null hypothesis

Reject H0: sample = 4

t distribution

tobs = 2.63


If tobs does not fall in critical area do not reject the null hypothesis

Do not reject H0: sample = 4

Decision

• Since tobs falls in the critical region we reject Ho and accept H1

• It is statistically significant, students tend to think favorably about living in the dorms.

• p < .05

Example• You wonder if the average IQ score of students at

Villanova significantly different (at alpha = .05)than the average IQ of the population (which is 100). You sample the students in this room.

• N = 54

• X = 130

• s = 18.4

The Steps

• Try to always follow these steps!

Step 1: Write out Hypotheses


• Null hypothesis– H0: sample = 100

Step 2: Calculate the Critical t

• N = 54

• df = 53 = .05

• tcrit = 2.0

Step 3: Draw Critical Region


Step 4: Calculate t observed

tobs = (X - ) / Sx


tobs = (X - ) / Sx

Sx = S / N


tobs = (X - ) / Sx

2.5 = 18.4 / 54


tobs = (X - ) / Sx

12 = (130 - 100) / 2.52.5 = 18.4 / 54

Step 5: See if tobs falls in the critical region




tobs = 12

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.

• The average IQ of students at Villanova is statistically different ( = .05) than the average IQ of the population.

Practice

• You recently finished giving 5 of your friends the MMPI paranoia measure. Is your friends average average paranoia score significantly ( = .10) different than the average paranoia of the population ( = 56.1)?

Scores

Person Score

Charlie 55

Lucy 49

Sally 58

Schroeder 60

Franklin 54


• Alternative hypothesis– H1: sample = 56.1

• Null hypothesis– H0: sample = 56.1


• N = 5

• df =4 = .10

• tcrit = 2.132


tcrit = 2.132tcrit = -2.132


tobs = (X - ) / Sx

-.48 = (55.2 - 56.1) / 1.88 1.88 = 4.21/ 5


tcrit = 2.132tcrit = -2.132

tobs = -.48

Step 6: Decision






• We fail to reject H0

• The average paranoia of your friends is not statistically different ( = .10) than the average paranoia of the population.

SPSS

5 55.2000 4.2071 1.8815MMPIN Mean

Std.Deviation

Std. ErrorMean

One-Sample Statistics

-.478 4 .657 -.9000 -6.1239 4.3239MMPIt df

Sig.(2-tailed)

MeanDifference Lower Upper

95% ConfidenceInterval of the Difference

Test Value = 56.1

One-Sample Test

One-tailed test

• In the examples given so far we have only examined if a sample mean is different than some value

• What if we want to see if the sample mean is higher or lower than some value

• This is called a one-tailed test

Remember

• You recently finished giving 5 of your friends the MMPI paranoia measure. Is your friends average paranoia score significantly ( = .10) different than the average paranoia of the population ( = 56.1)?

Hypotheses

• Alternative hypothesis– H1: sample = 56.1

• Null hypothesis– H0: sample = 56.1

What if. . .

• You recently finished giving 5 of your friends the MMPI paranoia measure. Is your friends average paranoia score significantly ( = .10) lower than the average paranoia of the population ( = 56.1)?

Hypotheses

• Alternative hypothesis– H1: sample < 56.1

• Null hypothesis– H0: sample = or > 56.1


• N = 5• df =4 = .10• Since this is a “one-tail” test use the one-tailed

column– Note: one-tail = directional test

• tcrit = -1.533– If H1 is < then tcrit = negative– If H1 is > then tcrit = positive


tcrit = -1.533


tobs = (X - ) / Sx


tobs = (X - ) / Sx

-.48 = (55.2 - 56.1) / 1.88 1.88 = 4.21/ 5


tcrit = -1.533


tcrit = -1.533

tobs = -.48

Step 6: Decision






• We fail to reject H0

• The average paranoia of your friends is not statistically less then ( = .10) the average paranoia of the population.

Practice• You just created a “Smart Pill” and you gave it to

150 subjects. Below are the results you found. Did your “Smart Pill” significantly ( = .05) increase the average IQ scores over the average IQ of the population ( = 100)?

• X = 103• s = 14.4


• Alternative hypothesis– H1: sample > 100

• Null hypothesis– H0: sample < or = 100


• N = 150

• df = 149 = .05

• tcrit = 1.645


tcrit = 1.645


tobs = (X - ) / Sx

2.54 = (103 - 100) / 1.181.18=14.4 / 150


tcrit = 1.645

tobs = 2.54

Step 6: Decision






• We reject H0 and accept H1.

• The average IQ of the people who took your “Smart Pill” is statistically greater ( = .05) than the average IQ of the population.

So far. . .

• We have been doing hypothesis testing with a single sample

• We find the mean of a sample and determine if it is statistically different than the mean of a population

Basic logic of research

Start with two equivalent groups of subjects

D ep en d en t V ariab leIf p e rson lives

E xp erim en ta l G rou pG ive m ed ica tion

S u b jec ts


C on tro l G rou pD o n o t g ive m ed ica tion

S u b jec ts

Treat them alike except for one thing



S u b jec ts



S u b jec ts

See if both groups are different at the end



S u b jec ts



S u b jec ts

Notice

• This means that we need to see if two samples are statistically different from each other

• We can use the same logic we learned earlier with single sample hypothesis testing

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

will increase test scores.

• You give the pill to 4 subjects and another 4 subjects get no pill

• You then examine their final exam grades

HypothesisTwo-tailed

• Alternative hypothesis– H1: pill = nopill

– In other words, the means of the two groups will be significantly different

• Null hypothesis– H0: pill = nopill

– In other words, the means of the two groups will not be significantly different

HypothesisOne-tailed

• Alternative hypothesis– H1: pill > nopill

– In other words, the pill group will score higher than the no pill group

• Null hypothesis– H0: pill < or = nopill

– In other words, the pill group will be lower or equal to the no pill group

For current example, lets just see if there is a difference

• Alternative hypothesis– H1: pill = nopill

– In other words, the means of the two groups will be significantly different

• Null hypothesis– H0: pill = nopill

– In other words, the means of the two groups will not be significantly different

Results

Pill Group

5

3

4

3

No Pill Group

1

2

4

3

Remember before. . . Step 2: Calculate the Critical t

• df = N -1

NowStep 2: Calculate the Critical t

• df = N1 + N2 - 2

• df = 4 + 4 - 2 = 6 = .05

• t critical = 2.447


tcrit = 2.447tcrit = -2.447

Remember before. . .Step 4: Calculate t observed

tobs = (X - ) / Sx

NowStep 4: Calculate t observed

tobs = (X1 - X2) / Sx1 - x2


tobs = (X1 - X2) / Sx1 - x2

• X1 = 3.75

• X2 = 2.50


tobs = (X1 - X2) / Sx1 - x2

Standard Error of a Difference

Sx1 - x2

When the N of both samples are equal

If N1 = N2:

Sx1 - x2 = Sx12 + Sx2

2

Results

Pill Group

5

3

4

3

No Pill Group

1

2

4

3

Standard Deviation

S =-1

Standard Deviation

Pill Group

5

3

4

3

No Pill Group

1

2

4

3

X1= 15

X12= 59

X2= 10

X22= 30

Standard Deviation

Pill Group

5

3

4

3

No Pill Group

1

2

4

3

S = .96 S = 1.29

X1= 15

X12= 59

X2= 10

X22= 30

Standard Deviation

Pill Group

5

3

4

3

No Pill Group

1

2

4

3

S = .96 S = 1.29

Sx= .48 Sx= . 645

X1= 15

X12= 59

X2= 10

X22= 30


Sx1 - x2


If N1 = N2:

Sx1 - x2 = Sx12 + Sx2

2


Sx1 - x2


If N1 = N2:

Sx1 - x2 = (.48)2 + (.645)2


Sx1 - x2


If N1 = N2:

Sx1 - x2 = (.48)2 + (.645)2= .80

Standard Error of a Difference Raw Score Formula


If N1 = N2:

Sx1 - x2 =

Sx1 - x2 =

X1= 15

X12= 59

N1 = 4

X2= 10

X22= 30

N2 = 4

Sx1 - x2 =

X1= 15

X12= 59

N1 = 4

X2= 10

X22= 30

N2 = 4

15 10

Sx1 - x2 =

X1= 15

X12= 59

N1 = 4

X2= 10

X22= 30

N2 = 4

15 1059 30

Sx1 - x2 =

X1= 15

X12= 59

N1 = 4

X2= 10

X22= 30

N2 = 4

15 1059 304 4

4 (4 - 1)

Sx1 - x2 =

X1= 15

X12= 59

N1 = 4

X2= 10

X22= 30

N2 = 4

15 1059 304 4

12

56.25 25

X1= 15

X12= 59

N1 = 4

X2= 10

X22= 30

N2 = 4

15 1059 304 4

12

56.25 257.75.80 =


tobs = (X1 - X2) / Sx1 - x2

Sx1 - x2 = .80X1 = 3.75

X2 = 2.50


tobs = (3.75 - 2.50) / .80

Sx1 - x2 = .80X1 = 3.75

X2 = 2.50


1.56 = (3.75 - 2.50) / .80

Sx1 - x2 = .80X1 = 3.75

X2 = 2.50


tcrit = 2.447tcrit = -2.447


tcrit = 2.447tcrit = -2.447

tobs = 1.56

Step 6: Decision






• We fail to reject H0.

• The final exam grades of the “pill group” were not statistically different ( = .05) than the final exam grades of the “no pill” group.

SPSS

4 2.5000 1.2910 .6455

4 3.7500 .9574 .4787

PILL.00

1.00

SCOREN Mean

Std.Deviation

Std. ErrorMean

Group Statistics

.500 .506 -1.555 6 .171 -1.2500 .8036 -3.2164 .7164

-1.555 5.534 .175 -1.2500 .8036 -3.2573 .7573

Equalvariancesassumed

Equalvariancesnotassumed

SCOREF 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

Practice• You wonder if psychology majors have

higher IQs than sociology majors ( = .05)

• You give an IQ test to 4 psychology majors and 4 sociology majors

Results

Psychology

110

150

140

135

Sociology

90

95

80

98

Step 1: Hypotheses

• Alternative hypothesis– H1: psychology > sociology

• Null hypothesis– H0: psychology = or < sociology


• df = N1 + N2 - 2

• df = 4 + 4 - 2 = 6 = .05

• One-tailed

• t critical = 1.943


tcrit = 1.943


tobs = (X1 - X2) / Sx1 - x2

9.38 =

X1= 535

X12=

72425

N1 = 4

X1 = 133.75

X2= 363

X22=

33129

N2 = 4

X2 = 90.75

535 36372425 331294 4

4 (4 - 1)


4.58 = (133.75 - 90.75) / 9.38

Sx1 - x2 = 9.38X1 = 133.75

X2 = 90.75


tcrit = 1.943

tobs = 4.58

Step 6: Decision






• We Reject H0, and accept H1

• Psychology majors have significantly ( = .05) higher IQs than sociology majors.

Practice

SPSS Problem #2

• 7.37

• 7.11

Example You give 100 random students a questionnaire designed to measure attitudes toward living in...

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