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Practice. Which is more likely: at least one ace with 4 throws of a fair die or at least one double ace in 24 throws of two fair dice? This is known as DeMere's problem, named after Chevalier De Mere. Blaise Pascal later solved this problem. Binomial Distribution. p = .482 of zero aces - PowerPoint PPT Presentation
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Practice
• Which is more likely: at least one ace with 4 throws of a fair die or at least one double ace in 24 throws of two fair dice? This is known as DeMere's problem, named after Chevalier De Mere.
• Blaise Pascal later solved this problem.
Binomial Distribution
)04(0 8333.1667.)!04(!0
!4)(
Xp
p = .482 of zero aces
1 - .482 = .518 at least one ace will occur
Binomial Distribution
)024(0 9722.0278.)!024(!0
!24)(
Xp
p = .508 of zero double aces
1 - .508 = .492 at least one double ace will occur
Practice
• Which is more likely: at least one ace with 4 throws of a fair die or at least one double ace in 24 throws of two fair dice? This is known as DeMere's problem, named after Chevalier De Mere.
• More likely at least one ace with 4 throws will occur
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
• Alternative hypothesis– H1: sample = 4
• 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
tcrit = 1.98tcrit = -1.98
t distribution
tobs = 2.63
tcrit = 1.98tcrit = -1.98
If tobs fall in critical area reject the null hypothesis
Reject H0: sample = 4
t distribution
tobs = 2.63
tcrit = 1.98tcrit = -1.98
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 ratings of the dorms is different than 4.
• 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
• Alternative hypothesis– H1: sample = 100
• Null hypothesis– H0: sample = 100
Step 2: Calculate the Critical t
• N = 54
• df = 53 = .05
• tcrit = 2.0
Step 3: Draw Critical Region
tcrit = 2.00tcrit = -2.00
Step 4: Calculate t observed
tobs = (X - ) / Sx
Step 4: Calculate t observed
tobs = (X - ) / Sx
Sx = S / N
Step 4: Calculate t observed
tobs = (X - ) / Sx
2.5 = 18.4 / 54
Step 4: Calculate t observed
tobs = (X - ) / Sx
12 = (130 - 100) / 2.52.5 = 18.4 / 54
Step 5: See if tobs falls in the critical region
tcrit = 2.00tcrit = -2.00
Step 5: See if tobs falls in the critical region
tcrit = 2.00tcrit = -2.00
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
Step 1: Write out Hypotheses
• Alternative hypothesis– H1: sample = 56.1
• Null hypothesis– H0: sample = 56.1
Step 2: Calculate the Critical t
• N = 5
• df =4 = .10
• tcrit = 2.132
Step 3: Draw Critical Region
tcrit = 2.132tcrit = -2.132
Step 4: Calculate t observed
tobs = (X - ) / Sx
-.48 = (55.2 - 56.1) / 1.88 1.88 = 4.21/ 5
Step 5: See if tobs falls in the critical region
tcrit = 2.132tcrit = -2.132
tobs = -.48
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 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
Step 2: Calculate the Critical t
• 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
Step 3: Draw Critical Region
tcrit = -1.533
Step 4: Calculate t observed
tobs = (X - ) / Sx
Step 4: Calculate t observed
tobs = (X - ) / Sx
-.48 = (55.2 - 56.1) / 1.88 1.88 = 4.21/ 5
Step 5: See if tobs falls in the critical region
tcrit = -1.533
Step 5: See if tobs falls in the critical region
tcrit = -1.533
tobs = -.48
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 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
Step 1: Write out Hypotheses
• Alternative hypothesis– H1: sample > 100
• Null hypothesis– H0: sample < or = 100
Step 2: Calculate the Critical t
• N = 150
• df = 149 = .05
• tcrit = 1.645
Step 3: Draw Critical Region
tcrit = 1.645
Step 4: Calculate t observed
tobs = (X - ) / Sx
2.54 = (103 - 100) / 1.181.18=14.4 / 150
Step 5: See if tobs falls in the critical region
tcrit = 1.645
tobs = 2.54
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 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
D ep en d en t V ariab leIf p e rson lives
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
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
D ep en d en t V ariab leIf p e rson lives
C on tro l G rou pD o n o t g ive m ed ica tion
S u b jec ts
See if both groups are different at the end
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
D ep en d en t V ariab leIf p e rson lives
C on tro l G rou pD o n o t g ive m ed ica tion
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
Step 3: Draw Critical Region
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
NowStep 4: Calculate t observed
tobs = (X1 - X2) / Sx1 - x2
NowStep 4: Calculate t observed
tobs = (X1 - X2) / Sx1 - x2
• X1 = 3.75
• X2 = 2.50
NowStep 4: Calculate t observed
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
Standard Error of a Difference
Sx1 - x2
When the N of both samples are equal
If N1 = N2:
Sx1 - x2 = Sx12 + Sx2
2
Standard Error of a Difference
Sx1 - x2
When the N of both samples are equal
If N1 = N2:
Sx1 - x2 = (.48)2 + (.645)2
Standard Error of a Difference
Sx1 - x2
When the N of both samples are equal
If N1 = N2:
Sx1 - x2 = (.48)2 + (.645)2= .80
Standard Error of a Difference Raw Score Formula
When the N of both samples are equal
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 =
NowStep 4: Calculate t observed
tobs = (X1 - X2) / Sx1 - x2
Sx1 - x2 = .80X1 = 3.75
X2 = 2.50
NowStep 4: Calculate t observed
tobs = (3.75 - 2.50) / .80
Sx1 - x2 = .80X1 = 3.75
X2 = 2.50
NowStep 4: Calculate t observed
1.56 = (3.75 - 2.50) / .80
Sx1 - x2 = .80X1 = 3.75
X2 = 2.50
Step 5: See if tobs falls in the critical region
tcrit = 2.447tcrit = -2.447
Step 5: See if tobs falls in the critical region
tcrit = 2.447tcrit = -2.447
tobs = 1.56
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 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
Step 2: Calculate the Critical t
• df = N1 + N2 - 2
• df = 4 + 4 - 2 = 6 = .05
• One-tailed
• t critical = 1.943
Step 3: Draw Critical Region
tcrit = 1.943
NowStep 4: Calculate t observed
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)
Step 4: Calculate t observed
4.58 = (133.75 - 90.75) / 9.38
Sx1 - x2 = 9.38X1 = 133.75
X2 = 90.75
Step 5: See if tobs falls in the critical region
tcrit = 1.943
tobs = 4.58
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
• Psychology majors have significantly ( = .05) higher IQs than sociology majors.
SPSS Problem #2
• 7.37 (TAT)
• 7.11 (Anorexia)