Transcript Test 2 - Interpersonal Research Laboratory

Practice
• A research study was conducted to examine the 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. Determine if age is
related to life satisfaction.
Older Adults
45
38
52
48
25
39
51
46
55
46
Younger Adults
34
22
15
27
37
41
24
19
26
36
Older
Younger
Mean = 44.5
Mean = 28.1
S = 8.682677518
S = 8.543353492
S2 = 75.388888888
S2 = 72.988888888
tobs = 4.257; t crit = 2.101
Age is related to life satisfaction.
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
X1= 535
Psychology
110
150
140
135
Sociology
90
95
80
X2= 265
X12= 72425
X22= 23525
N1 = 4
N2 = 3
X1= 535
X2= 265
X12= 72425
X22= 23525
N1 = 4
N2 = 3
N1 + N2 - 2
X1= 535
X2= 265
X12= 72425
X22= 23525
N1 = 4
N2 = 3
265
535
N1 + N2 - 2
72425
X1= 535
X2= 265
X12= 72425
X22= 23525
N1 = 4
N2 = 3
535
23525
N1 + N2 - 2
265
72425
X1= 535
X2= 265
X12= 72425
X22= 23525
N1 = 4
N2 = 3
535
4
23525
265
3
4
4+3-2
3
72425
X1= 535
X2= 265
X12= 72425
X22= 23525
N1 = 4
N2 = 3
535
71556.25
4
23525
5
265
23408.33
3
.25+.33
4
3
72425
X1= 535
X2= 265
X12= 72425
X22= 23525
N1 = 4
N2 = 3
535
71556.25
4
265
23525
23408.33
197.08 (.58) 3
5
.25+.33
4
3
72425
X1= 535
X2= 265
X12= 72425
X22= 23525
N1 = 4
N2 = 3
535
71556.25
4
23525
114.31
5
= 10.69
265
23408.33
3
.25+.33
4
3
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
Now
Step 4: Calculate t observed
tobs = (X1 - X2) / Sx1 - x2
15175
X1= 275
X2= 219
X12= 15175
X22= 16043
N1 = 5
N2 = 3
X1 = 55
X2 = 73
275
15125
5
16043
6
= 3.05
219
15987
3
.2 + .33
5
3
Step 4: Calculate t observed
-5.90 = (55 - 73) / 3.05
Sx1 - x2 = 3.05
X1 = 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
Group Statistics
TEMP
PHILLY
1.00
.00
N
5
3
Mean
55.0000
73.0000
Std.
Deviation
3.5355
5.2915
Std. Error
Mean
1.5811
3.0551
Independent Sam ple s Te st
Levene's Test for
Equality of Varianc es
F
TE MP
Equal
varianc es
as sumed
Equal
varianc es
not
as sumed
.986
Sig.
.359
t-t est for E quality of Means
t
df
Sig.
(2-tailed)
Mean
Difference
St d. E rror
Difference
95% Confidenc e
Int erval of t he Mean
Lower
Upper
-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
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)
Psychology (X)
Sociology (Y)
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
Psychology (X)
Sociology (Y)
Joe
100
Joe
90
Al
110
Al
89
Jay
105
Jay
86
Matched Samples Design
• 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
Matched Samples Design
• 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?
Hypothesis
One-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
Step 2: Calculate the Critical t
• N = Number of pairs
• df = N - 1
• 4-1=3
•  = .05
• t critical = 2.353
Step 3: Draw Critical Region
tcrit = 2.353
Step 4: Calculate t observed
tobs = (X - Y) / SD
Step 4: Calculate t observed
tobs = (X - Y) / SD
Step 4: Calculate t observed
tobs = (X - Y) / SD
X = 3.75
Y = 2.00
Step 4: Calculate t observed
tobs = (X - Y) / SD
Standard error of a difference
Step 4: Calculate t observed
tobs = (X - Y) / SD
SD = SD / N
N = number of pairs
S=
Test 1 w/ Pill (X)
Mel
3
Alice
5
Vera
4
Flo
3
S=
Test 2 w/o Pill (Y)
1
3
2
2
Test 1 w/ Pill (X)
Mel
3
Alice
5
Vera
4
Flo
3
S=
Difference
Test 2 w/o Pill (Y)
(D)
1
2
3
2
2
2
2
1
Test 1 w/ Pill (X)
Mel
3
Alice
5
Vera
4
Flo
3
Difference
Test 2 w/o Pill (Y)
(D)
1
2
3
2
2
2
2
1
D = 7
D2 =13
S=
N=4
Test 1 w/ Pill (X)
Mel
3
Alice
5
Vera
4
Flo
3
Difference
Test 2 w/o Pill (Y)
(D)
1
2
3
2
2
2
2
1
D = 7
7
S=
D2 =13
N=4
Test 1 w/ Pill (X)
Mel
3
Alice
5
Vera
4
Flo
3
Difference
Test 2 w/o Pill (Y)
(D)
1
2
3
2
2
2
2
1
D = 7
S=
13
7
D2 =13
N=4
Test 1 w/ Pill (X)
Mel
3
Alice
5
Vera
4
Flo
3
Difference
Test 2 w/o Pill (Y)
(D)
1
2
3
2
2
2
2
1
D = 7
S=
13
4-1
7
D2 =13
4
N=4
Test 1 w/ Pill (X)
Mel
3
Alice
5
Vera
4
Flo
3
Difference
Test 2 w/o Pill (Y)
(D)
1
2
3
2
2
2
2
1
D = 7
S=
7
12.25
4
13
3
D2 =13
N=4
Test 1 w/ Pill (X)
Mel
3
Alice
5
Vera
4
Flo
3
Difference
Test 2 w/o Pill (Y)
(D)
1
2
3
2
2
2
2
1
D = 7
.5 =
.75
3
7
D2 =13
4
N=4
Step 4: Calculate t observed
tobs = (X - Y) / SD
SD = SD / N
N = number of pairs
Step 4: Calculate t observed
tobs = (X - Y) / SD
.25=.5 / 4
N = number of pairs
Step 4: Calculate t observed
7.0 = (3.75 - 2.00) / .25
Step 5: See if tobs falls in the
critical region
tcrit = 2.353
Step 5: See if tobs falls in the
critical region
tcrit = 2.353
tobs = 7.0
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
• 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
Pa ired Sa mples Stati stics
Pair 1
TIME1
TIME2
Mean
3.7500
2.0000
N
4
4
St d.
Deviation
.9574
.8165
St d. Error
Mean
.4787
.4082
Paired Samples Correlations
N
Pair 1
Correlation
TIME1 &
TIME2
4
.853
Sig.
.147
Paired Samples Test
Paired Differences
Mean
Pair 1
TIME1 TIME2
1.7500
Std.
Deviation
Std. Error
Mean
.5000
.2500
95% Confidence
Interval of the Difference
Lower
Upper
.9544
2.5456
t
7.000
Sig.
(2-tailed)
df
3
.006
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
Hypothesis
One-tailed
• Alternative hypothesis
– H1: time1 > time2
• Null hypothesis
– H0: time1 < or = time2
Step 2: Calculate the Critical t
• N = Number of pairs
• df = N - 1
• 6-1=5
•  = .05
• t critical = 2.015
Step 4: Calculate t observed
tobs = (X - Y) / SD
Time 1 (X)
Child1
18
Child2
11
Child3
19
Child4
6
Child5
10
Child6
14
(D)
2
1
2
2
-1
2
Test 2 (Y)
16
10
17
4
11
12
D = 8
1.21 =
18
6-1
8
D2 =18
6
N=6
Step 4: Calculate t observed
tobs = (X - Y) / SD
.49=1.21 / 6
N = number of pairs
Step 4: Calculate t observed
2.73 = (13 - 11.66) / .49
X = 13
Y = 11.66
SD = .49
Step 5: See if tobs falls in the
critical region
tcrit = 2.015
tobs = 2.73
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
• Reject H0, and accept H1
• The program significantly ( = .05) lowered
the number of aggressive behaviors a child
performed.
SPSS
Pa ired Sa mples Stati stics
Pair 1
Mean
13.0000
11.6667
CTIME1
CTIME2
N
6
6
St d.
Deviation
4.9800
4.6762
St d. Error
Mean
2.0331
1.9090
Paired Samples Correlations
N
Pair 1
CTIME1
& CTIME2
Correlation
6
Sig.
.970
.001
Paired Samples Test
Paired Differences
Mean
Pair 1
CTIME1 CTIME2
1.3333
Std.
Deviation
Std. Error
Mean
1.2111
.4944
95% Confidence
Interval of the Difference
Lower
Upper
t
6.240E-02
2.697
2.6043
Sig.
(2-tailed)
df
5
.043
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
• Does drinking milkshakes affect (alpha =
.05) your weight?
• To see if milkshakes affect a persons weight
you collected data from 5 sets of twins.
You randomly had one twin drink water and
the other twin drank milkshakes. After 3
months you weighed them.
Results
Twin A
Twin B
Twin C
Twin D
Twin E
Water
186
200
190
162
175
Milkshakes
195
202
196
165
183
Hypothesis
Two-tailed
• Alternative hypothesis
– H1: water = milkshake
• Null hypothesis
– H0: water = milkshake
Step 2: Calculate the Critical t
• N = Number of pairs
• df = N - 1
• 5-1=4
•  = .05
• t critical = 2.776
Step 3: Draw Critical Region
tcrit = -2.776
tcrit = 2.776
Step 4: Calculate t observed
tobs = (X - Y) / SD
(D)
-9
-2
-6
-3
-8
D = -28
3.04 =
194
5-1
-28
D2 =194
5
N=6
Step 4: Calculate t observed
tobs = (X - Y) / SD
1.36=3.04 / 5
N = number of pairs
Step 4: Calculate t observed
-4.11 = (182.6 – 188.2) / 1.36
X = 182.6
Y = 188.2
SD = 1.36
Step 5: See if tobs falls in the critical
region
tcrit = -2.776
tobs = -4.11
tcrit = 2.776
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
• Reject H0, and accept H1
• Milkshakes significantly ( = .05) affect a
persons weight.
Practice
•
Sleep researchers decide to test the impact of REM sleep deprivation on a
computerized assembly line task. Subjects are required to participate in two
nights of testing. On each night of testing the subject is allowed a total of four
hours of sleep. However, on one of the nights, the subject is awakened
immediately upon achieving REM sleep. Subjects then took a cognitive test
which assessed errors in judgment. Did sleep deprivation lower the subjects
cognitive ability?
REM Deprived
26
15
8
44
26
13
38
24
17
29
Control Condition
20
4
9
36
20
3
25
10
6
14
• tobs = 6.175
• tcrit = 1.83
• Sleep deprivation lowered their cognitive
abilities.