Types of TwoSample Tests Independent test two conditions are comprised of different elements or subjects Example comparing children and adults on a task Paired test two conditions are comprised of the same elements or subjects same observation is measured under different testing ID: 816575
Download The PPT/PDF document "Paired t-tests Why can’t we all just r..." is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
Slide1
Paired t-tests
Why can’t we all just reduce variation due to chance!!
Slide2Types of Two-Sample Tests
Independent test
–
two conditions are comprised of different elements or subjects
Example: comparing children and adults on a task
Paired test
– two conditions are comprised of the same elements or subjects; same observation is measured under different testing conditions.
Example: pre-post observations of the same subjects
Slide3What is a paired t-test?
Paired t-test
: compares two samples that are comprised of the exact same observations (subjects).
-often used for pre-post studies
Used for
repeated measures
or
within-subjects research designs
: DV measured two of more times for each individual
Examples
:
pre-stress → yoga → post-stress
Reaction time sober and reaction time drunk
Student’s sleep on weekends and weekdays
Slide4Paired Data-Factory Safety Program
# of accidents per month
Factory
Before
After
Amherst
45
36
S. Hadley
73
60
Northampton
46
44
Belchertown
124
119
Springfield
33
35
W. Springfield
57
51
Westfield
83
77
Holyoke
34
29
Agawam
26
24
Chicopee
17
11
M
B
= 53.8
M
A
= 48.6
s
B
= 32.1
s
A
= 31.0
Slide5Hypothesis Testing for Paired Data
We are interested in difference scores (based on each pair of scores)
This tells us how much
change
there was
We are testing whether the difference between pairs of scores is likely just due to sampling error or represents a true change.
We want to make a conclusion about the difference between the two treatment conditions in the general population (
D
).Ho: D = 0Ha: D 0
Slide6Factory
Before
After
D (
Af-Bef
)
Amh
45
36
-9
S.Ha
73
60-13Noho4644-2Belch124119-5S’field33352WSpring5751-6Wes8377-6Holy3429-5Aga2624-2Chic1711-6 MB = 53.8MA = 48.6MD = -5.2
Compute D = X2-X1
Paired Data-Factory Safety Program
Slide7Null Hypothesis: Paired samples
D
= 0
Sampling distribution of mean differences
Slide8Hypothesis Testing for Paired Data
Large Sample; σ is known
Small sample; σ is unknown
df
= n-1 (where n = # pairs)
D
0
= null hypothesis mean difference (zero)
σ
D
or S
D
= standard deviation of difference scoresnD = # of pairs of scores
Slide9Factory Safety Program
Step 1:
two-tailed test
Step 2
: Ho:
D
= 0
Step 3
: Ha: D 0Step 4: = .05Step 5: df = n-1 = 10-1 = 9Critical t (9)= 2.262
Factory
Before
AfterAmherst4536S. Hadley7360Noho4644Btown124119Springfield3335W. Spring5751Westfield8377Holyoke3429Agawam2624Chicopee1711 MB= 53.8MA = 48.6 sB = 32.1sA = 31.0
Slide10Step 6a
: find the Average Difference Score
Factory
Before
After
D
Amh
45
36
-9
S.Ha
73
60-13Noho4644-2Belch124119-5S’field33352WSpring5751-6Wes8377-6Holy3429-5Aga2624-2Chic1711-6 = -52 MB = 53.8MA = 48.6MD = -5.2
s
B
= 32.1
s
A
= 31.0
Slide11Factory
Before
After
D
Amh
45
36
-9
S.Ha
73
60
-13
Noho4644-2Belch124119-5S’field33352WSpring5751-6Wes8377-6Holy3429-5Aga2624-2Chic1711-6 = -52 MB = 53.8MA = 48.6MD = -5.2
sB = 32.1
s
A
= 31.0
REMEMBER:
The difference scores are the data we are focusing on
Step 6a
: find the Average Difference Score
Slide12Factory
Before
After
D
D
2
Amh
45
36
-9
81
S.Ha
7360-13169Noho4644-24Belch124119-525S’field333524WSprg5751-636Wes8377-636Holy3429-525Aga2624-24Chic1711-636
= -52
= 420
M
B
= 53.8
M
A
= 48.6
M
D
= -5.2
s
B
= 32.1
s
A
= 31.0
s
D
=
Step 6b
: find S
D
(needed to compute SE and the t statistic)
Slide13Factory
Before
After
D
D
2
Amh
45
36
-9
81
S.Ha
7360-13169Noho4644-24Belch124119-525S’field333524WSprg5751-636Wes8377-636Holy34
29
-5
25
Aga
26
24
-2
4
Chic
17
11
-6
36
= -52
= 420
M
B
= 53.8
M
A
= 48.6
M
D = -5.2
s
B
= 32.1
s
A
= 31.0
s
D
=
Step 6b
:
find S
D
(needed to compute SE and the t statistic)
Slide14M
D
= -5.2
S
D
= 4.08
# pairs: n = 10
t
crit
=
2.26
Before
AfterDifferenceMB = 53.8MA = 48.6MD = -5.2sB = 32.1sA = 31.0sD = 4.08Step 6c: Calculate observed t
Slide15Null hypothesis sampling distribution of differences between means
D
= 0
+2.26
-2.26
t
obs
= -4.02
SE =
1.3
Slide16Factory Safety Program Effective
Step 7
:
reject the null: t(9) = -4.02, p < .05
Step 8:
Interpretation??
Accidents before program: M = 53.8
Accidents after program: M = 48.6
The safety program
significantly reduced the number of accidents in the factories, t(9) = -4.02, p < .05Or…there were significantly fewer accidents after the safety program was implemented than there were prior to its implementation, t(9) = -4.02, p < .05.
Slide1795% Confidence Interval
95% CI = [-8.12, -2.28]
“We are 95% sure that the population mean difference in the number of accidents before and after the intervention is between -8.12 and -2.28”
Slide18Effect sizes for paired t-tests
Cohen’s d =
Factory Safety example:
Cohen’s d =
Slide19Ice Cream and Exams
Does eating ice cream improve test performance? Last year I randomly selected 10 students from my stats course at Forks High School and gave each one an ice cream cone prior to the second exam and compared their scores on the first exam with their scores on the second exam.
My research question: Does eating ice cream change exam performance? Set
= .01.
Slide20Ice cream and exams
Student
Exam #1
Exam #2
Diff Score
D
2
Sam A
75
77
Sammy
P8081 Olivia A8587 Olivia B9091 Olivia C8587 Dylan 17576 Dylan 27072 Danny6566 Donna7577 Emery8081 Set = .01Determine the Ho and Ha, tcrit, tobs
and make a conclusion about my study
Slide21Advantages of repeated
measures design
Fewer subjects needed: because the same subjects are in both conditions
Allows one to study changes over time (practice effects)
Removes individual difference issues between groups
Protects against some third variable issues
Slide22Independent vs. Paired: Are y’all Liars?
An anonymous statistics teacher wants to assess the honesty of his students. At the beginning of the semester, he asks them to write down their actual GPA, and/or the GPA that they have reported to their parents. 77 subjects participated. The mean actual GPA was 3.21. The mean GPA reported to parents was 3.29. Determine whether the data provides enough evidence to conclude that students in general are liars. Set alpha = .01.
How will results compare if he used a repeated measures design vs. two independent groups?
Slide23Are Y’all Liars?
PAIRED
Actual: M
A
= 3.21
Parent
:
M
P
= 3.29MD = .08 sD = .19n=77 tcrit = t (76)=2.642Step 6: Calculate tobs
INDEPENDENTActual: MA = 3.21; s = .50; n=77
Parent
: MP = 3.29; s = .46; n=77MD = .08 tcrit = t (152)=2.609Step 6: Calculate tobs
Slide24Disadvantages of repeated
measures designs
Time related factors may influence the results
Health or mood may change
Outside factors may change
Participation in the first condition may influence participation in the second condition
Order effects
Slide25Assumptions of paired t-tests
The observations
within
each treatment condition must be independent
Obviously those between conditions are not (it’s the same subject!)
The population distribution of mean difference scores must be normal
Only a major concern with small samples (n > 30)