Download Paired Samples versus Independent Samples

Paired Samples versus Independent Samples
Paired Design 1
With paired data, we are interested in
comparing the responses within each
pair. We will analyze the differences of
the responses that form each pair.
Paired Data: Response = Annual Salary (in $1000s)
W ife Res pons e
Hus band Res pons e
Difference = W ife - Hus band
15
20
15 - 20 = -5
45
31
45 - 31 = 14
50
50
50 - 50 = 0
16
30
16 - 30 = -14
56
72
56 - 72 = -16
Mean Difference = -4.2
Paired Design 2
DEFINITION:
We have paired or matched samples when we know, in advance, that an
observation in one data set is directly related to a specific observation in
the other data set. It may be that the related sets of units are each
measured once (Paired Design 1), or that the same unit is measured twice
(Paired Design 2). In a paired design, the two sets of data must have the
same number of observations.
Independent Samples Design
Independent Samples Data:
Response = Annual Salary (in $1000s)
W om en Res pons e
M en Res pons e
15
20
45
31
50
50
16
30
56
72
M ean for W om en = 36.4
M ean for M en = 40.6
Di fference i n the m eans = 36.4 - 40.6 = -4.2
In the two independent samples scenario, we will compare the
responses of one treatment group as a whole to the responses of the
other treatment group as a whole. We will calculate summary measures
for the observations from one treatment group and compare them to
similar summary measures calculated from the observations from the
other treatment group.
DEFINITION:
We have two independent samples when two unrelated sets of units are
measured, one sample from each population, as in Independent Samples
Design 11.3. In a design with two independent samples, although the
same sample size is often preferable, the sample sizes might be different.
Let’s Do It! Paired Samples versus Independent Samples
(a) Three hundred registered voters were selected at random, 30 from
each of 10 midwestern counties, to participate in a study on
attitudes about how well the president is performing his job.
They were each asked to answer a short multiple-choice
questionnaire and then they watched a 20-minute video that
presented information about the job description of the
president. After watching the video, the same 300 selected
voters were asked to answer a follow-up multiple-choice
questionnaire. The investigator of this study will have two sets
of data: the initial questionnaire scores and the follow-up
questionnaire scores. Is this a paired or independent samples
design?
Circle one:
Paired
Independent
Explain:
(b) Thirty dogs were selected at random from those residing at the
humane society last month. The 30 dogs were split at random into
two groups. The first group of 15 dogs was trained to perform a
certain task using a reward method. The second group of 15 dogs
was trained to perform the same task using a reward-punishment
method. The investigator of this study will have two sets of data: the
learning times for the dogs trained with the reward method and the
learning times for the dogs trained with the reward-punishment
method. Is this a paired or independent samples design?
Circle one:
Paired
Independent
Explain:
Let’s Do It! 2 Design a Study
For each of the following research questions, briefly describe how you
might design a study to address the question (discuss whether paired or
independent samples would be obtained):
(a)
Do freshmen students use the library to study more often than
senior students?
(b)
Do books cost more on average at the local bookstore or through
Amazon.com?
(c)
Will taking summer school improve reading levels for
Kindergarteners going into first grade?
Paired Samples
In a paired design, units in each par are alike (in fact, they may be the
same unit), whereas units in different pairs may be quite dissimilar.
observation for
treatment 1
observation for
treatment 2
Population of Paired Observations
D = difference = treatment 1 - treatment 2
Since we are interested in the difference for each pair, the differences
are what we analyze in paired designs.
Example Weight Change
A study was conducted to estimate the mean weight change of a female
adult who quits smoking. The weights of eight female adults before they
stopped smoking and five weeks after they stopped smoking were
recorded. The differences, computed as “after -before,” are given below.
Subject
After
Before
Difference
1
154
148
6
2
181
176
5
3
151
153
-2
4
120
116
4
5
131
129
2
6
130
128
2
7
121
120
1
8
128
132
-4
Here we have another example of a paired design.
(a)
Compute the sample mean difference in weight.
(b)
Compute the sample standard deviation of the differences.
Solution
(a) The sample mean difference is d =1.75 pounds. Note that the
differences computed as “after - before” represent the weight gain
for a subject. A positive value indicates weight gain and a negative
value indicates a weight loss.
(b)
The sample standard deviation is SD =3.412 pounds
The Paired T-Test
Paired t-Test
Hypotheses:
H0 : D  0 versus H1 : D  0 or
H0 :  D  0
H0 :  D  0
Data:
H1 : D  0 or
versus H1 : D  0 .
versus
The sample of n differences, generically written as
d1 , d 2 ,, d n from which the sample mean difference
d and the sample standard deviation of the differences s D
can be computed.
bserved Test Statistic: T  d  0 and the null distribution for the T variable
sD
n
is a t(n-1) distribution.
p-value: We find the p-value for the test using the t(n - 1) distribution.
The direction of extreme will depend on how the alternative
hypothesis is expressed. 
Decision: A p-value less than or equal to  leads to rejection of H0

Notes:
 If we are interested in assessing if D is equal to some
hypothesized value that is not 0, we would replace 0 in the test
statistic expression with this other null value.
 The test statistic is the same no matter how the alternative
hypothesis is expressed.
Example Comparing Test Scores
A group of 10 randomly selected children of elementary school age
among those in the Mankato County who were recently diagnosed with
asthma was tested to see if a new children’s educational video is
effective in increasing the children’s knowledge about asthma. A nurse
gave the children an oral test containing questions about asthma before
and after seeing the animated video. The test scores are given below:
Child:
Before:
After:
1 2 3 4 5 6 7 8 9 10
61 60 52 74 64 75 42 63 53 56
67 62 54 83 60 89 44 67 62 57
Mean = 60
Mean =64.5
(a)
Explain why we have paired data here and not two independent
samples.
(b)
We are interested in examining the differences in the scores for
each child. Compute the differences and find the sample mean
difference and the sample standard deviation of the differences.
(c)
The researchers wish to assess if the data provide sufficient
evidence to conclude that the mean score after viewing the
educational video is significantly higher than the mean score before
the viewing. The test will be conduced at the 5% level of
significance. State the appropriate hypotheses to be tested in terms
of the population mean difference in test scores  d .
(d) Compute the observed t-test statistic value.
(e)
Find the corresponding p-value.
(f)
State the decision and conclusion using a 5% significance level.
Solution
(a) Since we have two observations from the same child, we have
paired data.
(b)
The observed differences computed here as are as follows: “afterbefore”.
Child:
d = After - Before
1
6
2
2
3
2
4
9
5
-4
6 7
14 2
8
4
9 10
9 1
Mean diff =4.5
The first observed difference is 6 and is represented by d1, and the last
difference is also positive and is represented by d10 = 1. The observed
sample mean difference is d  4.5 , which is our estimate of the
unknown mean difference,  D . The observed sample standard deviation
. , which is our estimate of the unknown
of the differences is sD  5126
population standard deviation  D .
(c)
Since we defined our differences as diff =after - before, it is positive
differences that would show some support that the video is
effective in improving the mean test score. Thus the corresponding
hypotheses to be tested are H0 : D  0 versus H1:  D  0.
4.5  0
t

 2.78 .
(d) The observed t-test statistic is given by
5126
.
10
This means we observed a sample mean difference that was about 2.78
standard errors above the hypothesized mean difference of zero.
Is this large enough (that is, far enough above zero) to reject the null
hypothesis?
t(9)
(e)
The p-value is the probability of getting a
test statistic as large as or larger than
the observed test statistic of 2.78,
computed using a t-distribution
with nine degrees of freedom.
Area=p-value
0
2.78
With the TI Using the tcdf: p-value = PT  2.78 = tcdf(2.78, E99, 9) =
0.0107
Using the T-Test function under the STAT TESTS menu.
In the TESTS menu located under the STAT button, we select the 2:TTest option. With the sample mean of 4.5, the sample standard deviation
of 5.126, and the sample size of n = 10, we can use the Stats option of
this test. The corresponding input and output screens are shown.
Notice that the null or hypothesized value is zero.
p-value = PT  2.78 = 0.01077.
(f) Decision and Conclusion
Since our p-value is less than 0.05, at the   0.05 significance level we
would reject H 0 , and conclude there is sufficient evidence to say that the
mean score after viewing the educational video is significantly higher
than the mean score before the viewing.
Let’s Do It!
Two creams are available by prescription for treating moderate skin
burns. A study to compare the effectiveness of the two creams is
conducted using 15 patients with moderate burns on their arms. Two
spots of the same size and degree of burn are marked on each patient’s
arm. One of the two creams is selected at random and applied to the
first spot, while the remaining spot is treated with the other cream. The
number of days until the burn has healed is recorded for each spot.
These data are provided with the difference in healing time (in days).
Consider the data and interval estimate for comparing the two burn
cream treatments in
Patient Number
Cream1= C1
Cream2= C2
Diff =C1- C2
1
16
14
2
2
2
4
-2
3
10
10
0
4
7
4
3
5
6
5
1
6
10
12
-2
7
5
5
0
8
4
6
-2
9
19
23
-4
10
7
10
-3
11
12
12
0
12
9
7
2
13
10
11
-1
14
20
24
-4
We wish to test the claim that there is no difference between the
two creams at the 5% significance level.
Homework page 344: 67, 68, 69, 72, 73, 76, 77
15
12
10
2