Chapter 25
Asking and Answering
Questions About the
Difference Between Two
Population Means: Paired
Samples
Estimating the Difference
Between Two Population Means:
Paired-Samples t Confidence
Interval for a Difference in
Population Means
The Paired-Samples t Confidence Interval
for a Difference in Population Means
Appropriate when the following conditions are
met:
1.
The samples are paired (for example, before/after
data.
2. The n sample differences can be viewed as a random
sample from a population of differences (or it is
reasonable to regard the sample of differences as
representative of the population of differences).
3. The number of sample differences is large (n ≥ 30) or
the population distribution of differences is
approximately normal.
The Paired-Samples t Confidence Interval
for a Difference in Population Means
When these conditions are met, a confidence interval for
the difference in population means is
𝑥𝑑 ± 𝑡 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
𝑠𝑑
𝑛
Where
n = the number of sample differences,
𝑥𝑑 = mean of the n sample differences,
𝑠𝑑 = standard deviation of the n sample differences.
The t critical value is based on df = n – 1.
The Paired-Samples t Confidence
Interval for a Difference in Population
Means
𝑥𝑑 ± 𝑡 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
𝑠𝑑
𝑛
Interpretation of Confidence Level
The confidence level specifies the long-run proportion of
the time that this method is successful in capturing the
true, but unknown, difference d = 1 - 2 in population
means.
Interpretation of Confidence Interval
You can be confident that the computed interval contains
the true, but unknown, value of the difference d = 1 - 2
in the population. This statement should be worded in
context.
Benefits of Ultrasound
In a study to investigate the effect of ultrasound therapy
on knee
extension,
range of
was
forto do
Because
the samples
aremotion
paired,
themeasured
first thing
people in a representative
sample
of physical
therapy
is to compute the
sample
differences.
patients both before and after ultrasound therapy.
Range of Motion
Patient
1
2
3
4
5
6
7
Before Ultrasound
31
53
45
57
50
43
32
After Ultrasound
Differences
32
59
46
64
49
45
40
-1
-6
-1
-7
1
-2
-8
The mean and standard deviation computed
from these sample differences are 𝑥𝑑 = -3.43
and sd = 3.51
Range of Motion Continued . . .
Range of Motion
Patient
1
2
3
4
5
6
7
Before Ultrasound
31
53
45
57
50
43
32
After Ultrasound
32
59
46
64
49
45
40
Differences
-1
-6
-1
-7
1
-2
-8
The mean and standard deviation computed from these
sample differences are 𝑥𝑑 = -3.43 and sd = 3.51
You want to estimate
d = 1 – 2 = mean difference in knee random of motion
where
1 = mean knee range of motion before ultrasound
and
2 = mean knee range of motion after ultrasound
Range of Motion Continued . . .
Calculate
The t critical value for df = 6 and 95% confidence level is 2.45.
The method used to construct this interval
𝑠𝑑
3.51
is successful
the
actual
𝑥𝑑 ± 𝑡 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙
𝑣𝑎𝑙𝑢𝑒 in capturing
= −3.43
± 2.45
𝑛
7
difference in population
means about 95%
= (−6.68,
of the−0.18)
time.
Communicate Results
We are 95% confident that the interval (-6.68, -0.18)
contains the true, but unknown difference in mean range of
motion. Because both endpoints are negative, you would
estimate that knee range of motion after ultrasound is
greater than the mean range of motion before ultrasound by
somewhere between 0.18 and 6.68 degrees.
The Paired t Test
Appropriate when the following conditions are met:
1.
The samples are paired.
2. The n sample differences can be viewed as a random
sample from a population of differences (or it is
reasonable to regard the sample of differences as
representative of the population of differences).
3. The number of sample differences is large (n ≥ 30) or
the population distribution of differences is
approximately normal.
Summary of the Paired t test for Comparing
Two Population Means Continued
When these conditions are met, the following test
statistic can be used:
𝑥𝑑 − 𝜇0
𝑡= 𝑠
𝑑
𝑛
Where 0 is the hypothesized value of the population mean
difference from the null hypothesis, n is the number of
sample differences, and 𝑥𝑑 and 𝑠𝑑 are the mean and
standard deviation of the sample differences.
Summary of the Paired t test for Comparing
Two Population Means Continued
Form of the null hypothesis:
H0: d = 0
When the conditions are met and the null hypothesis is true, this t test
statistic has a t distribution with df = n – 1.
When the Alternative
Hypothesis Is . . .
The P-value Is . . .
Ha: d > 0
Area under the t curve to the right of
the calculated value of the test statistic
Ha: d < 0
Area under the t curve to the left of
the calculated value of the test statistic
Ha: d ≠ 0
2·(area to the right of t) if t is positive
Or
2·(area to the left of t) if t is negative
In a study to investigate the effect of ultrasound therapy
on knee extension, range of motion was measured for
people
in a representative
sample
of physical
Because
the samples are
paired,
the firsttherapy
thing to do
patients both
and after
ultrasound
therapy.
is before
to compute
the sample
differences.
Range of Motion
Patient
1
2
3
4
5
6
7
Before Ultrasound
31
53
45
57
50
43
32
After Ultrasound
32
59
46
64
49
45
40
Differences
-1
-6
-1
-7
1
-2
-8
Is there evidence that the ultrasound therapy increases
range of motion?
The mean and standard deviation computed
from these sample differences are
𝑥𝑑 = -3.43 and sd = 3.51
Range of Motion Continued . . .
Hypotheses:
The population characteristics of interest are
1 = mean range of motion for physical therapy patients before ultrasound
2 = mean range of motion for physical therapy patients after ultrasound
Because the samples are paired, you should also define d :
d = 1 – 2 = mean difference in range of motion (before – after)
Translating the question of interest into hypotheses gives:
H0: d = 0 versus H0: d < 0
Range of Motion Continued . . .
Calculate
Test Statistic:
𝑡=
−3.43−0
3.51
7
= −2.59
P-value:
P-value = area under t distribution with 6df to the left of -2.59
= 0.02
Communicate Results
Because the P-value (0.02) is less than 0.05, reject H0.
There is convincing evidence that mean knee range of
motion for physical therapy patients before ultrasound is
less than the mean range of motion after ultrasound.
Avoid These Common
Mistakes
Avoid These Common Mistakes
1. Remember that the results of a hypothesis
test can never show strong support for the null
hypothesis. In two-sample situations, this
means that you shouldn’t be convinced that
there is not difference between two population
means based on the outcome of a hypothesis
test.
Avoid These Common Mistakes
2. If you have complete information (a census)
for both populations, there is no need to carry
out a hypothesis test or to construct a
confidence interval – in fact, it would be
inappropriate to do so.
Avoid These Common Mistakes
3. Don’t confuse statistical significance with
practical significance. In the two-sample
setting, it is possible to be convinced that two
population means are not equal even in
situations where the actual difference
between them is small enough that it is of no
practical use.
After rejecting a null hypothesis of no difference,
it is useful to look at a confidence interval
estimate of the difference to get a sense of
practical significance.
Avoid These Common Mistakes
4. Correctly interpreting confidence intervals in
the two-sample case is more difficult than in
the one-sample case, so take particular care
when providing two-sample confidence
interval interpretations.
Because the two-sample confidence interval
estimates a difference (1 – 2), the most
important thing to note is whether or not
the interval includes 0.