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Comparing Two Populations or Treatments

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Title: Comparing Two Populations or Treatments


1
Chapter 11
  • Comparing Two Populations or Treatments

2
Two Means Using Independent Samples
  • An example to compare two populations A
    university financial aid director wants to
    determine whether the mean cost of textbooks is
    different for students in the engineering college
    than for students in the liberal arts college.
  • Population 1 all students enrolled in the
    engineering college.
  • Population 2 all students enrolled in the
    liberal arts college.
  • Goal To compare their respective mean textbook
    costs
  • Two random samples one from each population.
  • An example to compare two treatments An
    agriculture scientist wish to compare weight
    gains for animals placed on two different diets.

3
11.1 Inferences Concerning the Difference Between
Two Populations
4
Two Means Using Independent Samples
  • Two samples are said to be independent samples if
    the selection of the individuals or objects that
    make up one sample does not influence the
    selection of individuals or objects in the other
    sample.
  • A comparison of means focuses on the difference
    µ1-µ2
  • µ1 - µ2 0 is equivalent to µ1 µ2
  • µ1 - µ2 gt 0 is equivalent to µ1 gt µ2
  • µ1 - µ2 lt 0 is equivalent to µ1lt µ2
  • is the point estimate of µ1 - µ2 .
  • is calculated from sample values,
    and therefore, it has a sampling distribution.

5
Properties of the Sampling Distribution of
  • If the random samples on which and are
    based are selected independently of one another,
    then
  • If n1 and n2 are both large or if the population
    distributions are (at least approximately)
    normal, then the sampling distribution of
    are also normal or approximately normal.

6
The Two-Sample t Test for Comparing Two
Population Means
  • Null hypothesis H0 µ1 µ2 hypothesized value
  • Alternative Hypothesis
  • (1) Ha µ1 µ2 gt hypothesized value
  • P-Value Area to the right of the computed t
  • (2) Ha µ1 µ2 lt hypothesized value
  • P-Value Area to the left of the computed t
  • (3) Ha µ1 µ2 ? hypothesized value
  • P-Value
  • 2(area to the right of the computed t
    ) if t gt 0, OR 2(area to the left of the
    computed t ) if t lt 0.

7
The Two-Sample t Test for Comparing Two
Population Means
Test Statistics
Degrees of Freedom for the Two-Sample t Test
where
Truncate (round down) df to obtain an integer
value.
8
Comments for Two-Sample t Test
  • Assumption 1 The two samples are independently
    selected random samples.
  • Assumption 2 The sample sizes are large (in
    general, 30 or larger), or the population
    distributions are (at least approximately)
    normal.
  • If s is known, two-sample z test can also be
    used.

9
  • Example Brain Size Do children diagnosed with
    ADHD (attention deficit/hyperactivity disorder)
    have smaller brains than children without this
    condition? Brain scans were completed for 152
    children with ADHD and 139 children without ADHD.
    Summary values are given below.

Is there evidence that the mean brain
volume of children with ADHD is smaller than the
mean for children with ADHD? Use a
.05? Solution The two samples are
independently selected and n1 and n2 are both
large, so it is reasonable to proceed with the
two sample t test for comparing the two
population means. Let µ1 true mean brain
volume for children with ADHD, and µ2
true mean brain volume for children without ADHD
Then µ1 -µ2 difference in mean brain volume.
Continued on next slide
10
Solution to the Example of Brain Size
H0 µ1 - µ2 0 Ha µ1 - µ2 lt 0 Significance
level a .05.
Appendix Table 4 shows that the area under the t
curve with 288 df to the left of -3.36 is 0.
Therefore, P-value 0 lt .05. Conclusion We
reject H0. The is convincing evidence that the
mean brain volume for children with ADHD is
smaller than the mean for children without ADHD.
11
  • Example (Oral Contraceptive Use and Bone Mineral
    Density)
  • To assess the impact of oral contraceptive use
    on bone mineral density (BMD), researchers
    carried out a study comparing BMD (in grams per
    centimeter) for women who had used oral
    contraceptives for at least 3 months to BMD for
    women who have never used oral contraceptives
  • Never used oral contraceptives
  • 0.82, 0.94, 0.96, 1.31, 0.94, 1.21, 1.26, 1.09,
    1.13, 1.14
  • Used oral contraceptives
  • 0.94, 1.09, 0.97, 0.98, 1.14, 0.85, 1.30, 0.89,
    0.87, 1.01
  • Is there evidence that women who use oral
    contraceptives have a lower mean BMD? Use a
    .05.
  • Solution The two samples are independently
    selected, so it is reasonable to proceed with the
    two sample t test for comparing the two
    population means.
  • Assumption The sample size are small . We must
    be willing to assume that the two populations can
    be viewed as at least approximately normally
    distributed

Continued on next slide
12
Solution to the Example Oral Contraceptive and
Bone Mineral Density
The boxplots based on the two samples are
reasonably symmetric and because there are no
outliers, it is plausible to assume the
populations are approximately normal.
  • Let
  • µ1 true mean bone mineral density for women
    who never used oral contraceptives, and
  • µ2 true mean bone mineral density for women
    who used oral contraceptives
  • Then µ1 -µ2 difference in mean bone mineral
    density.

Continued on next slide
13
Solution to the Example of Oral Contraceptive and
Bone Mineral Density
H0 µ1 - µ2 0 Ha µ1 - µ2 gt 0 Significance
level a .05.
Appendix Table 4 shows that the area under the t
curve with 17 df to the right of 1.1 is 0.143.
Therefore, P-value .143 gt .05. Conclusion We
fail to reject H0. The is no convincing evidence
to support the claim that the mean bone mineral
density is lower for women who use oral
contraceptives.
14
  • Exercise Comparing Two Treatments
  • First year nursing students enrolled in a
    science course were randomly assigned to one of
    the two groups the control group in which
    students were provided with conventional
    study-skill guidelines, and the treatment group,
    in which students were provided homework
    prescription based on their identified learning
    style. Does the treatment improve science grades?
    Use a significance level of .01.

Answer P-value 0. There is evidence that the
mean science score for the treatment group is
higher.
15
The Two-Sample t Confidence Interval for the
Difference Between Two Population or Treatment
Means
The confidence interval for µ1 - µ2 is
The t critical value is based on the degree of
freedom
where
Truncate (round down) df to obtain an integer
value.
16
Assumptions for the Two Sample t Confidence
Interval for µ1 µ2
  • The two samples are independently chosen random
    samples, and
  • The sample sizes are both large (in general, n1 ?
    30 and n2 ? 30) or the population distributions
    are approximately normal,

17
  • Example Effect of Talking on Blood Pressure
  • Does talking elevate blood pressure?
    Patients with high blood pressure were randomly
    assigned to one of two groups. Those in the first
    group (the talking group) were asked questions
    about their medical history and sources of stress
    before their blood pressure was measured. Those
    in the second group (the counting group) were
    asked to count from 1 to 100 four times before
    their blood pressure was measured. The following
    data values are the diastolic blood pressure (in
    mm Hg) for the two groups.
  • Talking 104 110 107 112 108 103 108 118
  • n1 8, mean 108.75, standard deviation
    4.74
  • Counting 110 96 103 98 100 109 97 105
  • n2 8, mean 102.25, standard deviation
    5.39
  • Construct a 95 confidence interval to estimate
    µ1 µ2, the difference in mean diastolic blood
    pressure for the two treatments.

18
Solution to the Example Effect of Talking on
Blood Pressure To estimate µ1 µ2, the
difference in mean diastolic pressure for the two
treatments, we calculate a 95 confidence
interval
In the 13-df row of Appendix Table 3, the t
critical value for a 95 confidence level is
2.16. The 95 interval is then
The 95 confidence for µ1 µ2 is (1.02,
11.98). We estimate that the mean diastolic blood
pressure when talking is higher than the mean
when counting by somewhere between 1.02 and 11.98
mm Hg.
19
11.2 Inference Concerning the Difference Between
Two Population or Treatment Means Using Paired
Samples
  • Independent samples Two samples for which the
    individuals or objects in the first sample are
    selected independently from those in the second
    sample.
  • Paired samples Two samples for which each
    observation in one sample is paired in a
    meaningful way with a particular observation in
    the second sample.

20
  • An Example of Paired Samples Alcohol Content
    of Wine
  • Are wine drinkers being misled by bottle
    labels? The London Times reported the results of
    a study that compared the actual alcohol content
    on wine to the alcohol content printed on the
    label. A representative subset (six bottles) of
    the resulting data is given in the following
    table

The two samples are paired rather than
independent because both samples are composed of
observations on the same six bottles. Let µ1
the mean actual alcohol content for the
population of all wines and µ2 the mean labeled
alcohol content for the population of all wines.
Hypotheses of interest might be
H0 µ1 - µ2 0 versus Ha µ1 - µ2 gt 0.
21
Paired t Test
µd the mean value of the difference
population µd µ1 µ2 sd the standard
deviation of the difference population Let µ0 be
hypothesized value.
  • Equivalent Hypothesis when samples are paired
  • H0 µd µ0
  • Ha µd gt µ0
  • Ha µd lt µ0
  • Ha µd ? µ0
  • Hypothesis
  • H0 µ1 µ2 µ0
  • Ha µ1 µ2 gt µ0
  • Ha µ1 µ2 lt µ0
  • Ha µ1 µ2 ? µ0

22
Paired t Test
  • Assumptions
  • The samples are paired.
  • The n sample differences can be viewed as a
    random sample from a population of differences.
    (Because inferences about µd can be based on the
    n observed sample differences, the original
    two-sample problem becomes a familiar one-sample
    problem.)
  • The number of sample differences is large (in
    general, at least 30) or the population
    distribution of differences is approximately
    normal.

23
Paired t Test for Computing Two Population Means
  • H0 µ0hypothesized value
  • Test statistic
  • where and sd are the sample mean and
    standard deviation of the differences,
    respectively.
  • Alternative Hypothesis P-Value
  • Ha µd gt hypothesized value area in upper tail

  • Ha µd lt hypothesized value area in lower tail
  • Ha µd ? hypothesized value sum of area in two
    tails

24
  • Example Improve Memory by Playing Chess?
  • Can taking chess lessons and playing chess
    daily improve memory? Twelve sixth-grade students
    who have never previously played chess
    participated in a program where they took chess
    lessons and played chess daily for 9 months.
    Their memory test (the Test of Cognitive Skills)
    scores before and after the program are given in
    the table below.
  • Do the students who participated in the
    chess program tend to achieve higher memory
    scores after completion of the program?
  • Use a significance level a .05.

25
Solution to the Example Improve memory by
playing chess
Let µ1 mean memory score for six-graders with
no chess training µ2 mean memory score for
six-graders after chess training and µd µ1 -
µ2 mean memory score difference between
students with no chess training and students who
have completed chess training
  • The researchers wants to know if students
    who have completed chess training have higher
    memory scores ( µ1 lt µ2, or µd lt 0 ), so a
    lower-tailed paired t-test with a .05 is used.
  • Null and Alternative hypotheses H0 µd 0
    Ha µd lt 0
  • Using the 12 differences in the table
    we compute

P-value From the df 12 - 1 11 column of
Appendix Table 4, we find P-value area
to the left of - 4.56 area to the right of 4.56
0 Conclusion Reject H0 because P-value lt a.
The data support the theory that mean memory
score is higher for six-graders after completion
of the chess training.
26
Using Excel for Paired t-Test Input the
data of two samples in two columns, and then go
to Data Analysis like we did mny times before. In
the Data Analysis dialog box, choose t-test
Paired Two Sample for Means.
27
In the dialog box, enter Variable 1 range
(A1A13) and Variable 2 range (B1B13). Enter
also the hypothesized value (0 in this problem),
and significance level (0.05) for Alpha (a)
28
The results are shown in the following outcome
box. The P-value for one-tailed test is
0.000405682, which is almost 0.
29
  • Exercise Spatial Ability
  • A study discussed the effects of prenatal
    exposure to DES (a drug banned by FDA in 1971) on
    spatial ability in males born to mothers treated
    with DES. In the study 10 males exposed to DES
    and their unexposed brothers underwent a spatial
    ability test, and the results are

Is the mean spatial ability score higher for
those unexposed to DES? Using significance level
a.05.
Answer P-value .020. There is evidence that
the mean spatial ability score is lower for those
who were exposed to DES than those who were
unexposed.
30
Paired t Confidence Interval for µd
  • When
  • The samples are paired,
  • The n sample differences can be viewed as a
    random sample from a population of differences,
    and
  • The number of sample differences is large (in
    general, at least 30) or the population
    distribution of differences is approximately
    normal,
  • The paired t confidence interval for µd is
  • For a specified confidence level, the (n-1) df
    row of the table inside the back cover gives the
    appropriate t critical value.

31
Example Lactic Acid in the Blood After Exercise
  • To exam the effect of exercise on the amount of
    lactic acid in blood, eight males were selected
    at random from those attending a week-long
    training camp. Blood lactate levels were measured
    before and after playing three racquetball games
    and the results were shown in the table.
  • Construct a 95 confidence interval.

32
  • Solution to Example Lactic Acid in the Blood
    After Exercise
  • Solution Because n 8, df 8 - 1 7, and
    the t critical value for 95 confidence level is
    2.37. Use the 8 differences we compute

The 95 confidence interval is (-20.568, -6.692).
We are 95 confident that the mean increase in
blood lactate level is somewhere between -20.568,
-6.692.
Analysis The 8 men were selected at random
from a training camp. The box plot of the 8
sample differences is consistent with a
difference population that is approximately
normal, so the paired t confidence interval is
appropriate.
33
In Chapter 9, we learned how to use Excel to find
confidence intervals. This slide and next will
help us to review the process. Just need to
remember that for paired t confidence intervals
we should use the differences.
34
The 95 confidence interval for the difference in
mean blood lactate level is -13.63 (mean)
6.922 (-20.5, -6.7).
35
  • Exercise Benefits of Ultrasound Revisited.
  • Ultrasound is often used in the treatment of
    soft tissue injuries. In an experiment of
    investigate the effect of an ultrasound and
    stretch therapy on knee extension, range of
    motion was measured both before and after
    treatment for a sample of physical therapy
    patients. A subset of the data is given in the
    accompanying table.
  • Construct a 95 confidence interval.

Answer 95 confidence interval (-6.60, -0.18)
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