Download Chapter 9. Inferences from Two Samples

Chapter 9. Inferences from Two Samples
Chapter Problem: Is the “Freshman 15” real, or is it a myth?
1. Popular belief: college students gain 15 lb during freshmen year
2. Journal of American College Health, Vol.55, No.1.
3. Data set 3 in Appendix B (weights in kg, 15 lb = 6.8 kg)
4. Two weight for each of 67 students (obtained in Sept. and Apr.)
5. Use weight change: April weight – September weight
6. The published paper has some limitations:
i.
ii.
All subjects volunteered for the study
All the subjects were attending Rutgers
7. The “Freshmen 15” constitutes a Claim made about the
population of college students
8. Let d denote the mean of “April weight – September weight”
9. H0: d = 15
1
9-1 Review and Preview
1. Chapter 7: Construction CI
2. Chapter 8: methods of testing claims made about population
3. This Chapter:
i.
ii.
Extend the methods for estimating values of population parameters
Extend the methods for testing hypotheses to situations involving two
sets of sample data instead of one.
4. Methods for using sample data from two populations so that
inferences can be made about those populations
Test that H0: d = 15
Test the claim that the proportion of children who contract polio is less
for children given the Silk Vaccine than for children given placebo
iii. Test the claim that subjects treated with Lipitor have a mean cholesterol
level that is lower than the mean cholesterol level for subjects given a
placebo
i.
ii.
2
9-2 Inference About Two Populations
1. Testing a claim made about the two population proportions
2. Constructing a CI estimating the difference between the two
population proportions
(can be also applied to claims about probabilities, percentages)
3
9-2 Inference About Two Populations
1. Testing a claim made about the two population proportions
Objective: Test a claim about two population proportions or
construct a CI of the difference between two population proportions.
Notation:
p1 = population proportion for population 1
n1 = sample size for population 1
x1 = number of success in the sample for population 1
x
pˆ 1  1
n1
qˆ1  1  pˆ1
(similar for population 2, change index from 1 to 2)
4
9-2 Inference About Two Populations
1. Testing a claim made about the two population proportions
Pooled Sample Proportion:
the pooled sample proportion is denoted by p and is given by:
p
x1  x2
n1  n2
q  1 p
Requirements:
1) The sample proportions are from two simple random samples that are
independent. (two samples are independent if the sample values selected from
one population are not related to or somewhat naturally paired or matched
with the sample values selected from the other sample.)
2) np ≥ 5, nq ≥ 5 (in words, the number of success is at least 5, the number of
failure is also at least 5)
5
9-2 Inference About Two Populations
1. Testing a claim made about the two population proportions
Test statistic for two proportions (with H0 : p1 = p2 )
( pˆ 1  pˆ 2 )  ( p1  p2 )
where p1 – p2 = 0 (assumed in null hypothesis)
pq p q

n1
n2
x
x
pˆ 1  1 and pˆ 2  2 ,
n1
n2
z
p is the pooled sample proportion, and q  1  p.
P-Value: use Table A-2. Follow the procedure in Figure 8 – 5.
Critical Value: use Table A-2. (Based on , follow steps in
section 8-2.
6
9-2 Inference About Two Populations
1. Testing a claim made about the two population proportions
Confidence Interval Estimate of p1 – p2
The CI estimate of the difference p1 – p2 is:
ˆ1  p
ˆ 2 )  E  ( p1  p2 )  ( p
ˆ1  p
ˆ2)  E
(p
where the margin of error E is given by E  z

2
pˆ 1qˆ1 pˆ 2 qˆ 2
 .
n1
n2
Rounding: rounding the confidence interval limits to three
significant digits.
7
9-2 Inference About Two Populations
1. Testing a claim made about the two population proportions
e.g.1 Do Airbags Save Lives? The table below lists results from a
simple random sample of front-seat occupants involved in car
crashes (real data, Chance, Vol.18, No. 2). Use 0.05 significance
level to test the claim that the fatality rate of occupants is lower for
those in the cars equipped with airbags.
Airbags Available
No Airbags Available
Occupants Fatalities
41
52
Total Number of Occupants 11,541
9,853
Sol. Requirements. 1) both are simple random samples, and two
samples are independent. 2) in case 1, #success = 41, # of failure =
11,500; in case 2, # success = 52, # failure = 9,801. All > 5.
8
9-2 Inference About Two Populations
1. Testing a claim made about the two population proportions
P-Value Method.
Step 1. p1 < p2
Step 2. Alternative: p1 ≥ p2
Step 3. H0: p1 = p2
H1: p1 < p2
Step 4.  = 0.05
Step 5. Use normal distribution. Use pooled sample:
p
x1  x2
41  52

 0.004347
n1  n2 11,541  9,853
q  1  p  1  0.004347  0.995653
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9-2 Inference About Two Populations
1. Testing a claim made about the two population proportions
Step 6. Test statistic. Find the value of the test statistic:
( pˆ 1  pˆ 2 )  ( p1  p2 )
z
pq p q

n1
n2
z
52 
 41


0
 11,541 9,853 
 1.91
(0.004347)(0.995653) (0.004347)(0.995653)

11,541
9,853
This is a left-tailed test, so P-value = 0.0281
Step 7. P-value < 0.05, we reject the null hypothesis of p1 = p2 .
10
9-2 Inference About Two Populations
1. Testing a claim made about the two population proportions
Interpretation. See Figure 8-7 for wording) Because we reject the
null hypothesis, we conclude that there is sufficient evidence to
support the claim that the proportion of accident fatalities for
occupants with airbags is less than the proportion of fatalities
for occupants in cars without airbags. Based on these result, it
appears that airbags are effective in saving lives.
11
9-2 Inference About Two Populations
1. Testing a claim made about the two population proportions
Traditional Method.
Step 6. For  = 0.05, left-tailed test, the critical value is z = –1.645.
The test statistic has a value of –1.91 falls in the critical region. So
we reject Null hypothesis.
12
9-2 Inference About Two Populations
2. CI method
CI Method. With 90% confidence interval, z/2 = 1.645.
E  z
2
pˆ 1qˆ1 pˆ 2 qˆ 2

n1
n2
 41  11,500   52  9,801 


 


11
,
541
11
,
541
9
,
853
9
,
853




 1.645 
11,541
9,853
 0.001507
p̂1 = 41/11,541 = 0.003553,
p̂2 = 52/9,853 = 0.005278, so
Lower limit  ( pˆ1  pˆ 2 )  E  0.00323;
Upperlimit  ( pˆ1  pˆ 2 )  E  0.000218;
13
9-2 Inference About Two Populations
2. CI Method
CI Method.
So, the CI is: –0.00323 < (p1 – p2) < –0.000218
Interpretation. The CI do not contain 0, implying that the significant
difference between two proportions. The CI method suggest
that the fatality rate is lower for occupants in cars with airbags
than for occupants in cars without airbags. The CI also provides
an estimate of the amount of the difference between the two
fatality rates.
14
9-2 Inference About Two Populations
Rationale: why do the procedures of this section work?
p̂1 can be approximated by normal distribution with mean p and
1
standard deviation p1q1 / n1
p̂2 can be approximated by normal distribution with mean p2 and
standard deviation p2 q2 / n2
pˆ1  pˆ 2 can be approximated by normal distribution with mean p1 – p2
and variance
 (2pˆ  pˆ )   p2ˆ   p2ˆ 
1
2
1
2
p1q1 p2 q2

n1
n2
15
9-2 Inference About Two Populations
Rationale: why do the procedures of this section work?
Use the pooled estimate. The pooled estimate of the common value
of p1 and p2 is
x1  x2
p
n1  n2
.
If we replace p1 and p2 by p, and replace q1 and q2 by q , the
variance leads to the following standard deviation:
 ( pˆ  pˆ ) 
1
2
pq p q

n1
n2
This leads to the z test statistic introduced earlier.
16
9-2 Inference About Two Populations
Rationale: why do the procedures of this section work?
The form of the confidence interval requires an expression for the
variance different from the one given above. When constructing a CI
estimate of the difference between two different proportions, we
don’t assume that the two proportions are equal, and we estimate the
standard deviation as
 ( pˆ  pˆ ) 
1
In the test statistic,
z
2
pˆ 1qˆ1 pˆ 2 qˆ 2

n1
n2
( pˆ 1  pˆ 2 )  ( p1  p2 )
,
pˆ 1qˆ1 pˆ 2 qˆ 2

n1
n2
Use the positive and negative values of z (for two tails) and solve for
p1 – p2.
17
9-3 Inference About Two Means: Independent Samples
Part 1. 1 and 2 are unknown, are NOT assumed to be equal
Part 2.
1. 1 and 2 are known
2. 1 and 2 are unknown, 1 = 2
Typically,  is unknown
18
9-3 Inference About Two Means: Independent Samples
Part 1. 1 and 2 are unknown, are NOT assumed to be equal
Definition.
Two samples are independent if the sample values from one
population are not related to or somehow naturally paired or
matched with the same sample values from the other population.
Two samples are dependent if sample values are paired. (that is,
each pair of sample values consists of two measurements from the
same subject (such as before/after data), or each pair of sample
values consists of matched pairs (such as husband/wife data), where
the matching is based on some inherent relationship.)
19
9-3 Inference About Two Means: Independent Samples
Part 1. 1 and 2 are unknown, are NOT assumed to be equal
e.g.1 Independent Samples. University of Arizona psychologists
conducted a study in which 210 women and 186 men wore
microphones so that the numbers of words that they spoke could be
recorded. The sample word counts for men and the sample word
counts for women are two independent samples.
E.g.2. Dependent Samples. Rutgers University researchers
conducted a study in which 67 students were weighed in September
of their freshmen year and again in April of their Freshmen year.
The two samples are dependent, because each September weight is
paired with the April weight for the same student.
20
9-3 Inference About Two Means: Independent Samples
Part 1. 1 and 2 are unknown, are NOT assumed to be equal
e.g.3 Clinical Experiment. In an experiment designed to study the
effectiveness of treatments for viral group,
• 46 children were treated with low humidity
• 46 other were treated with high humidity
• Used The Westley Group Score to assess the results after 1 hour.
• Both samples have the same number of subjects and the sample
scores can be listed in adjacent columns of the same length;
• Two samples are independent. the scores are from two different
groups of subjects.
21
9-3 Inference About Two Means: Independent Samples
Part 1. 1 and 2 are unknown, are NOT assumed to be equal
Objective: Test a claim about two independent means or construct a
CI estimate of the difference between two population means.
Notation:
1 = population mean for population 1
1 = population standard deviation for population 1
n1 = size of the sample for population 1
x1 = sample mean
s1 = sample standard deviation
(similar for population 2, change index from 1 to 2)
22
9-3 Inference About Two Means: Independent Samples
Part 1. 1 and 2 are unknown, are NOT assumed to be equal
Requirement:
1) 1 and 2 are unknown, and it is not assumed that 1 = 2
2) Two samples are independent
3) Both samples are simple random samples
4) Either or both of these conditions are satisfied
– The two samples are both larger (n1 > 30, n2 > 30)
– Both samples come from populations having normal distribution
23
9-3 Inference About Two Means: Independent Samples
Part 1. 1 and 2 are unknown, are NOT assumed to be equal
Hypothesis Test Statistic for Two Means: Independent Samples
t
( x1  x2 )  ( 1   2 )
s12 s22

n1 n2
( where 1 – 2 is often assumed to be 0 )
Degree of Freedom (denoted by df): when finding critical values or
P-Values, use the following:
1) In this book, we use: df = smaller of n1 – 1 and n2 – 1 (simple
and conservative estimate)
2) More accurate but more difficult estimate:
( A  B) 2
df 
A2
B2

n1  1 n2  1
where A =
s12
n1
and B =
s22
n2
24
9-3 Inference About Two Means: Independent Samples
Part 1. 1 and 2 are unknown, are NOT assumed to be equal
P-Values: Refer to the t distribution in Table A-3, use the
procedure in Figure 8-5
Critical values: Refer to the t distributions in Table A-3.
CI Estimate of 1 – 2 : Independent Samples
The CI estimate of the difference 1 – 2 is:
( x1  x2 )  E  (1  2 )  ( x1  x2 )  E
where the margin of error E is given by E  t

2
s12 s22

n1 n2
.
25
9-3 Inference About Two Means: Independent Samples
Part 1. 1 and 2 are unknown, are NOT assumed to be equal
e.g.4 Are Men and Women Equal Talkers? USA-Today proclaim:
“Men, women are equal talkers”. Research report, “Are Women
Really More Talkative Than Men”, Science, Vol.317, No. 5834.
Data set 8 in Appendix B.
Number of Words Spoken In a Day
Men
Women
n1 = 186
n2 = 210
x1 = 15,668.5
x2 = 16,215.0
s1 = 8632.5
s2 = 7301.2
Sol. Requirements. 1) 1 and 2 are unknown, and it is not assumed
that 1 = 1; 2) two samples are independent; 3) simple random
samples; 4) Both samples are large (can also verify the normality
using Statdisk)
26
9-3 Inference About Two Means: Independent Samples
Part 1. 1 and 2 are unknown, are NOT assumed to be equal
Traditional Method
Step 1. 1 = 2
Step 2. Alternative: 1  2
Step 3. H0: 1 = 2 (original claim)
H1: 1  2
We now proceed with the assumption that 1 – 2 = 0
Step 4.  = 0.05
Step 5. We have two independent samples, and we are testing a
claim about two population means, we use t distribution.
27
9-3 Inference About Two Means: Independent Samples
Part 1. 1 and 2 are unknown, are NOT assumed to be equal
Step 6. Test statistic:
t
( x1  x2 )  ( 1   2 )
2
1
2
2
s
s

n1 n2

(15,668.5  16,215)  0
2
8632.5
7301.2

186
210
2
 0.676
We find the critical values are t = 1.972 (A-3, df = 185, use df =
200). (use Calculator, can find P-value = 0.4998)
Step 7. The test statistic is not in the critical region. We fail to reject null
hypothesis.
Interpretation: There is not sufficient evidence to warrant rejection of the claim
that men and women speak the same number of words in a day. There does
not appear to be a significant difference between the two means.
28
9-3 Inference About Two Means: Independent Samples
Part 1. 1 and 2 are unknown, are NOT assumed to be equal
e.g.5 Using the sample data in e.g.4, construct a 95% CI of the
difference between the mean number of words spoken by men
and the mean number of words spoken by women.
CI Method
Step 1. Requirement check done in e.g.4. t/2 = 1.972. is found in
e.g.4. So the margin of error:
E  t
2
s12 s22
8632.52 7301.2 2

 1.972

 1595.4
n1 n2
186
210
Use x1 = 15,668.5, x2 = 16,215.0, and E = 1595.4, we get
–2141.9 < (1 – 2) < 1048.9
29
9-3 Inference About Two Means: Independent Samples
Part 1. 1 and 2 are unknown, are NOT assumed to be equal
Interpretation. We are 95% confident that the limit of –2141.9 words
and 1048.9 words actually do contain the difference between
the two population means. Because this CI contains 0, this
suggest that it is very possible that the two population means
are equal. So there is no significant difference between the two
means.
Rationale: why do the Test Statistic and CI have the particular form
presented here?
x1  x2 can be approximated by t distribution with mean
2
2
standard deviation equals to s1  s2
n1 n2
1 – 2 and
30
9-3 Inference About Two Means: Independent Samples
Part 2. 1 and 2 are known
Requirement:
1) 1 and 2 are known
2) Two samples are independent
3) Both samples are simple random samples
4) Either or both of these conditions are satisfied
– The two samples are both larger (n1 > 30, n2 > 30)
– Both samples come from populations having normal distribution
31
9-3 Inference About Two Means: Independent Samples
Part 2. 1 and 2 are known (not realistic)
Hypothesis Test Statistic for Two Means
z
( x1  x2 )  ( 1   2 )
 12
n1

 22
n2
P-Values and Critical Values: refer to Table A-2
CI Estimate of 1 – 2 : Independent Samples
The CI estimate of the difference 1 – 2 is:
( x1  x2 )  E  (1  2 )  ( x1  x2 )  E
where the margin of error E is given by E  z

2
 12
n1

 22
n2
32
9-3 Inference About Two Means: Independent Samples
Part 2. Alternative Method: 1 = 2 and Pool the Sample Variances
Requirement:
1) 1 and 2 are NOT known, but 1 = 2
2) Two samples are independent
3) Both samples are simple random samples
4) Either or both of these conditions are satisfied
– The two samples are both larger (n1 > 30, n2 > 30)
– Both samples come from populations having normal distribution
33
9-3 Inference About Two Means: Independent Samples
Part 2. Alternative Method: 1 = 2 and Pool the Sample Variances
Hypothesis Test Statistic for Two Means
t
Where
( x1  x2 )  ( 1   2 )
s 2p s 2p

n1 n2
(n1  1) s12  (n2  1) s22
s 
(n1  1)  (n2  1)
2
p
(pooled variance)
And df = n1 + n2 – 2.
34
9-3 Inference About Two Means: Independent Samples
Part 2. Alternative Method: 1 = 2 and Pool the Sample Variances
CI Estimate of 1 – 2 : Independent Samples
The CI estimate of the difference 1 – 2 is:
( x1  x2 )  E  (1  2 )  ( x1  x2 )  E
where the margin of error E is given by E  t
s 2p

2
n1

s 2p
n2
And df = n1 + n2 – 2.
35
9-3 Inference About Two Means: Independent Samples
Part 2. 1 and 2 are known
Figure 9-3
36
9-3 Inference About Two Means: Independent Samples
Part 2. Alternative Method: 1 = 2 and Pool the Sample Variances
Why Alternative method?
• Df is little higher, so the hypothesis test has more power
• CI littler narrower
• That’s why statistician use this method.
37
9-4 Inference from Dependent Samples
Introduction.
Methods for testing hypothesis and constructing CI involving mean
of the differences of the values from two dependent populations.
Typical examples of dependent samples:
• Each pair of sample values consists of two measurements from
the same subject
• Each pair of sample consists of a matched pair.
38
9-4 Inference from Dependent Samples
Objective: Test a claim about mean of the differences from
dependent samples or construct CI estimate of the mean of the
differences from dependent samples.
Notation:
d = individual difference between the two values in a single matched pair
d = mean value of the different d for the population of all pairs of data
d = mean value of the difference d for the paired sample data
sd = standard deviation of the differences d for the paired sample data
n = number of pairs of data
39
9-4 Inference from Dependent Samples
Requirements:
1) Sample data are dependent
2) The samples are simple random sample
3) Either or both of these conditions are satisfied
– The number of pairs of sample data is large (n > 30)
– Or the pairs of values have differences that are from a population having a
distribution that is approximately normal.
Hypothesis Test Statistic for Dependent Samples
t
d  d
sd
n
Where degree of freedom df = n – 1
P-values and Critical values: Table A-3 (t-distribution)
40
9-4 Inference from Dependent Samples
Confidence Intervals for Dependent Samples
d  E  d  d  E
where:
E  t
2
sd
n
41
9-4 Inference from Dependent Samples
e.g.1 Hypothesis Test of Claimed Freshman Weight Gain. Data set 8
in Appendix B includes measured weights of college students in
September and April of their freshman year. Table 9-1 lists a small
portion of those sample values. (we use the small portion of the data
so that we can better illustrate the method of hypothesis testing.) Use
the sample data in Table 9-1 to test the claim that for the population
of students, the mean change in weight from September to April is
equal to zero.
Table 9-1 Weight (kg) Measurements of Students in Their Freshman Year
April weight
66 52 68 69 71
September weight
67 53 64 71 70
Difference d = (April weight) – (September weight)
–1 –1
4
–2
1
42
9-4 Inference from Dependent Samples
Requirement:
1) Dependent samples (because the values are paired)
2) NOT simple random sample (because all subjects volunteered
for the study. We will proceed as if it is simple random sample.
3) 5 pairs of data values. Histogram by Statdisk shows that the data
is approximately normal.
43
9-4 Inference from Dependent Samples
Traditional Method
Step 1. d = 0 kg
Step 2. Alternative: d  0 kg
Step 3. H0: d = 0 kg (original claim)
H1: d  0 kg
We now proceed with the assumption that d = 0
Step 4.  = 0.05
Step 5. We use t distribution.
Step 6. d = 0.2; sd = 2.4, so the statistic
t
d  d
sd
n

0.2  0
2.4
5
 0.186
44
9-4 Inference from Dependent Samples
Step 6. Continue: so the t-statistic = 0.186
For  = 0.05, df = 5 – 1 = 4, use the column for 0.05 (Area in two
tails) in table A-3, we find critical values t = 2.776.
Step 7. Because the test statistic does NOT fall in the critical region,
we fail to reject the null hypothesis.
Interpretation. There is not sufficient evidence to warrant rejection
of the claim that for the population of students, the mean change in
weight from September to April is equal to zero. Based on the
sample results listed in table 9-1, there does not appear to be a
significant weight gain from September to April.
45
9-4 Inference from Dependent Samples
P-Value Method
Step 6. Use technology, P-value = 0.8605 which is greater than 0.05.
So we fail to reject the null hypothesis.
46
9-4 Inference from Dependent Samples
e.g.2 Hypothesis Test of Claimed Freshman Weight Gain. Data
values in Data set 8 in Appendix B includes measured weights of
college students in September and April of their freshman year. Use
all 67 data values to test the claim that for the population of students,
the mean change in weight from September to April is equal to zero.
Sol. Test statistic = 2.48. P-value is 0.016. We now reject the null
hypothesis, and conclude that there is sufficient evidence to warrant
rejection of the claim that the mean difference is equal to zero. (there
is a significant change)
Next two examples will construct the CI
47
9-4 Inference from Dependent Samples
e.g.3 CI for Estimating the Mean Weight Change. Using data in
Table 9-1, construct a 95% CI estimate of d , which is the mean of
the “April-September” weight differences of college students in their
freshman year.
Sol. Requirement check see e.g.1. d = 0.2; sd = 2.4; n = 5;
t/2 = 2.776. We find the margin of error
E  t
2
So the CI is
sd
2.4
 2.776 
 3.0
n
5
0.2 – 3.0 < d < 0.2 + 3.0, Or –2.8 < d < 3.2
Interpretation. We have 95% confidence that the limits of –2.8 kg
and 3.2 kg contains the true value of the mean weight change from
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September to April.
9-4 Inference from Dependent Samples
e.g.4 CI for Estimating the Mean Weight Change. Using all data
values in Data set 3 in Appendix B, construct a 95% CI estimate of
d , which is the mean of the “April-September” weight differences
of college students in their freshman year.
Sol. Repeat e.g.3 using Statdisk with all 67 pair of data values, we
obtain:
CI:
0.2306722 < d < 2.127537
Interpretation. The CI suggests that the mean weight gain is likely to
be between 0.2 kg and 2.1 kg. The CI does NOT include zero, so the
larger data set suggest that the typical college student does gain
some weight during the freshman year.
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9-4 Inference from Dependent Samples
e.g.5 Is the Freshman 15 a Myth? Freshman 15 is the claim that d =
6.8 kg (note that 6.8 kg = 15 lb). Using all data values in Data set 3
in Appendix B, to test the claim that d = 6.8 kg, with  = 0.05.
Minitab shows that the statistic = –11.83, and the P-value is 0.000
(round to three decimal places). Because P-value < 0.05, we reject
the null hypotheses, and there is sufficient evidence to warrant
rejection of the claim that the mean weight change is equal to 6.8 kg.
e.g.4 Shows that the mean weight gain is likely between 0.2 kg and
2.1 kg, so the claim of a mean weight gain of 15 lb appears to be
unfounded.
These results shows that the “Freshman 15” is a myth.
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9-4 Inference from Dependent Samples
Experimental Design.
Design principle:
When designing an experiment or planning an observational study,
using dependent samples with paired data is generally better than
using two independent samples.
• E.g. compare the effectiveness of two different types of fertilizer
(one organic and another chemical)
• The fertilizers are to be used on 20 plots of land with equal area
• Divide each of the 20 plots in half. One half treated with organic
fertilizer, another half treated with chemical fertilizer
• Resulted in paired data
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