Download Ch9a

Chapter 9
Inferences from
Two Samples
9.2 Inferences About Two Proportions
9.3 Inferences About Two Means (Independent)
9.4 Inferences About Two Means (Matched Pairs)
9.5 Comparing Variation in Two Samples
1
Objective
Compare the parameters of two populations
using two samples from each population.
Use Confidence Intervals and Hypothesis Tests
For the first population use index 1
For the second population use index 2
9.2
9.3
9.4
9.5
Compare p1 , p2
Compare µ1 , µ2 (Independent)
Compare µ1 , µ2 (Matched Pairs)
Compare σ12 , σ22
2
Section 9.2
Inferences About Two Proportions
Objective
Compare the proportions of two populations
using two samples from each population.
Hypothesis Tests and Confidence Intervals
of two proportions use the z-distribution
3
Notation
First Population
p1
First population proportion
n1
First sample size
x1
Number of successes in first sample
p1
First sample proportion
4
Notation
Second Population
p2
Second population proportion
n2
Second sample size
x2
Number of successes in second sample
p2
Second sample proportion
5
Definition
The pooled sample proportion p
x1 + x2
p= n +n
1
2
q =1–p
6
Requirements
(1) Have two independent random samples
(2) For each sample:
The number of successes is at least 5
The number of failures is at least 5
Both requirements must be satisfied to make a
Hypothesis Test or to find a Confidence Interval
7
Tests for Two Proportions
The goal is to compare the two proportions
H0 : p1 = p2
H1 : p1  p2
Two tailed
H0 : p 1 = p 2
H0 : p1 = p2
H1 : p 1 < p 2
H1 : p1 > p2
Left tailed
Right tailed
Note: We only test the relation between p1 and p2
(not the actual numerical values)
8
Finding the Test Statistic
z=
^ )–(p –p )
( p^1 – p
2
1
2
Note: p1 –
pq
pq
n 1 + n2
p2 =0 according to H0
This equation is an altered form of the test
statistic for a single proportion (see Ch. 8-3)
9
Test Statistic
Note: Hypothesis Tests are done in same way as
in Ch.8 (but with different test statistics)
10
Steps for Performing a Hypothesis
Test on Two Proportions
•
Write what we know
•
State H0 and H1
•
Draw a diagram
•
Calculate the sample and pooled proportions
•
Find the Test Statistic
•
Find the Critical Value(s)
•
State the Initial Conclusion and Final Conclusion
Note: Same process as in Chapter 8
11
Example 1
The table below lists results from a simple random sample
of front-seat occupants involved in car crashes.
Use a 0.05 significance level to test the claim that the
fatality rate of occupants is lower for those in cars
equipped with airbags.
p1 : Proportion of fatalities with airbags
p2 : Proportion of fatalities with no airbags
What we know:
x1 = 41
n1 = 11541
x2 = 52
n2 = 9853
Claim
p1 < p2
α = 0.05
Claim: p1 < p2
Note: Each sample has more than 5 successes and failures, thus fulfilling the requirements 12
Given:
Example 1
x2 = 52
n2 = 9853
α = 0.05
Claim: p1 < p2
Diagram
H0 : p1 = p2
H1 : p1 < p2
x1 = 41
n1 = 11541
Left-Tailed
H1 = Claim
Sample Proportions
z = –1.9116
z-dist.
–zα = –1.645
Pooled Proportion
Test Statistic
Critical Value
(Using StatCrunch)
Initial Conclusion: Since z is in the critical region, reject H0
Final Conclusion: We Accept the claim the fatality rate of occupants is
lower for those who wear seatbelts
13
Example 1
Given:
α = 0.05
Claim: p1 < p2
x2 = 52
n2 = 9853
Diagram
H0 : p1 = p2
H1 : p1 < p2
x1 = 41
n1 = 11541
Left-Tailed
H1 = Claim
z = –1.9116
z-dist.
–zα = –1.645
Using StatCrunch
Stat → Proportions → Two sample → With summary
Sample 1: Number of successes: . 41
● Hypothesis Test
Number of observations: 11541
Null: prop. diff.=
Sample 2: Number of successes: . 52
Alternative
Number of observations: 9853
0
<
P-value = 0.028
Initial Conclusion: Since P-value is less than α (with α = 0.05), reject H0
Final Conclusion: We Accept the claim the fatality rate of occupants is
lower for those who wear seatbelts
14
Confidence Interval Estimate
We can observe how the two proportions relate by
looking at the Confidence Interval Estimate of p1–p2
CI = ( (p1–p2) – E, (p1–p2) + E )
Where
15
Example 2
Use the same sample data in Example 1 to construct a
90% Confidence Interval Estimate of the difference
between the two population proportions (p1–p2)
x1 = 41
n1 = 11541
x2 = 52
n2 = 9853
p1 = 0.003553
p2 = 0.005278
CI = (-0.003232, -0.000218 )
Note: CI negative implies p1–p2 is negative. This implies p1<p2
16
Example 2
Use the same sample data in Example 1 to construct a
90% Confidence Interval Estimate of the difference
between the two population proportions (p1–p2)
x1 = 41
n1 = 11541
x2 = 52
n2 = 9853
p1 = 0.003553
p2 = 0.005278
CI = (-0.003232, -0.000218 )
Note: CI negative implies p1–p2 is negative. This implies p1<p2
17
Example 2
Use the same sample data in Example 1 to construct a
90% Confidence Interval Estimate of the difference
between the two population proportions (p1–p2)
x1 = 41
n1 = 11541
x2 = 52
n2 = 9853
Using StatCrunch
Stat → Proportions → Two sample → With summary
Sample 1: Number of successes: . 41
● Confidence Interval
Number of observations: 11541
Level 0.9
Sample 2: Number of successes: . 52
Number of observations: 9853
CI = (-0.003232, -0.000218 )
Note: CI negative implies p1–p2 is negative. This implies p1<p2
18
Interpreting Confidence Intervals
If a confidence interval limits does not contain 0, it
implies there is a significant difference between
the two proportions (i.e. p1 ≠ p2).
Thus, we can interpret a relation between the two
proportions from the confidence interval.
In general:
• If p1 = p2 then the CI should contain 0
• If p1 > p2 then the CI should be mostly positive
• If p1 > p2 then the CI should be mostly negative
19
Example 3
Drug Clinical Trial
Chantix is a drug used as an aid to stop smoking. The number of
subjects experiencing insomnia for each of two treatment groups in
a clinical trial of the drug Chantix are given below:
Chantix Treatment
Placebo
Number in group
129
805
Number experiencing insomnia
19
13
(a) Use a 0.01 significance level to test the claim proportions of
subjects experiencing insomnia is the same for both groups.
(b) Find the 99% confidence level estimate of the difference of the
two proportions. Does it support the result of the test?
What we know:
x1 = 41
n1 = 129
x2 = 52
n2 = 9853
α = 0.01
Claim: p1= p2
Note: Each sample has more than 5 successes and failures, thus fulfilling the requirements 20
Example 3a
Given: x1 = 19
n1 = 129
Diagram
H0 : p1 = p2
H1 : p1 ≠ p2
x2 = 13
n2 = 805
Two-Tailed
H0 = Claim
Sample Proportions
-zα/2 = -2.576
α = 0.01
Claim: p1= p2
z-dist.
z = 7.602
zα/2 = 2.576
Pooled Proportion
Test Statistic
Critical Value
(Using StatCrunch)
Initial Conclusion: Since z is in the critical region, reject H0
Final Conclusion: We Reject the claim the proportions of the subjects
experiencing insomnia is the same in both groups.
21
Example 3a
Given: x1 = 19
n1 = 129
Diagram
H0 : p1 = p2
H1 : p1 ≠ p2
α = 0.01
Claim: p1= p2
x2 = 13
n2 = 805
z-dist.
Two-Tailed
H0 = Claim
Using StatCrunch
Stat → Proportions → Two sample → With summary
Sample 1: Number of successes: . 19
Number of observations: 129
Sample 2: Number of successes: . 13
Number of observations: 805
● Hypothesis Test
Null: prop. diff.=
Alternative
0
≠
P-value < 0.0001
i.e. the P-value
is very small
Initial Conclusion: Since the P-value is less than α (0.01), reject H0
Final Conclusion: We Reject the claim the proportions of the subjects
experiencing insomnia is the same in both groups.
22
Example 3b
Use the same sample data in Example 3 to construct a
99% Confidence Interval Estimate of the difference
between the two population proportions (p1–p2)
x1 = 19
n1 = 129
x2 = 13
n2 = 805
p1 = 0.14729
p2 = 0.01615
CI = (0.0500, 0.2123 )
Note: CI does not contain 0 implies p1 and p2 have significant difference. 23
Example 3b
Use the same sample data in Example 3 to construct a
99% Confidence Interval Estimate of the difference
between the two population proportions (p1–p2)
x1 = 19
n1 = 129
x2 = 13
n2 = 805
Using StatCrunch
Stat → Proportions → Two sample → With summary
Sample 1: Number of successes: . 19
Number of observations: 129
Sample 2: Number of successes: . 13
Number of observations: 805
● Confidence Interval
Level
0.9
CI = (0.0500, 0.2123 )
Note: CI does not contain 0 implies p1 and p2 have significant difference. 24
25
Section 9.3
Inferences About Two Means
(Independent)
Objective
Compare the proportions of two independent
means using two samples from each population.
Hypothesis Tests and Confidence Intervals of
two proportions use the t-distribution
26
Definitions
Two samples are independent if the sample
values selected from one population are not
related to or somehow paired or matched with
the sample values from the other population
Examples:
Flipping two coins (Independent)
Drawing two cards (not independent)
27
Notation
First Population
μ1
First population mean
σ1
First population standard deviation
n1
First sample size
x1
First sample mean
s1
First sample standard deviation
28
Notation
Second Population
μ2
Second population mean
σ2
Second population standard deviation
n2
Second sample size
x2
Second sample mean
s2
Second sample standard deviation
29
Requirements
(1) Have two independent random samples
(2) σ1 and σ2 are unknown and no assumption is
made about their equality
(3) Either or both the following holds:
Both sample sizes are large (n1>30, n2>30)
or
Both populations have normal distributions
All requirements must be satisfied to make a
Hypothesis Test or to find a Confidence Interval
30
Tests for Two Independent Means
The goal is to compare the two Means
H0 : μ1 = μ2
H0 : μ1 = μ2
H0 : μ1 = μ2
H1 : μ1 ≠ μ2
H1 : μ1 < μ2
H1 : μ1 > μ2
Two tailed
Left tailed
Right tailed
Note: We only test the relation between μ1 and μ2
(not the actual numerical values)
31
Finding the Test Statistic
x  x  

t
1
2
1
2
1
 2

2
2
s
s

n1 n2
Note: 1 –
2 =0 according to H0
Degrees of freedom: df = smaller of n1 – 1 and n2 – 1.
This equation is an altered form of the test statistic
for a single mean when σ unknown (see Ch. 8-5)
32
Test Statistic
Degrees of freedom
df = min(n1 – 1, n2 – 1)
Note: Hypothesis Tests are done in same way as
in Ch.8 (but with different test statistics)
33
Steps for Performing a Hypothesis
Test on Two Independent Means
•
Write what we know
•
State H0 and H1
•
Draw a diagram
•
Find the Test Statistic
•
Find the Degrees of Freedom
•
Find the Critical Value(s)
•
State the Initial Conclusion and Final Conclusion
Note: Same process as in Chapter 8
34
Example 1
A headline in USA Today proclaimed that “Men,
women are equal talkers.” That headline referred to
a study of the numbers of words that men and
women spoke in a day.
Use a 0.05 significance level to test the claim that
men and women speak the same mean number of
words in a day.
35
n1 = 186
x1 = 15668.5
s1 = 8632.5
Example 1
H0 : µ1 = µ2
H1 : µ1 ≠ µ2
Two-Tailed
H0 = Claim
n2 = 210
x2 = 16215.0
s2 = 7301.2
t-dist.
df = 185
t = 7.602
-tα/2 = -1.97
Test Statistic
α = 0.05
Claim: μ1 = μ2
tα/2 = 1.97
Degrees of Freedom
df = min(n1 – 1, n2 – 1) = min(185, 209) = 185
Critical Value
tα/2 = t0.025 = 1.97
(Using StatCrunch)
Initial Conclusion: Since t is not in the critical region, accept H0
Final Conclusion: We accept the claim that men and women speak the
same average number of words a day.
36
Example 1
H0 : µ1 = µ2
H1 : µ1 ≠ µ2
n1 = 186
x1 = 15668.5
s1 = 8632.5
Two-Tailed
H0 = Claim
α = 0.05
Claim: μ1 = μ2
Stat → T statistics → Two sample → With summary
Sample 1:
Using StatCrunch
(Be sure to not use pooled variance)
n2 = 210
x2 = 16215.0
s2 = 7301.2
Sample 2:
Mean
Std. Dev.
Size
Mean
Std. Dev.
Size
15668.5
8632.5
186
16215.0
7301.2
210
● Hypothesis Test
Null: prop. diff.=
Alternative
0
≠
(No pooled variance)
P-value = 0.4998
Initial Conclusion: Since P-value > α (0.05), accept H0
Final Conclusion: We accept the claim that men and women speak the
same average number of words a day.
37
Confidence Interval Estimate
We can observe how the two proportions relate by
looking at the Confidence Interval Estimate of μ1–μ2
CI = ( (x1–x2) – E, (x1–x2) + E )
2
2
Where
df = min(n1–1, n2–1)
38
Example 2
Use the same sample data in Example 1 to construct a
95% Confidence Interval Estimate of the difference
between the two population proportions (µ1–µ2)
n1 = 186
n2 = 210
df = min(n1–1, n2–1) = min(185, 210) = 185
df = min(n1–1, n2–1) = min(185, 210) = 185
x1 = 15668.5
x2 = 16215.0
tα/2 = t0.05/2 = t0.025 = 1.973
tα/2 = t0.1/2 = t0.05 = 1.973
s1 = 8632.5
s2 = 7301.2
x1 - x2 = 15668.5 – 16215.0 = -546.5
x1 - x2 = 15668.5 – 16215.0 = -546.5
(x1 - x2) + E = -546.5 + 1596.17 = 1049.67
(x1 - x2) – E = -546.5 – 1596.17 = -2142.67
CI = (-2142.7, 1049.7)
39
Example 2
Use the same sample data in Example 1 to construct a
95% Confidence Interval Estimate of the difference
between the two population proportions (µ1–µ2)
n1 = 186
x1 = 15668.5
s1 = 8632.5
n2 = 210
x2 = 16215.0
s2 = 7301.2
Stat → T statistics → Two sample → With summary
Using StatCrunch
CI = (-2137.4, 1044.4)
Sample 1:
Sample 2:
Mean
Std. Dev.
Size
Mean
Std. Dev.
Size
15668.5
8632.5
186
16215.0
7301.2
210
● Confidence Interval
Level:
0.95
(No pooled variance)
Note: slightly different because
of rounding errors
40
Example 3
Consider two different classes. The students in the first
class are thought to generally be older than those in the
second. The students’ ages for this semester are summed
as follows:
n1 = 93
x1 = 21.2
s1 = 2.42
n2 = 67
x2 = 19.8
s2 = 4.77
(a) Use a 0.1 significance level to test the claim that the
average age of students in the first class is greater than the
average age of students in the second class.
(b) Construct a 90% confidence interval estimate of the
difference in average ages.
41
Example 3a
H0 : µ1 = µ2
H1 : µ1 > µ2
n1 = 93
x1 = 21.2
s1 = 2.42
n2 = 67
x2 = 19.8
s2 = 4.77
α = 0.1
Claim: µ1 > µ2
Right-Tailed
H1 = Claim
t-dist.
df = 66
Test Statistic
tα/2 = 1.668
t = 7.602
Degrees of Freedom
df = min(n1 – 1, n2 – 1) = min(92, 66) = 66
Critical Value
tα/2 = t0.05 = 1.668
(Using StatCrunch)
Initial Conclusion: Since t is in the critical region, reject H0
Final Conclusion: We accept the claim that the average age of students
in the first class is greater than that in the second.
42
Example 3a
H0 : µ1 = µ2
H1 : µ1 > µ2
n1 = 93
x1 = 21.2
s1 = 2.42
Right-Tailed
H1 = Claim
α = 0.1
Claim: µ1 > µ2
Stat → T statistics → Two sample → With summary
Sample 1:
Using StatCrunch
(Be sure to not use pooled variance)
n2 = 67
x2 = 19.8
s2 = 4.77
Sample 2:
Mean
Std. Dev.
Size
Mean
Std. Dev.
Size
21.2
2.42
93
19.8
4.77
67
● Hypothesis Test
Null: prop. diff.=
Alternative
0
≠
(No pooled variance)
P-value = 0.0299
Initial Conclusion: Since P-value < α (0.1), reject H0
Final Conclusion: We accept the claim that the average age of students
in the first class is greater than that in the second.
43
Example 3b
(90% Confidence Interval)
n1 = 93
x1 = 21.2
s1 = 2.42
n2 = 67
x2 = 19.8
s2 = 4.77
α = 0.1
df = min(n1–1, n2–1) = min(92, 66) = 66
tα/2 = t0.1/2 = t0.05 = 1.668
x1 - x2 = 21.2 – 19.8 = 1.4
mp
(x1 - x2) + E = 1.4 + 1.058 = 2.458
(x1 - x2) – E = 1.4 – 1.058 = 0.342
CI = (0.34, 2.46)
44
Example 3b
(90% Confidence Interval)
n1 = 93
x1 = 21.2
s1 = 2.42
n2 = 67
x2 = 19.8
s2 = 4.77
α = 0.1
Stat → T statistics → Two sample → With summary
Sample 1:
Using StatCrunch
(Be sure to not use pooled variance)
Sample 2:
Mean
Std. Dev.
Size
Mean
Std. Dev.
Size
21.2
2.42
93
19.8
4.77
67
● Hypothesis Test
Null: prop. diff.=
Alternative
0
≠
(No pooled variance)
CI = (0.35, 2.45)
45