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Definitions
Section 9.3
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
Inferences About Two Means
(Independent)
Objective
Examples:
Compare the proportions of two independent
means using two samples from each population.
Flipping two coins (Independent)
Hypothesis Tests and Confidence Intervals of
two proportions use the t-distribution
Drawing two cards (not independent)
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Notation
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
Second Population
1
μ2
Second population mean
σ2
Second population standard deviation
n2
Second sample size
x2
Second sample mean
s2
Second sample standard deviation
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Requirements
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Tests for Two Independent Means
The goal is to compare the two Means
(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)
H0 : μ1 = μ2
H0 : μ1 = μ2
H0 : μ1 = μ2
H1 : μ1 ≠ μ2
H1 : μ1 < μ2
H1 : μ1 > μ2
Two tailed
or
Left tailed
Right tailed
Both populations have normal distributions
Note: We only test the relation between μ1 and μ2
(not the actual numerical values)
All requirements must be satisfied to make a
Hypothesis Test or to find a Confidence Interval
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Test Statistic
Finding the Test Statistic
t
x  x  
1
2
1
 2

s12 s22

n1 n2
Note: 1 –
2 =0 according to H0
Degrees of freedom
df = min(n1 – 1, n2 – 1)
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)
Note: Hypothesis Tests are done in same way as
in Ch.8 (but with different test statistics)
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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
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Example 1
2
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.
Note: Same process as in Chapter 8
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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
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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 15668.5
Std. Dev. 8632.5
Size
186
Mean 16215.0
Std. Dev. 7301.2
Size
210
● Hypothesis Test
Null: prop. diff.=
Alternative
0
≠
(No pooled variance)
P-value = 0.4998
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
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.
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Final Conclusion: We accept the claim that men and women speak the
same average number of words a day.
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Example 2
Confidence Interval Estimate
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)
We can observe how the two proportions relate by
looking at the Confidence Interval Estimate of 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
CI = ( (x1–x2) – E, (x1–x2) + E )
2
s2 = 7301.2
x1 - x2 = 15668.5 – 16215.0 = -546.5
x1 - x2 = 15668.5 – 16215.0 = -546.5
2
Where
(x1 - x2) + E = -546.5 + 1596.17 = 1049.67
(x1 - x2) – E = -546.5 – 1596.17 = -2142.67
df = min(n1–1, n2–1)
CI = (-2142.7, 1049.7)
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Example 2
Example 3
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)
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 = 186
x1 = 15668.5
s1 = 8632.5
n2 = 210
x2 = 16215.0
s2 = 7301.2
Stat → T statistics → Two sample → With summary
Sample 1:
Sample 2:
Using StatCrunch
Mean 15668.5
Std. Dev. 8632.5
Size
186
Mean 16215.0
Std. Dev. 7301.2
Size
210
n1 = 93
x1 = 21.2
s1 = 2.42
● Confidence Interval
Level:
3
0.95
(No pooled variance)
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.
Note: slightly different because
of rounding errors
CI = (-2137.4, 1044.4)
Example 3a
H0 : µ1 = µ2
H1 : µ1 > µ2
n1 = 93
x1 = 21.2
n2 = 67
x2 = 19.8
s1 = 2.42
s2 = 4.77
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α = 0.1
Claim: µ1 > µ2
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Example 3a
H0 : µ1 = µ2
Right-Tailed
H1 = Claim
t-dist.
df = 66
H1 : µ1 > µ2
n1 = 93
x1 = 21.2
n2 = 67
x2 = 19.8
s1 = 2.42
s2 = 4.77
Right-Tailed
H1 = Claim
Stat → T statistics → Two sample → With summary
Sample 1:
Mean
Std. Dev.
Size
Sample 2:
Mean
Std. Dev.
Size
Using StatCrunch
Test Statistic
𝒕=
𝒙1 −𝒙𝟐
𝒔1 𝒔2
+
𝒏1 𝒏2
=
21.2−19.8
(2.42)𝟐 (4.77)𝟐
+
93
67
= 2.207
tα/2 = 1.668
t = 7.602
(Be sure to not use pooled variance)
α = 0.1
Claim: µ1 > µ2
21.2
2.42
93
19.8
4.77
67
● Hypothesis Test
Null: prop. diff.=
Alternative
0
≠
(No pooled variance)
P-value = 0.0299
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
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.
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Final Conclusion: We accept the claim that the average age of students
in the first class is greater than that in the second.
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Example 3b
(90% Confidence Interval)
n1 = 93
x1 = 21.2
s1 = 2.42
n2 = 67
x2 = 19.8
s2 = 4.77
α = 0.1
Example 3b
(90% Confidence Interval)
df = min(n1–1, n2–1) = min(92, 66) = 66
𝒔1
𝒏1
Sample 1:
Mean
Std. Dev.
Size
Sample 2:
Mean
Std. Dev.
Size
Using StatCrunch
x1 - x2 = 21.2 – 19.8 = 1.4
2
𝒔
mp
+ 𝒏1 = 1.668
1
n2 = 67
x2 = 19.8
s2 = 4.77
α = 0.1
Stat → T statistics → Two sample → With summary
tα/2 = t0.1/2 = t0.05 = 1.668
𝑬 = 𝒕𝜶
n1 = 93
x1 = 21.2
s1 = 2.42
(Be sure to not use pooled variance)
2.42 𝟐
93
+
4.77 𝟐
67
21.2
2.42
93
19.8
4.77
67
● Hypothesis Test
Null: prop. diff.=
Alternative
0
≠
(No pooled variance)
= 1.058
(x1 - x2) + E = 1.4 + 1.058 = 2.458
(x1 - x2) – E = 1.4 – 1.058 = 0.342
CI = (0.34, 2.46)
CI = (0.35, 2.45)
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