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
1
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)
2
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
3
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
4
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
5
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)
6
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)
7
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)
8
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
9
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.
10
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.
11
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.
12
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)
13
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)
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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
15
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.
16
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.
17
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.
18
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)
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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
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)
20