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Section 9.4
Inferences About Two Means
(Matched Pairs)
Objective
Compare of two matched-paired means using
two samples from each population.
Hypothesis Tests and Confidence Intervals of
two dependent means use the t-distribution
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Definition
Two samples are dependent if there is some
relationship between the two samples so that
each value in one sample is paired with a
corresponding value in the other sample.
Two samples can be treated as the matched
pairs of values.
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Examples
• Blood pressure of patients before they are
given medicine and after they take it.
• Predicted temperature (by Weather
Forecast) and the actual temperature.
• Heights of selected people in the morning
and their heights by night time.
• Test scores of selected students in Calculus-I
and their scores in Calculus-II.
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Example 1
First sample: weights of 5 students in April
Second sample: their weights in September
These weights make 5 matched pairs
Third line: differences between April weights and
September weights (net change in weight for each
student, separately)
In our calculations we only use differences (d),
not the values in the two samples.
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Notation
d
Individual difference between two matched
paired values
μd
Population mean for the difference of the
two values.
n
Number of paired values in sample
d
Mean value of the differences in sample
sd
Standard deviation of differences in sample
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Requirements
(1) The sample data are dependent
(i.e. they make matched pairs)
(2) Either or both the following holds:
The number of matched pairs is large (n>30)
or
The differences have a normal distribution
All requirements must be satisfied to make a
Hypothesis Test or to find a Confidence Interval
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Tests for Two Dependent Means
Goal: Compare the mean of the differences
H0 : μd = 0
H0 : μd = 0
H0 : μd = 0
H1 : μd ≠ 0
H1 : μd < 0
H1 : μd > 0
Two tailed
Left tailed
Right tailed
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Finding the Test Statistic
t=
d – µd
sd
n
Note: md = 0 according to H0
degrees of freedom: df = n – 1
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Test Statistic
Degrees of freedom
df = n – 1
Note: Hypothesis Tests are done in same way as
in Ch.8-5
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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
•
Calculate the Sample Stats
•
Find the Test Statistic
•
Find the Critical Value(s)
•
State the Initial Conclusion and Final Conclusion
Note: Same process as in Chapter 8
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Example 1
Assume the differences in weight form a normal
distribution.
Use a 0.05 significance level to test the claim that for the
population of students, the mean change in weight from
September to April is 0 kg
(i.e. on average, there is no change)
Claim: μd = 0
using α = 0.05
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Example 1
d Data: -1 -1 4 -2 1
H0 : µd = 0
H1 : µd ≠ 0
Two-Tailed
H0 = Claim
n=5
d = 0.2
t = 0.186
-tα/2 = -2.78
Sample Stats
t-dist.
df = 4
tα/2 = 2.78
sd = 2.387
Use StatCrunch: Stat – Summary Stats – Columns
Test Statistic
Critical Value
tα/2 = t0.025 = 2.78
(Using StatCrunch, df = 4)
Initial Conclusion: Since t is not in the critical region, accept H0
Final Conclusion: We accept the claim that mean change in weight from
September to April is 0 kg.
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Example 1
d Data: -1 -1 4 -2 1
Sample Stats
H0 : µd = 0
H1 : µd ≠ 0
n=5
Two-Tailed
H0 = Claim
d = 0.2
sd = 2.387
Use StatCrunch: Stat – Summary Stats – Columns
Stat → T statistics→ One sample → With summary
Sample mean:
0.2
Sample std. dev.: 2.387
Sample size:
5
● Hypothesis Test
Null: proportion=
0
Alternative
≠
P-value = 0.8605
Initial Conclusion: Since P-value is greater than α (0.05), accept H0
Final Conclusion: We accept the claim that mean change in weight from
September to April is 0 kg.
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Confidence Interval Estimate
We can observe how the two proportions relate by
looking at the Confidence Interval Estimate of μ1–μ2
CI = ( d – E, d + E )
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Example 2
Sample Stats
n=5
d = 0.2
tα/2 = t0.025 = 2.78
Find the 95% Confidence Interval Estimate
of μd from the data in Example 1
sd = 2.387
(Using StatCrunch, df = 4)
CI = (-2.8, 3.2)
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Example 2
Sample Stats
n=5
d = 0.2
Find the 95% Confidence Interval Estimate
of μd from the data in Example 1
sd = 2.387
Stat → T statistics→ One sample → With summary
Sample mean:
0.2
Sample std. dev.: 2.387
Sample size:
5
● Confidence Interval
Level:
0.95
CI = (-2.8, 3.2)
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Section 9.5
Comparing Variation in Two Samples
Objective
Compare of two population variances using two
samples from each population.
Hypothesis Tests and Confidence Intervals of
two variances use the F-distribution
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Requirements
(1) The two populations are independent
(2) The two samples are random samples
(3) The two populations are normally distributed
(Very strict!)
All requirements must be satisfied to make a
Hypothesis Test or to find a Confidence Interval
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Important
The first sample must have a larger sample
standard deviation s1 than the second
sample. i.e. we must have
s1 ≥ s2
If this is not so, i.e. if s1 < s2 , then we will
need to switch the indices 1 and 2
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Notation
σ1
First population standard deviation
s1
First sample standard deviation
n1
First sample size
σ2
Second population standard deviation
s2
Second sample standard deviation
n2
Second sample size
Note: Use index 1 on sample/population with the
larger sample standard deviation (s)
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Tests for Two Proportions
The goal is to compare the two population
variances (or standard deviations)
H0 : σ1 = σ2
H0 : σ1 = σ2
H1 : σ1 ≠ σ2
H1 : σ1 > σ2
Two tailed
Right tailed
Note: We do not consider σ1 < σ2
(since we used indexes 1 and 2 such that s1 is larger)
Note: We only test the relation between σ1 and σ2
(not the actual numerical values)
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The F-Distribution
Similar to the χ2-dist.
• Not symmetric
• Non-negative values (F ≥ 0)
• Depends on two degrees of freedom
df1 = n1 – 1
(Numerator df )
df2 = n2 – 1
(Denominator df )
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The F-Distribution
df1 = n1 – 1
df2 = n2 – 1
On StatCrunch: Stat – Calculators – F
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Test Statistic for Hypothesis
Tests with Two Variances
F=
s
s
2
1
2
2
Where s12 is the first (larger) of the two
sample variances
Because of this, we will always have F ≥ 1
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Use of the F Distribution
If the two populations have equal variances,
then F = s12/s22 will be close to 1
(Since s12 and s22 will be close in value)
If the two populations have different variances,
then F = s12/s22 will be greater than 1
(Since s12 will be larger than s22)
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Conclusions from the F-Distribution
Values of F close to 1 are evidence in favor
of the claim that the two variances are equal.
Large values of F, are evidence against this
claim (i.e. it suggest there is some difference
between the two)
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Steps for Performing a Hypothesis
Test on Two Independent Means
•
Write what we know
•
Index the variables such that s1 ≥ s2 (important!)
•
State H0 and H1
•
Draw a diagram
•
Find the Test Statistic
•
Find the two degrees of freedom
•
Find the Critical Value(s)
•
State the Initial Conclusion and Final Conclusion
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Example 1
Below are sample weights (in g) of quarters made before
1964 and weights of quarters made after 1964.
When designing coin vending machines, we must
consider the standard deviations of pre-1964 quarters and
post-1964 quarters.
Use a 0.05 significance level to test the claim that the
weights of pre-1964 quarters and the weights of post-1964
quarters are from populations with the same standard
deviation.
Claim: σ1 = σ2 using α = 0.05
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Example 1
H0 : σ1 = σ2
H1 : σ1 ≠ σ2
n1 = 40
n2 = 40
s1 = 0.08700 s2 = 0.06194
α = 0.05
(Note: s1≥s2)
Two-Tailed
H0 = Claim
F = 1.973
Test Statistic
2
Fα/2 = 1.891
2
Degrees of Freedom
df1 = n1 – 1 = 39
df2 = n2 – 1 = 39
Critical Value
F is in the critical region
Using StatCrunch: Stat – Calculators – F
Fα/2 = F0.025 = 1.891
Initial Conclusion: Since F is in the critical region, reject H0
Final Conclusion: We reject the claim that the weights of the pre-1964
and post-1964 quarters have the same standard deviation
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Example 1
H0 : σ1 = σ2
H1 : σ1 ≠ σ2
n1 = 40
n2 = 40
s1 = 0.08700 s2 = 0.06194
s12 = 0.007569
Two-Tailed
H0 = Claim
α = 0.05
(Note: s1≥s2)
s22 = 0.003837
Stat → Variance → Two sample → With summary
Sample 1: Variance: 0.007569
40
Size:
Sample 2: Variance: 0.003837
Size:
● Hypothesis Test
Null: variance ratio=
1
Alternative
≠
40
P-value = 0.0368
Initial Conclusion: Since P-value is less than α (0.05), reject H0
Final Conclusion: We reject the claim that the weights of the pre-1964
and post-1964 quarters have the same standard deviation
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