Transcript Review PowerPoint Show - paired groups - RIT

Two Sample Hypothesis Testing
for Means from Paired or
Dependent Groups
© 2010 Pearson Prentice Hall. All rights reserved
A sampling method is independent when
the individuals selected for one sample
do not dictate which individuals are to be
in a second sample. A sampling method
is dependent when the individuals
selected to be in one sample are used to
determine the individuals to be in the
second sample.
Dependent samples are often referred to
as matched-pairs samples.
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Statistical inference methods on matched-pairs
data use the same methods as inference on a
single population mean with  unknown,
except that the differences are analyzed.
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Testing Hypotheses Regarding the Difference
of Two Means Using a Matched-Pairs Design
To test hypotheses regarding the mean difference of
matched-pairs data, the following must be
satisfied:
1. the sample is obtained using simple random
sampling
2. the sample data are matched pairs,
3. the differences are normally distributed with no
outliers or the sample size, n, is large (n ≥ 30).
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Step 1: Determine the null and alternative
hypotheses. The hypotheses can be
structured in one of three ways, where
d is the population mean difference of
the matched-pairs data.
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Step 2: Select a level of significance, , based
on the seriousness of making a
Type I error.
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Step 3: Compute the test statistic
d
t0 
sd
n
which approximately follows Student’s
t-distribution with n-1 degrees of
freedom. The values of d and sd are

the mean and standard deviation of the
differenced data.

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P-Value Approach
Step 4: Use Table VI to determine the P-value
using n-1 degrees of freedom.
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P-Value Approach
Two-Tailed
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P-Value Approach
Left-Tailed
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P-Value Approach
Right-Tailed
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P-Value Approach
Step 5: If the P-value < , reject the null
hypothesis.
If the P-value ≥ α, fail to reject the null
hypothesis
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Step 6: State the conclusion in the
context of the problem.
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These procedures are robust, which means that
minor departures from normality will not
adversely affect the results. However, if the
data have outliers, the procedure should not
be used.
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Parallel Example 2: Testing a Claim Regarding
Matched-Pairs Data
The following data represent the cost of a one night stay
in Hampton Inn Hotels and La Quinta Inn Hotels for a
random sample of 10 cities. Test the claim that
Hampton Inn Hotels are priced differently than La
Quinta Hotels at the =0.05 level of significance.
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City
Dallas
Tampa Bay
St. Louis
Seattle
San Diego
Chicago
New Orleans
Phoenix
Atlanta
Orlando
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Hampton Inn
129
149
149
189
109
160
149
129
129
119
La Quinta
105
96
49
149
119
89
72
59
90
69
Solution
This is a matched-pairs design since the hotel prices
come from the same ten cities. To test the
hypothesis, we first compute the differences and
then verify that the differences come from a
population that is approximately normally
distributed with no outliers because the sample size
is small.
The differences (Hampton - La Quinta) are:
24 53 100 40 -10 71 77 70 39 50
with d = 51.4 and sd = 30.8336.
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Solution
No violation of normality assumption.
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Solution
No outliers.
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Solution
Step 1: We want to determine if the prices differ:
H0: d = 0 versus
H1: d  0
Step 2: The level of significance is =0.05.
Step 3: The test statistic is
51.4
t0 
30.8336
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 5.2716.
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Solution: P-Value Approach
Step 4: Because this is a two-tailed test, the P-value
is two times the area under the t-distribution
with n-1=10-1=9 degrees of freedom to the
right of the test statistic t0=5.27.
That is, P-value = 2P(t > 5.27) ≈
2(0.00026)=0.00052 (using technology).
Approximately 5 samples in 10,000 will yield
results as extreme as we obtained if the null
hypothesis is true.
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Solution: P-Value Approach
Step 5: Since the P-value is less than the level of
significance =0.05, we reject the null
hypothesis.
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Solution
Step 6: There is sufficient evidence to conclude that
Hampton Inn hotels and La Quinta hotels are
priced differently at the =0.05 level of
significance.
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Objective 3
• Construct and Interpret Confidence Intervals
for the Population Mean Difference of
Matched-Pairs Data
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Confidence Interval for Matched-Pairs Data
A (1-)100% confidence interval for d is given
by
sd
Lower bound: d  t  
n
2
s
Upper bound: d  t   d
n
2
The critical 
value t/2 is determined using n-1
degrees of freedom.
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
Confidence Interval for Matched-Pairs Data
Note: The interval is exact when the population
is normally distributed and approximately
correct for nonnormal populations, provided
that n is large.
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Parallel Example 4: Constructing a Confidence Interval for
Matched-Pairs Data
Construct a 90% confidence interval for the mean
difference in price of Hampton Inn versus La
Quinta hotel rooms.
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Solution
•
We have already verified that the differenced
data come from a population that is
approximately normal with no outliers.
•
Recall d= 51.4 and sd = 30.8336.
•
From Table VI with  = 0.10 and 9 degrees of
freedom, we find t/2 = 1.833.

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Solution
Thus,
•
30.8336 
 33.53
Lower bound = 51.4 1.833
 10 
•
30.8336 
Upper bound = 51.4 1.833
 69.27
 10 

We are 90% confident that the mean difference in hotel room
price for Ramada Inn versus La Quinta Inn is between
$33.53 and$69.27.
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