Download Section 11.3 - Navidi/Monk

HYPOTHESIS TESTS FOR THE
DIFFERENCE BETWEEN TWO MEANS:
PAIRED SAMPLES
Section 11.3
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Objectives
1.
2.
Perform a hypothesis test with matched pairs using the P-value
method
Perform a hypothesis test with matched pairs using the critical
value method
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OBJECTIVE 1
Perform a hypothesis test with matched pairs
using the P-value method
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Paired Samples
A sample of eight automobiles were run to determine their mileage, in
miles per gallon. Then each car was given a tune-up, and run again to
measure the mileage a second time.
The sample mean mileage was
higher after tune-up. We would
like to determine how strong the
evidence is that the population
mean mileage is higher after
tune-up.
Automobile
1
2
3
4
5
6
7
8
After
35.44
35.17
31.07
31.57
26.48
23.11
25.18
32.39
Before
33.76
34.30
29.55
30.90
24.92
21.78
24.30
31.25
These are paired samples, because each value before tune-up is
paired with the value from the same car after tune-up.
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Matched Pairs
When we have paired samples, the pairs are called matched pairs. By
computing the difference between the values in each matched pair, we
construct a sample of differences:
If we denote the population mean
mileage before tune-up by 𝜇1 , and
the population mean mileage after
tune-up by 𝜇2 , then we are interested
in the difference 𝜇1 − 𝜇2 . Because
these are paired samples, the
population mean of the differences,
𝜇𝑑 , is the same as 𝜇1 − 𝜇2 . Therefore,
performing a hypothesis test on 𝜇𝑑 is
the same as performing a hypothesis
test on the difference of the population
means 𝜇1 − 𝜇2 .
Automobile
1
2
3
4
5
6
7
8
After
35.44
35.17
31.07
31.57
26.48
23.11
25.18
32.39
Before Difference
33.76
1.68
34.30
0.87
29.55
1.52
30.90
0.67
24.92
1.56
21.78
1.33
24.30
0.88
31.25
1.14
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Notation
We use the following notation:
• 𝑑 is the sample mean of the differences between the values in the
matched pairs.
• 𝑠𝑑 is the sample standard deviation of the differences between the
values in the matched pairs.
• 𝜇𝑑 is the population mean difference for the matched pairs.
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Assumptions
The method just described requires the following assumptions:
Assumptions:
1. We have a simple random sample of matched pairs.
2. Either the sample size is large (𝑛 > 30), or the differences between
items in the matched pairs show no evidence of strong skewness and
no outliers. This is required to be sure that 𝑑 will be approximately
normally distributed.
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Hypothesis Test with Matched-Pair Data Using
the P-Value Method
Step 1:
Step 2:
State the null and alternate hypotheses.
If making a decision, choose a significance level 𝛼.
Step 3:
Compute the test statistic 𝑡 =
Step 4:
Compute the P-value. The P-value is an area under the 𝑡 curve with
𝑛 − 1 degrees of freedom.
Step 5:
Interpret the P-value. If making a decision, reject 𝐻0 if the P-value is less than
or equal to the significance level 𝛼.
State a conclusion.
Step 6:
𝑑 − 𝜇0
.
𝑠𝑑 𝑛
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Example
Using the data about a tune-up improving car
engine gas mileage, test 𝐻0 : 𝜇d = 0 versus
𝐻1 : 𝜇d > 0. Use the 𝛼 = 0.01 significance level.
Solution:
We have a simple random sample of differences.
Because the sample size is small (𝑛 = 8), we must
check for signs of strong skewness or outliers.
Following is a dotplot of the differences.
Automobile Difference
1
1.68
2
0.87
3
1.52
4
0.67
5
1.56
6
1.33
7
0.88
8
1.14
The dotplot does not reveal any outliers or strong skewness. Therefore we
may proceed.
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Solution
The null and alternate hypotheses are 𝐻0 : 𝜇d = 0 versus 𝐻1 : 𝜇d > 0.
We compute the sample mean and sample standard deviation of the differences.
These are
𝑑 = 1.20625
𝑠𝑑 = 0.37317
Under the assumption that 𝐻0 is true, 𝜇d = 𝜇0 = 0, and the value of the test
statistic is
𝑡=
𝑑 − 𝜇0
𝑠𝑑 𝑛
1.20625 −0
8
= 0.37317
= 9.1427.
Also, the test statistic has a 𝑡 distribution with 𝑛 − 1 = 8 − 7 degrees of freedom.
Since this is a right tailed test, the P-value is the area to the right of the observed
value of 𝑡 = 9.1427. Using Table A.3 or technology, we find that P = 0.0000193.
The P-value is nearly 0, which is very strong evidence against 𝐻0 . Because P <
0.01, we reject 𝐻0 at the 𝛼 = 0.01 level. We conclude that the gas mileage
increased after a tune-up.
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OBJECTIVE 2
Perform a hypothesis test with matched pairs
using the critical value method
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Testing a Hypothesis with Matched-Pair Data
Using the Critical Value Method
The critical value method for matched-pair data is essentially the same as that for a
population mean with 𝜎 unknown. The assumptions for the critical value method are the
same as for the P-value method.
Step 1: State the null and alternate hypotheses.
Step 2: Choose a significance level 𝛼 and find the critical value or values.
𝑑 − 𝜇0
.
𝑠𝑑 𝑛
Step 3:
Compute the test statistic 𝑡 =
Step 4:
Determine whether to reject 𝐻0 , as follows:
Step 5:
State a conclusion.
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Example
For a sample of nine automobiles,
the mileage (in 1000s of miles) at
which the original front brake pads
were worn to 10% of their original
thickness was measured, as was
the mileage at which the original
rear brake pads were worn to 10%
of their original thickness. The
results are given.
Automobile
1
2
3
4
5
6
7
8
9
Rear
42.7
36.7
46.1
46.0
39.9
51.7
51.6
46.1
47.3
Front Difference
32.8
9.9
26.6
10.1
35.6
10.5
36.4
9.6
29.2
10.7
40.9
10.8
40.9
10.7
34.8
11.3
36.6
10.7
The differences in the last column of the table are Rear − Front. Can you
conclude that the mean time for the rear brake pads to wear out is longer than
the mean time for the front pads? Use the α = 0.05 significance level.
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Solution
Since the sample size is small, we construct a dotplot.
The dotplot shows no evidence of outliers or extreme skewness, so we may
proceed.
We are interested in determining whether the mean time for the rear pads is
longer than for the front. Therefore, the hypotheses are
𝐻0 : 𝜇d = 0
𝐻1 : 𝜇d > 0
Because this is a right-tailed test, the critical value is the value for which the
area to the right is 0.05. The sample size is 𝑛 = 9, so there are 9 − 1 = 8
degrees of freedom. The critical value is 𝑡𝛼 = 1.860.
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Solution
We compute the sample mean and standard deviation of the differences
𝑑 = 10.1444
𝑠𝑑 = 1.0333
The test statistic is 𝑡 =
𝑑−0
𝑠𝑑 𝑛
10.478
= 0.5215
9
= 60.28.
This is a right-tailed test, so we reject 𝐻0 if 𝑡 ≥ 𝑡𝛼 . Because 𝑡 = 60.28 and 𝑡𝛼 =
1.860, we reject at 𝐻0 the 𝛼 = 0.05 level.
We conclude that the mean time for rear brake pads to wear out is longer than
the mean time for front brake pads.
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You Should Know…
•
•
How to perform a hypothesis test with matched pairs using the Pvalue method
How to perform a hypothesis test with matched pairs using the
critical value method
Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.