MTH120_Chapter11
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Transcript of MTH120_Chapter11
Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill/Irwin
11- 1
Chapter ElevenTwo Sample Tests of Two Sample Tests of HypothesisHypothesisGOALS
When you have completed this chapter, you will be able to:ONE
Understand the difference between dependent and independent samples.
THREE
Conduct a test of hypothesis about the difference between two independent population means when at least one sample has less than 30 observations.
TWO
Conduct a test of hypothesis about the difference between two independent population means when both samples have 30 or more observations.
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Chapter Eleven continuedTwo Sample Tests of Two Sample Tests of HypothesisHypothesisGOALS
When you have completed this chapter, you will be able to:
FIVEConduct a test of hypothesis regarding the difference in two population proportions.
FOUR
Conduct a test of hypothesis about the mean difference between paired or dependent observations.
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Comparing two populations Comparing two populations
We wish to know whether the distribution of the differences in sample means has a mean of 0.
If both samples contain at least 30 observations we use the z distribution as the test statistic.
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Comparing two populationsComparing two populations
No assumptions about the shape of the populations are required.
The samples are from independent populations.The formula for computing the value of z is:
2
22
1
21
21
n
s
n
s
XXz
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EXAMPLE 1EXAMPLE 1
Two cities, Bradford and Kane are separated only by the Conewango River. There is competition between the two cities. The local paper recently reported that the mean household income in Bradford is $38,000 with a standard deviation of $6,000 for a sample of 40 households. The same article reported the mean income in Kane is $35,000 with a standard deviation of $7,000 for a sample of 35 households. At the .01 significance level can we conclude the mean income in Bradford is more?
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EXAMPLE 1 EXAMPLE 1 continuedcontinued
Step 1: State the null and alternate hypotheses.
H0: µB µK ; H1: µB > µK
Step 2: State the level of significance. The .01 significance level is stated in the problem.Step 3: Find the appropriate test statistic. Because both samples are more than 30, we can use z as the test statistic.
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Example 1 Example 1 continuedcontinued
Step 4: State the decision rule.
The null hypothesis is rejected if z is greater than
2.33.
98.1
35
)000,7($
40
)000,6($
000,35$000,38$22
z
Step 5: Compute the value of z and make a decision.
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Example 1 Example 1 continuedcontinued
The decision is to not reject the null hypothesis. We cannot conclude that the mean household income in Bradford is larger.
The p-value is:
P(z > 1.98) = .5000 - .4761 = .0239
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Small Sample Tests of MeansSmall Sample Tests of Means
The t distribution is used as the test statistic if one or more of the samples have less than 30 observations.
The required assumptions are:1. Both populations must follow the
normal distribution.2. The populations must have equal
standard deviations.3. The samples are from independent
populations.
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Small sample test of means Small sample test of means continuedcontinued
Finding the value of the test statistic requires two steps.1. Pool the sample standard deviations.
2
)1()1(
21
222
2112
nn
snsns p
21
2
21
11
nns
XXt
p
Determine the value of t from the following formula.
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EXAMPLE 2EXAMPLE 2
A recent EPA study compared the highway fuel economy of domestic and imported passenger cars. A sample of 15 domestic cars revealed a mean of 33.7 mpg with a standard deviation of 2.4 mpg. A sample of 12 imported cars revealed a mean of 35.7 mpg with a standard deviation of 3.9. At the .05 significance level can the EPA conclude that the mpg is higher on the imported cars?
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Example 2 Example 2 continuedcontinued
Step 1: State the null and alternate hypotheses.
H0: µD µI ; H1: µD < µI
Step 2: State the level of significance.The .05 significance level is stated in the problem.Step 3: Find the appropriate test statistic. Both samples are less than 30, so we use the t distribution.
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EXAMPLE 2 EXAMPLE 2 continuedcontinued
Step 5: We compute the pooled variance:
918.921215
)9.3)(112()4.2)(115(
2
))(1())(1(
22
21
222
2112
nn
snsns p
Step 4: The decision rule is to reject H0 if t<-1.708.
There are 25 degrees of freedom.
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Example 2 Example 2 continuedcontinued
We compute the value of t as follows.
H0 is not rejected. There is insufficient sample evidence to claim a higher mpg on the imported cars.
640.1
12
1
15
1312.8
7.357.33
11
21
2
21
nns
XXt
p
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Hypothesis Testing Involving Paired Hypothesis Testing Involving Paired ObservationsObservations
Independent samples are samples that are not related in any way.
Dependent samples are samples that are paired or related in some fashion. For example:
If you wished to buy a car you would look at the same car at two (or more) different dealerships and compare the prices. If you wished to measure the effectiveness of a new diet you would weigh the dieters at the start and at the finish of the program.
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Hypothesis Testing Involving Paired Hypothesis Testing Involving Paired ObservationsObservations
Use the following test when the samples are dependent:
where is the mean of the differences is the standard deviation of the differencesn is the number of pairs (differences)
td
s nd
/
dsd
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EXAMPLE 3EXAMPLE 3
An independent testing agency is comparing the daily rental cost for renting a compact car from Hertz and Avis. A random sample of eight cities revealed the following information. At the .05 significance level can the testing agency conclude that there is a difference in the rental charged?
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EXAMPLE 3 EXAMPLE 3 continuedcontinued
City Hertz ($) Avis ($)Atlanta 42 40Chicago 56 52
Cleveland 45 43Denver 48 48
Honolulu 37 32Kansas City 45 48
Miami Seattle
41
46
39
50
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EXAMPLE 3 EXAMPLE 3 continuedcontinued
Step 1: H Hd d0 10 0: :
Step 2: H0 is rejected if t < -2.365 or t > 2.365. We use the t distribution with 7 degrees of freedom.
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Example 3 Example 3 continuedcontinued
City Hertz Avis d d2
Atlanta 42 40 2 4
Chicago 56 52 4 16
Cleveland 45 43 2 4
Denver 48 48 0 0
Honolulu 37 32 5 25
Kansas City 45 48 -3 9
Miami 41 39 2 4
Seattle 46 50 -4 16
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Example 3 Example 3 continuedcontinued
00.18
0.8
n
dd
1623.3
188
878
1
222
n
n
dd
sd
894.081623.3
00.1
ns
dt
d
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Example 3 Example 3 continuedcontinued
Step 3: Because 0.894 is less than the critical value, do not reject the null hypothesis. There is no difference in the mean amount charged by Hertz and Avis.
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Two Sample Tests of ProportionsTwo Sample Tests of Proportions
We investigate whether two samples came from populations with an equal proportion of successes.
The two samples are pooled using the following formula.
where X1 and X2 refer to the number of successes in the respective samples of n1 and n2.
21
21
nn
XXpc
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Two Sample Tests of Proportions Two Sample Tests of Proportions continuedcontinued
The value of the test statistic is computed from the following formula.
21
21
)1()1(
n
pp
n
pp
ppz
cccc
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Example 4Example 4
Are unmarried workers more likely to be absent from work than married workers? A sample of 250 married workers showed 22 missed more than 5 days last year, while a sample of 300 unmarried workers showed 35 missed more than five days. Use a .05 significance level.
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Example 4 Example 4 continued continued
The null and the alternate hypothesis are:
H0: U M H1: U > M
The null hypothesis is rejected if the computed value of z is greater than 1.65.
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Example 4 Example 4 continuedcontinued
The pooled proportion is
250300
2235
cp
The value of the teat statistic is
10.1
250
)1036.1(1036.
300
)1036.1(1036.250
22
300
35
z
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Example 4 Example 4 continuedcontinued
The null hypothesis is not rejected. We cannot conclude that a higher proportion of unmarried workers miss more days in a year than the married workers.
The p-value is:
P(z > 1.10) = .5000 - .3643 = .1457