# 9 Uji Hipotesa Dua Parameter
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Transcript of # 9 Uji Hipotesa Dua Parameter
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Chapter 10
Estimation and Hypothesis Testing
for Two Population Parameters
Business Statistics: A Decision-Making Approach
8th Edition
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Chapter Goals
After completing this chapter, you should be able to:
Test hypotheses or form interval estimates for
two independent population means
Standard deviations known
Standard deviations unknown
two means from paired samples
the difference between two population proportions
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Estimation for Two Populations
Estimating two population values
Population means,
independent samples
Paired samples
Population proportions
Group 1 vs. independent
Group 2
Same group before vs.
after treatment
Proportion 1
vs.
Proportion 2
Examples:
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Difference Between Two Means
Population means, independent
samples
σ1 and σ2 known
σ1 and σ2 unknown, n1 and n2 30
σ1 and σ2 unknown, n1 or n2 < 30
Goal: Form a confidence interval for the difference
between two population means, μ1 – μ2
The point estimate for the difference is
x1 – x2
*
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Independent Samples
Different data sources
Unrelated
Independent
Sample selected from one population has no effect on the sample selected from the other population
Use the difference between 2 sample means
Use z test or pooled variance t test
Population means, independent
samples
σ1 and σ2 known
σ1 and σ2 unknown, n1 and n2 30
σ1 and σ2 unknown, n1 or n2 < 30
*
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Population means, independent
samples
σ1 and σ2 known
σ1 and σ2 unknown, n1 and n2 30
σ1 and σ2 unknown, n1 or n2 < 30
σ1 and σ2 known
Assumptions:
Samples are randomly and
independently drawn
Population distributions
are normal or both sample
sizes are 30
Population standard
deviations are known
*
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Population means, independent
samples
σ1 and σ2 known
σ1 and σ2 unknown, n1 and n2 30
σ1 and σ2 unknown, n1 or n2 < 30 2
2
2
1
2
1
xx n
σ
n
σσ
21
(continued)
σ1 and σ2 known
When σ1 and σ2 are known and
both populations are normal or
both sample sizes are at least 30,
the test statistic is a z-value…
…and the standard error of
x1 – x2 is
*
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Population means, independent
samples
σ1 and σ2 known
σ1 and σ2 unknown, n1 and n2 30
σ1 and σ2 unknown, n1 or n2 < 30
2
2
2
1
2
1/221
n
σ
n
σzxx
The confidence interval for
μ1 – μ2 is:
σ1 and σ2 known
(continued)
*
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Population means, independent
samples
σ1 and σ2 known
σ1 and σ2 unknown, n1 and n2 30
σ1 and σ2 unknown, n1 or n2 < 30
σ1 and σ2 unknown, large samples
Assumptions:
Samples are randomly andindependently drawn
both sample sizes are 30
Population standarddeviations are unknown*
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Population means, independent
samples
σ1 and σ2 known
σ1 and σ2 unknown, n1 and n2 30
σ1 and σ2 unknown, n1 or n2 < 30
σ1 and σ2 unknown, large samples
Forming interval estimates:
use sample standard deviation s to estimate σ
the test statistic is a z value
(continued)
*
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Population means, independent
samples
σ1 and σ2 known
σ1 and σ2 unknown, n1 and n2 30
σ1 and σ2 unknown, n1 or n2 < 30
2
2
2
1
2
1/221
n
s
n
szxx
The confidence interval for
μ1 – μ2 is:
σ1 and σ2 unknown, large samples
(continued)
*
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Population means, independent
samples
σ1 and σ2 known
σ1 and σ2 unknown, n1 and n2 30
σ1 and σ2 unknown, n1 or n2 < 30
σ1 and σ2 unknown, small samples
Assumptions:
populations are normallydistributed
the populations have equalvariances
samples are independent
*
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Population means, independent
samples
σ1 and σ2 known
σ1 and σ2 unknown, n1 and n2 30
σ1 and σ2 unknown, n1 or n2 < 30
σ1 and σ2 unknown, small samples
Forming interval estimates:
The population variances are assumed equal, so usethe two sample standard deviations and pool them toestimate σ
the test statistic is a t valuewith (n1 + n2 – 2) degreesof freedom
(continued)
*
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Population means, independent
samples
σ1 and σ2 known
σ1 and σ2 unknown, n1 and n2 30
σ1 and σ2 unknown, n1 or n2 < 30
σ1 and σ2 unknown, small samples
The pooled standard deviation is :
(continued)
2nn
s1ns1ns
21
2
22
2
11p
*
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Population means, independent
samples
σ1 and σ2 known
σ1 and σ2 unknown, n1 and n2 30
σ1 and σ2 unknown, n1 or n2 < 30
21
p/221
n
1
n
1stxx
The confidence interval for
μ1 – μ2 is:
σ1 and σ2 unknown, small samples
(continued)
Where t/2 has (n1 + n2 – 2) d.f., and
2nn
s1ns1ns
21
2
22
2
11p
*
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Hypothesis Tests for the Difference Between Two Means
Testing Hypotheses about μ1 – μ2
Use the same situations discussed already:
Standard deviations known or unknown
Sample sizes 30 or not 30
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Hypothesis Tests for Two Population Proportions
Lower tail test:
H0: μ1 μ2
HA: μ1 < μ2
i.e.,
H0: μ1 – μ2 0HA: μ1 – μ2 < 0
Upper tail test:
H0: μ1 ≤ μ2
HA: μ1 > μ2
i.e.,
H0: μ1 – μ2 ≤ 0HA: μ1 – μ2 > 0
Two-tailed test:
H0: μ1 = μ2
HA: μ1 ≠ μ2
i.e.,
H0: μ1 – μ2 = 0HA: μ1 – μ2 ≠ 0
Two Population Means, Independent Samples
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Hypothesis tests for μ1 – μ2
Population means, independent samples
σ1 and σ2 known
σ1 and σ2 unknown, n1 and n2 30
σ1 and σ2 unknown, n1 or n2 < 30
Use a z test statistic
Use s to estimate unknown σ, approximate with a z test statistic
Use s to estimate unknown σ, use a t test statistic and pooled standard deviation
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Population means, independent
samples
σ1 and σ2 known
σ1 and σ2 unknown, n1 and n2 30
σ1 and σ2 unknown, n1 or n2 < 30
2
2
2
1
2
1
2121
n
σ
n
σ
μμxxz
The test statistic for
μ1 – μ2 is:
σ1 and σ2 known
*
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Population means, independent
samples
σ1 and σ2 known
σ1 and σ2 unknown, n1 and n2 30
σ1 and σ2 unknown, n1 or n2 < 30
σ1 and σ2 unknown, large samples
The test statistic for
μ1 – μ2 is:
2
2
2
1
2
1
2121
n
s
n
s
μμxxz
*
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Population means, independent
samples
σ1 and σ2 known
σ1 and σ2 unknown, n1 and n2 30
σ1 and σ2 unknown, n1 or n2 < 30
σ1 and σ2 unknown, small samples
Where t/2 has (n1 + n2 – 2) d.f.,
and
2nn
s1ns1ns
21
2
22
2
11p
21
p
2121
n
1
n
1s
μμxxt
The test statistic for
μ1 – μ2 is:
*
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Two Population Means, Independent Samples
Lower tail test:
H0: μ1 – μ2 0HA: μ1 – μ2 < 0
Upper tail test:
H0: μ1 – μ2 ≤ 0HA: μ1 – μ2 > 0
Two-tailed test:
H0: μ1 – μ2 = 0HA: μ1 – μ2 ≠ 0
a a/2 a/2a
-za -za/2za za/2
Reject H0 if z < -za Reject H0 if z > za Reject H0 if z < -za/2or z > za/2
Hypothesis tests for μ1 – μ2
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Pooled sp t Test: Example
You’re a financial analyst for a brokerage firm. Is there a
difference in dividend yield between stocks listed on the
NYSE & NASDAQ? You collect the following data:
NYSE NASDAQ
Number 21 25
Sample mean 3.27 2.53
Sample std dev 1.30 1.16
Assuming equal variances, is
there a difference in average
yield ( = 0.05)?
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Calculating the Test Statistic
1.2256
22521
1.161251.30121
2nn
s1ns1ns
22
21
2
22
2
11p
2.040
25
1
21
11.2256
02.533.27
n
1
n
1s
μμxxt
21
p
2121
The test statistic is:
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Solution
H0: μ1 - μ2 = 0 i.e. (μ1 = μ2)
HA: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)
= 0.05
df = 21 + 25 - 2 = 44
Critical Values: t = ± 2.0154
Test Statistic: Decision:
Conclusion:
Reject H0 at a = 0.05
There is evidence of a difference in means.
t0 2.0154-2.0154
.025
Reject H0 Reject H0
.025
2.040
040.2
25
1
21
12256.1
53.227.3
t
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Paired Samples
Tests Means of 2 Related Populations
Paired or matched samples
Repeated measures (before/after)
Use difference between paired values:
Eliminates Variation Among Subjects
Assumptions:
Both Populations Are Normally Distributed
Or, if Not Normal, use large samples
Paired samples
d = x1 - x2
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Paired Differences
The ith paired difference is di , where
Paired samples
di = x1i - x2i
The point estimate forthe population mean paired difference is d :
1n
)d(d
s
n
1i
2
i
d
n
d
d
n
1i
i
The sample standard deviation is
n is the number of pairs in the paired sample
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Paired Differences
The confidence interval for d is :Paired samples
1n
)d(d
s
n
1i
2
i
d
n
std d
/2
Where t/2 has
n - 1 d.f. and sd is:
(continued)
n is the number of pairs in the paired sample
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Hypothesis Testing for Paired Samples
The test statistic for d is :Paired
samples
1n
)d(d
s
n
1i
2
i
d
n
s
μdt
d
d
Where t/2 has n - 1 d.f.
and sd is:
n is the number of pairs in the paired sample
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Lower tail test:
H0: μd 0HA: μd < 0
Upper tail test:
H0: μd ≤ 0HA: μd > 0
Two-tailed test:
H0: μd = 0HA: μd ≠ 0
Paired Samples
Hypothesis Testing for Paired Samples
a a/2 a/2a
-ta -ta/2ta ta/2
Reject H0 if t < -ta Reject H0 if t > ta Reject H0 if t < -ta/2or t > ta/2
Where t has n - 1 d.f.
(continued)
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Paired Samples Example
Assume you send your sales people to a “customer service”
training workshop. Is the training effective? You collect the
following data:
Number of Complaints: (2) - (1)Sales person Before (1) After (2) Difference, di
C.B. 6 4 - 2
T.F. 20 6 -14
M.H. 3 2 - 1
R.K. 0 0 0
M.O. 4 0 - 4
-21
d =∑di
n
5.67
1n
)d(ds
2
i
d
= -4.2
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Paired Samples: Solution
Has the training made a difference in the number of complaints
(at the 0.01 level)?
- 4.2d =
1.6655.67/
04.2
n/s
μdt
d
d
H0: μd = 0HA: μd 0
Test Statistic:
Critical Value = ± 4.604 d.f. = n - 1 = 4
Reject
/2
- 4.604 4.604
Decision: Do not reject H0
(t stat is not in the reject
region)Conclusion: There is not a significant change in the number of complaints.
Reject
/2
- 1.66 = .01
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Two Population Proportions
Goal : Form a confidence interval for or test a hypothesis about the difference between two population proportions, p1 – p2
The point estimate for the difference is p1 – p2
Population proportions
Assumptions:
n1p1 5 , n1(1-p1) 5
n2p2 5 , n2(1-p2) 5
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Confidence Interval for Two Population Proportions
Population proportions
2
22
1
11/221
n
)p(1p
n
)p(1pzpp
The confidence interval for
p1 – p2 is:
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Hypothesis Tests for Two Population Proportions
Population proportions
Lower tail test:
H0: p1 p2
HA: p1 < p2
i.e.,
H0: p1 – p2 0HA: p1 – p2 < 0
Upper tail test:
H0: p1 ≤ p2
HA: p1 > p2
i.e.,
H0: p1 – p2 ≤ 0HA: p1 – p2 > 0
Two-tailed test:
H0: p1 = p2
HA: p1 ≠ p2
i.e.,
H0: p1 – p2 = 0HA: p1 – p2 ≠ 0
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Two Population Proportions
Population proportions
21
21
21
2211
nn
xx
nn
pnpnp
The pooled estimate for the overall proportion is:
where x1 and x2 are the numbers from
samples 1 and 2 with the characteristic of interest
Since we begin by assuming the null
hypothesis is true, we assume p1 = p2 and
pool the two p estimates
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Two Population Proportions
Population proportions
21
2121
n
1
n
1)p1(p
ppppz
The test statistic for
p1 – p2 is:
(continued)
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Hypothesis Tests for Two Population Proportions
Population proportions
Lower tail test:
H0: p1 – p2 0HA: p1 – p2 < 0
Upper tail test:
H0: p1 – p2 ≤ 0HA: p1 – p2 > 0
Two-tailed test:
H0: p1 – p2 = 0HA: p1 – p2 ≠ 0
a a/2 a/2a
-za -za/2za za/2
Reject H0 if z < -za Reject H0 if z > za Reject H0 if z < -za/2or z > za/2
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Example: Two population Proportions
Is there a significant difference between
the proportion of men and the proportion of
women who will vote Yes on Proposition A?
In a random sample, 36 of 72 men and 31 of 50 women
indicated they would vote Yes
Test at the .05 level of significance
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Example: Two population Proportions
The hypothesis test is:
H0: p1 – p2 = 0 (the two proportions are equal)HA: p1 – p2 ≠ 0 (there is a significant difference between proportions)
The sample proportions are:
Men: p1 = 36/72 = .50
Women: p2 = 31/50 = .62
.549122
67
5072
3136
nn
xxp
21
21
The pooled estimate for the overall proportion is:
(continued)
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The test statistic for p1 – p2 is:
Example: Two population Proportions
(continued)
.025
-1.96 1.96
.025
-1.31
Decision: Do not reject H0
Conclusion: There is not significant evidence of a difference in proportions who will vote yes between men and women.
1.31
50
1
72
1.549)(1.549
0.62.50
n
1
n
1p)(1p
ppppz
21
2121
Reject H0 Reject H0
Critical Values = ±1.96For = .05
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Two Sample Tests in EXCEL
For independent samples:
Independent sample Z test with variances known:
Tools | data analysis | z-test: two sample for means
Independent sample Z test with large sample
Tools | data analysis | z-test: two sample for means
If the population variances are unknown, use sample
variances
For paired samples (t test):
Tools | data analysis… | t-test: paired two sample for
means
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Chapter Summary
Compared two independent samples Formed confidence intervals for the differences between two
means
Performed Z test for the differences in two means
Performed t test for the differences in two means
Compared two related samples (paired samples) Formed confidence intervals for the paired difference
Performed paired sample t tests for the mean difference
Compared two population proportions Formed confidence intervals for the difference between two
population proportions
Performed Z-test for two population proportions
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Hypothesis Tests for One and Two Population
Variances
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Chapter Goals
After completing this chapter, you should be able to:
Formulate and complete hypothesis tests for a
single population variance
Find critical chi-square distribution values from
the chi-square table
Formulate and complete hypothesis tests for the
difference between two population variances
Use the F table to find critical F values
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Hypothesis Tests for Variances
Hypothesis Testsfor Variances
Tests for a SinglePopulation Variances
Tests for TwoPopulation Variances
Chi-Square test statistic F test statistic
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Single Population
Hypothesis Tests for Variances
Tests for a SinglePopulation Variances
Chi-Square test statistic
H0: σ2 = σ02
HA: σ2 ≠ σ02
H0: σ2 σ02
HA: σ2 < σ02
H0: σ2 ≤ σ02
HA: σ2 > σ02
*Two tailed test
Lower tail test
Upper tail test
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Chi-Square Test Statistic
Hypothesis Tests for Variances
Tests for a SinglePopulation Variances
Chi-Square test statistic *
The chi-squared test statistic for a Single Population Variance is:
2
22
σ
1)s(n
where
2 = standardized chi-square variable
n = sample size
s2 = sample variance
σ2 = hypothesized variance
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The Chi-square Distribution
The chi-square distribution is a family of distributions,
depending on degrees of freedom:
d.f. = n - 1
0 4 8 12 16 20 24 28 0 4 8 12 16 20 24 28 0 4 8 12 16 20 24 28
d.f. = 1 d.f. = 5 d.f. = 15
2 22
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Finding the Critical Value
The critical value, , is found from the chi-square table
Do not reject H0 Reject H0
2
2
2
H0: σ2 ≤ σ02
HA: σ2 > σ02
Upper tail test:
Upper Tail Chi-square
Tests
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Example
A commercial freezer must hold the selected
temperature with little variation. Specifications call for
a standard deviation of no more than 4 degrees (or
variance of 16 degrees2). A sample of 16 freezers is
tested and
yields a sample variance
of s2 = 24. Test to see
whether the standard
deviation specification
is exceeded. Use
= .05
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Finding the Critical Value
The the chi-square table to find the critical value:
Do not reject H0 Reject H0
= .05
2
2
2
= 24.9958
= 24.9958 ( = .05 and 16 – 1 = 15 d.f.)
22.516
1)24(16
σ
1)s(n2
22
The test statistic is:
Since 22.5 < 24.9958, do not reject H0
There is not significant evidence at the = .05 levelthat the standard deviation
specification is exceeded
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Lower Tail or Two Tailed Chi-square Tests
H0: σ2 = σ02
HA: σ2 ≠ σ02
H0: σ2 σ02
HA: σ2 < σ02
2/2
Do not reject H0Reject
21-
2
Do not reject H0
Reject
/2
21-
/2
2
/2
Reject
Lower tail test: Two tail test:
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Hypothesis Tests for Variances
Tests for TwoPopulation Variances
F test statistic
*
F Test for Difference in Two Population Variances
H0: σ12 – σ2
2 = 0HA: σ1
2 – σ22 ≠ 0 Two tailed test
Lower tail test
Upper tail test
H0: σ12 – σ2
2 0HA: σ1
2 – σ22 < 0
H0: σ12 – σ2
2 ≤ 0HA: σ1
2 – σ22 > 0
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Hypothesis Tests for Variances
F test statistic*
F Test for Difference in Two Population Variances
Tests for TwoPopulation Variances2
2
2
1
s
sF
The F test statistic is:
= Variance of Sample 1n1 - 1 = numerator degrees of freedom
n2 - 1 = denominator degrees of freedom= Variance of Sample 2
2
1s
2
2s
(Place the larger sample
variance in the numerator)
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The F Distribution
The F critical value is found from the F table
The are two appropriate degrees of freedom: numerator
and denominator
In the F table,
numerator degrees of freedom determine the row
denominator degrees of freedom determine the
column
where df1 = n1 – 1 ; df2 = n2 – 12
2
2
1
s
sF
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F0
rejection region for a one-tail test is
Finding the Critical Value
F0
2/2
2
2
1 Fs
sF F
s
sF
2
2
2
1
(when the larger sample variance in the numerator)
rejection region for a two-tailed test is
/2
F F/2
Reject H0Do not reject H0
Reject H0Do not reject H0
H0: σ12 – σ2
2 = 0HA: σ1
2 – σ22 ≠ 0
H0: σ12 – σ2
2 0HA: σ1
2 – σ22 < 0
H0: σ12 – σ2
2 ≤ 0HA: σ1
2 – σ22 > 0
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F Test: An Example
You are a financial analyst for a brokerage firm. You want
to compare dividend yields between stocks listed on the
NYSE & NASDAQ. You collect the following data:
NYSE NASDAQ
Number 21 25
Mean 3.27 2.53
Std dev 1.30 1.16
Is there a difference in the
variances between the NYSE
& NASDAQ at the = 0.05 level?
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F Test: Example Solution
Form the hypothesis test:
H0: σ21 – σ2
2 = 0 (there is no difference between variances)
HA: σ21 – σ2
2 ≠ 0 (there is a difference between variances)
Find the F critical value for = .05:
Numerator:
df1 = n1 – 1 = 21 – 1 = 20
Denominator:
df2 = n2 – 1 = 25 – 1 = 24
F.05/2, 20, 24 = 2.327
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F Test: Example Solution
The test statistic is:
0
256.116.1
30.1
s
sF
2
2
2
2
2
1
/2 = .025
F/2
=2.327
Reject H0Do not reject H0
H0: σ12 – σ2
2 = 0HA: σ1
2 – σ22 ≠ 0
F = 1.256 is not greater than
the critical F value of 2.327, so
we do not reject H0
(continued)
Conclusion: There is no evidence of a
difference in variances at = .05
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Chapter Summary
Performed chi-square tests for the variance
Used the chi-square table to find chi-square critical
values
Performed F tests for the difference between two
population variances
Used the F table to find F critical values