View
24
Download
4
Category
Tags:
Preview:
DESCRIPTION
statistika
Citation preview
Chapter 10
Estimation and Hypothesis Testing
for Two Population Parameters
Business Statistics: A Decision-Making Approach
8th Edition
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
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:
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
*
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
*
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
*
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
*
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)
*
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*
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)
*
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)
*
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
*
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)
*
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
*
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
*
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
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
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
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
*
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
*
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:
*
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
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)?
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:
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
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
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
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
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
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)
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
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
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
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:
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
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
Two Population Proportions
Population proportions
21
2121
n
1
n
1)p1(p
ppppz
The test statistic for
p1 – p2 is:
(continued)
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
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
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)
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
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
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
Hypothesis Tests for One and Two Population
Variances
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
Hypothesis Tests for Variances
Hypothesis Testsfor Variances
Tests for a SinglePopulation Variances
Tests for TwoPopulation Variances
Chi-Square test statistic F test statistic
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
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
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
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
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
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
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:
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
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)
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
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
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?
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
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
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
Recommended