# 9 uji hipotesa dua parameter

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


samples

σ1 and σ2 known



*


samples

σ1 and σ2 known



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

*


samples

σ1 and σ2 known


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

*


samples

σ1 and σ2 known



2

2

2

1

2

1/221

n

σ

n

σzxx

The confidence interval for

μ1 – μ2 is:

σ1 and σ2 known

(continued)

*


samples

σ1 and σ2 known



σ1 and σ2 unknown, large samples

Assumptions:

Samples are randomly andindependently drawn

both sample sizes are 30

Population standarddeviations are unknown*


samples

σ1 and σ2 known




Forming interval estimates:

use sample standard deviation s to estimate σ

the test statistic is a z value

(continued)

*


samples

σ1 and σ2 known



2

2

2

1

2

1/221

n

s

n

szxx


μ1 – μ2 is:


(continued)

*


samples

σ1 and σ2 known



σ1 and σ2 unknown, small samples

Assumptions:

populations are normallydistributed

the populations have equalvariances

samples are independent

*


samples

σ1 and σ2 known




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)

*


samples

σ1 and σ2 known




The pooled standard deviation is :

(continued)

2nn

s1ns1ns

21

2

22

2

11p

*


samples

σ1 and σ2 known



21

p/221

n

1

n

1stxx


μ1 – μ2 is:


(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



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


samples

σ1 and σ2 known



2

2

2

1

2

1

2121

n

σ

n

σ

μμxxz

The test statistic for

μ1 – μ2 is:

σ1 and σ2 known

*


samples

σ1 and σ2 known





μ1 – μ2 is:

2

2

2

1

2

1

2121

n

s

n

s

μμxxz

*


samples

σ1 and σ2 known




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


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)


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:


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


Assumptions:

n1p1 5 , n1(1-p1) 5

n2p2 5 , n2(1-p2) 5

Confidence Interval for Two Population Proportions


2

22

1

11/221

n

)p(1p

n

)p(1pzpp


p1 – p2 is:



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



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



21

2121

n

1

n

1)p1(p

ppppz


p1 – p2 is:

(continued)



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


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:


(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



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



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


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


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:


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


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


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


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


0

256.116.1

30.1

s

sF

2

2

2

2

2

1

/2 = .025

F/2

=2.327


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

# 9 uji hipotesa dua parameter

Documents

independent population

independent samples1

standard deviations

population standarddeviations

population valuespopulation

population variances

independent group

sample sizes