Pengujian Hipotesis Dua Populasi By. Nurvita Arumsari, Ssi, MSi

Statistik TP608377 A

Pengujian Hipotesis Dua PopulasiBy. Nurvita Arumsari, Ssi, MSi

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

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:

Two – Sample Mean Population Test

Hypothesis Tests forTwo Population Means

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

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/2

or z > za/2

Hypothesis tests for μ1 – μ2

Hypothesis tests for μ1 – μ2


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

Independent Samples


σ1 and σ2 known



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

*

σ1 and σ2 known


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

*

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…


σ1 and σ2 known



2

22

1

21

2121

nσ

nσ

μμxxz

The test statistic for μ1 – μ2 is:

σ1 and σ2 known

*


σ1 and σ2 known



2

22

1

21

/221n

σ

n

σzxx

The confidence interval for μ1 – μ2 is:

σ1 and σ2 known(continued)

*

σ1 and σ2 unknown, large samples


σ1 and σ2 known




Assumptions:

Samples are randomly and independently drawn

both sample sizes are 30

Population standard deviations are unknown*


σ1 and σ2 known




Forming interval estimates:

use sample standard deviation s to estimate σ

the test statistic is a z value

(continued)

*


σ1 and σ2 known



2

22

1

21

/221n

s

n

szxx


σ1 and σ2 unknown, large samples (continued)

*

σ1 and σ2 unknown, small samples


σ1 and σ2 known




Assumptions:

populations are normally distributed

the populations have equal variances

samples are independent

*


σ1 and σ2 known




Forming interval estimates:

The population variances are assumed equal, so use the two sample standard deviations and pool them to estimate σ

the test statistic is a t value with (n1 + n2 – 2) degrees of freedom

(continued)

*


σ1 and σ2 known




The pooled standard deviation is

(continued)

2nn

s1ns1ns

21

222

211

p

*


σ1 and σ2 known



21

p/221n

1

n

1stxx


σ1 and σ2 unknown, small samples (continued)

Where t/2 has (n1 + n2 – 2) d.f.,

and

2nn

s1ns1ns

21

222

211

p

*


σ1 and σ2 known




Where t/2 has (n1 + n2 – 2) d.f.,

and

2nn

s1ns1ns

21

222

211

p

21p

2121

n1

n1

s

μμxxt

The test statistic for μ1 – μ2 is:

*

Paired Samples

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 < -t /2a

or t > t /2a Where t has n - 1 d.f.

(continued)

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

d = x1 - x2

Paired Differences

The ith paired difference is di , wherePaired samples di = x1i - x2i

The point estimate for the population mean paired difference is d :

1n

)d(ds

n

1i

2i

d

n

dd

n

1ii

The sample standard deviation is

n is the number of pairs in the paired sample

Paired Differences

The confidence interval for d isPaired samples

1n

)d(ds

n

1i

2i

d

n

std d/2

Where t/2 has

n - 1 d.f. and sd

(continued)

n is the number of pairs in the paired sample

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


Hypothesis Tests forTwo 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 < -z /2a

or z > z /2a



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

n1

n1

)p1(p

ppppz

The test statistic for p1 – p2 is:

(continued)

Confidence Interval forTwo Population

Proportions


2

22

1

11/221 n

)p(1p

n

)p(1pzpp

The confidence interval for p1 – p2 is:

Contoh Kasus

Pooled sp t Test: Example

You’re a welding inspector. Is there a difference in level of precision between welding machine A & B? You collect the following data: A BNumber 21 25Sample mean 3.27 2.53Sample std dev 1.30 1.16

Assuming equal variances, isthere a difference in average yield ( = 0.05)?

Calculating the Test Statistic

1.2256

22521

1.161251.30121

2nn

s1ns1ns

22

21

222

211

p

2.040

251

211

1.2256

02.533.27

n1

n1

s

μμxxt

21p

2121

The test statistic is:

Solution

H0: μ1 - μ2 = 0 i.e. (μ1 = μ2)

HA: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)

= 0.05df = 21 + 25 - 2 = 44Critical Values: t = ± 2.0154

Test Statistic:

Decision:

Conclusion:

Reject H0 at a = 0.05There is evidence of a difference in means.

t0 2.0154-2.0154

.025

Reject H0 Reject H0

.025

2.040

040.2

251

211

2256.1

53.227.3

t

Assume you are a welding inspector. In the first inspection, you found some defects in welding result. Then, you got a welding training. After you got a training, you inspect the welding result, like in table. Is the training effective? You collect the following data:

Paired Samples Example

Welding ins.

Number of Defects Difference

Before After

C.B 20 3 -17

T.F 20 6 -14

M.H 15 4 -11

R.K 10 4 -6

M.O 17 7 -10

Assume you are a welding inspector. In the first inspection, you found some defects in welding result. Then, you got a welding training. After you got a training, you inspect the welding result, like in table. Is the training effective? You collect the following data:

d = di

n

159.41n

)d(ds

2i

d

= -11.2

Welding ins.

Number of Defects Difference

Before After

C.B 20 3 -17

T.F 20 6 -14

M.H 15 4 -11

R.K 10 4 -6

M.O 17 7 -10

Paired Samples Example

Example: Two population Proportions

Is there a significant difference between the proportion of boy and the proportion of girl who will prefer on Welding department?

In a random sample, 52 of 72 boys and 10 of 50 girls indicated they would prefer welding department

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:

Boys: p1 = 52/72 = .72

Girls: p2 = 10/50 = .2

.508122

62

5072

0125

nn

xxp

21

21

The pooled estimate for the overall proportion is:

Example: Two population

Proportions(continued)

Documents

Pengujian Hipotesis Dua Populasi By. Nurvita Arumsari, Ssi, MSi