t4_2 [compatibility mode].pdf

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ANALYSIS OF VARIANCE

(ANOVA)

Wah Mong Weh

Jabatan Matematik

IPG KSAH

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Test Statistic for One-Way ANOVA

A excessively large F test statistic isevidence against equal population means.Thus, the null will be rejected.

F = variance between samplesvariance within samples

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The variance is calculated in

two different ways and the

ratio of the two values is

formed.

W

B

MSMSF=

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MSB

, Mean Square Between, the variance

between samples, measures the differences

related to the treatment given to each sample.

2.MSW

Mean Square Within, the variance

within samples, measures the differences

related to entries within the same sample. The

variance within samples is due to sampling

error.

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Example 1 (Equal sample sizes)

A researcher wishes to try three different

techniques to lower the blood pressure of

individuals diagnosed with high blood

pressure. The subjects are randomly assigned

to three groups; the first group takes

medication, the second group exercises, and

the third group follows a special diet. After

four weeks, the reduction in each person's

blood pressure is recorded. At =0.05, test the

claim that there is no difference among the

means. The data are shown6

Medication

Exercise Diet

10 6 5

12 8 9

9 3 12

15 0 8

13 2 4

Sample means 11.8 3.8 7.6

sample sd 2.3874 3.193 3.209

sample var 5.7 10.2 10.3


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The hypotheses

H0: 1=2=3

H1: At least one mean is different from theothers

Find the critical value for F =0.05,

df (2, 12) = 3.89

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2

p

x

s

nsF =

8

(i) Compute the variance of the sample means ie

compute the variance of the three values: 11.8, 3.8,

7.6 with a calculator

std dev = 4.00 var = 16

(ii) Compute the mean of the sample variances ie

find the mean of the three values: 5.7, 10.2, 10.3

mean = 8.73

F = 5(16)/8.73

= 9.1638 = 9.16

(iii) Since 9.16 > 3.89, reject the null hypothesis

There is enough evidence to support the claim that

there is a difference among the means

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Example 2

A PLUS employee wishes to see if there is a

significant difference in the number of

employees at the PLUS offices in three states.

The data are shown. At =0.05, can it be

concluded that there is a significant difference

in the average number of employees at each

state?

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Kedah Penang Perak7 10 1

14 1 1232 1 119 0 910 11 111 1 11

mean 15.5 4 5.8std dev 9.05 5.06 5.38

ar 81.9 25.6 29.0

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The hypotheses are

H0: 1=2=3

H1: At least one mean is different from the other

Find the critical value for F based on =0.05,

df (2, 15) = 3.68

(i) Compute the variance of the sample means

compute the variance of the three values: 15.5,

4.0, 5.8 with a calculator

(ii) Compute the mean of the sample variances ie

find the mean of the three values: 81.9, 25.6,

29.0

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F = 6(38.32)/45.5

= 5.05319= 5.05

(iii) Since 5.05 > 3.68, rejectthe null hypothesis

There is enough evidence to support the claim thatthere is a difference among the means


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Example 1 (Equal sample sizes)Medica

tionExercis

e Diet

10 6 512 8 99 3 12

15 0 813 2 4

Sample totals 59 19 38

Grand mean = 7.73

x = 116x2 = 1162

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The hypotheses are

H0: 1=2=3

H1: At least one mean is different from theothers

Find the critical value for F =0.05, df (2, 12)

3.89

(i) Compute the total sum of squares SST

(ii) Compute the between sum of squares SSB

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Lets try solve example 2 with thismethod

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Example 1 : Output ExcelAnova: Single Factor

SUMMARY

Groups Count Sum Average Variance

Column 1 5 59 11.8 5.7

Column 2 5 19 3.8 10.2

Column 3 5 38 7.6 10.3

ANOVA

Source of

Variation SS df MS F P-value F crit Between

Groups 160.13 2.00 80.07 9.17 0.00 3.89

WithinGroups 104.80 12.00 8.73

Total 264.9333 14

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Q5AMei2012

Sebanyak 15 orang telah diagih secara rawakkepada 3 kumpulan untuk menjalani satukajian dengan 3 jenis kaedah pengajaranMatematik yang berbeza.

Pada penghujung semester, ujian diberikepada 15 pelajar itu. Jadual di bawahmenunjukkan markah yang diperolehi olehpelajar-pelajar dalam ketiga-tiga kumpulanitu.

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KAEDAH A KAEDAH B KAEDAH C

50 85 95

70 55 75

50 70 70

65 90 85

90 70 70

N1 = 5 N2 = 5 N3 = 5

T1 = 325 T2 = 370 T3 = 395

(a) Jalan ujian ANOVA pada aras keertian 0.01 dengan

(1) nyatakan hipotesis nol dan hipotesis alternatif yangberkaitan.


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(ii) tentukan nilai kritikal bagi F

(iii) Kirakan nilai bagi ujian statistik

(b) Apakah keputusan ujian tersebut ?

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Q5AMei2011

Bahagian pentadbiran akademik Universiti

Putra (UPM) berminat untuk mengkajipencapaian akademik bagi tiga programIjazah Sarjana Muda (BS). Data bagi puratamata gred terkumpul (PMGK) daripadaprogram BS Pertanian, BS Ekonomi dan BSEkologi telah dipungut dan direkod sepertidalam jadual dibawah.

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BS Pertanian BS Ekologi BS Ekonomi

2.3 2.5 3.1

2.2 2.7 3.3

2.4 2.7 3.2

2.3 2.9 2.9

2.5 2.6 3.0

(a) Nyatakan hipotesis alternatif dan hipotesis nol.

(b) Lengkapkan jadual bagi ringkasan ANOVA

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PuncaPuncaVariansVarians

Darjahkebebasan

(d,f)

Jumlahkuasa Dua

(SS)

Varians(MS) F

AntaraKumpulan

2

Dalamkumpulan

12

Jumlah 14

Gunakan = 0.05

( c) Apakah kesimpulan yang dapat dibuat oleh bahagian

pentadbiran pada aras = 0.05 ?

t4_2 [compatibility mode].pdf

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