mte3105 pengujian hipotesis khi kuasa dua 2

MTE 3105 STATISTIKPENGUJIAN HIPOTESIS 8Ujian Khi Kuasa Dua

Dr. Lam Kah Kei

Chi Square Tests

• to determine if a distribution of observed frequencies differs from the theoretical expected frequencies.

• Chi-square statistics use categorical data, • thus instead of using means and variances, this

test uses frequencies.

Jenis Ujian Khi Kuasa Dua

• Ujian Penyesuaian Terbaik (Test of Goodness of Fit)• Ujian Perkaitan / Ketakbersandaran (Test of Independence/ Test of Association)

Test for Independence

• The test of independence analyzes the relationship between two nominal variables.

• The procedure uses the special terms independent to mean not related/not associated, and not independent to mean related/associated.

• Two variables are independent if the occurrence of one of the variables does not affect the occurrence of the other.

• The two nominal variables form a contingency table of cells.

SW388R6Data Analysis and Computers I

Slide 6

Independence Defined

• When two variables are independent, there is no relationship between them. We would expect that the frequency breakdowns of the dependent variable to be similar for all groups.

SW388R6Data Analysis and Computers I

Slide 7

Independence Demonstrated• Suppose we are interested in the relationship between gender

and attending college.

• If there is no relationship between gender and attending college, and 40% of our total sample attend college, we would expect 40% of the males in our sample to attend college and 40% of the females to attend college.

• If there is a relationship between gender and attending college, we would expect a higher proportion of one group to attend college than the other group, e.g. 60% to 20%.

SW388R6Data Analysis and Computers I

Slide 8

Displaying Independent and Dependent Relationships

Independent Relationship between Gender and College

40% 40% 40%

0%

20%

40%

60%

80%

100%

Males Females TotalPo

po

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n A

tte

nd

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Co

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Dependent Relationship between Gender and College

60%

20%

40%

0%

20%

40%

60%

80%

100%

Males Females TotalPo

po

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When the variables are independent, the proportion in both groups is close to the same size as the proportion for the total sample.

When group membership makes a difference, the dependent relationship is indicated by one group having a higher proportion than the proportion for the total sample.

Table of Contingency• Example: Is Disease Associated With Exposure?• 2 x 2 contingency table• How many cells?

Column totals

Row totalscells

Total

Displaying Data in Contingency Table

Table of Contingency• Language preference and types of school• 3 x 2 Contingency table

• How many schools?• How many cells in the table?

Expected Cell Frequencies (Ei)

Test of Independence

χ

SW388R6Data Analysis and Computers I

Slide 14

Hypotheses• The null hypothesis is that the two variables are independent.

This will be true if the observed counts in the sample are equal to the expected counts.

• The research hypothesis states that the two variables are dependent or related. This will be true if the observed counts for the categories of the variables in the sample are different from the expected counts.

• The decision rule for the chi-square test of independence is the same as our other statistical tests.

Chi-Square Test of Independence

Example: A study examined whether Schools for the Deaf that identified giftedness in their students is related to the type of language preferences. Table below shows observed number of schools which did not identify students as “gifted,” (Level I) and number of schools of the deaf which did (Level II) according to language preference. Test at 5% significance level whether relationship exists between the types of schools and language preference.

H0: There is no association between types of schools and language preference

or There is independence between types of schools and

language preference.H1: There is association between types of schools and

language preference or There is no independence between types of schools and

language preference

1. Determine the hypotheses.

Test of Independence

2. Calculate expected frequencies for each cell category (contingency table).

3. Specify the distribution, df, level of significance.

• Although not all expected frequencies E > 5, some books state assumptions as every cell at least 1 and at least 20%, of E>5 then χ2 distribution,

• df = (r-1)(c-1) = (3-1)(2-1) = 2, and α = .05

4. Determine the critical value and rejection rule.• From table, Critical value χ2

2,.05 = 5.991• Reject null if computed χ2 >5.991

205.991

df = 2

Critical X2 = = 5.991

5. Compute the value of the test statistic.

857.2Calculated 2

The test statistic 2.857 < 5.991 does not fall in the rejection region, so fail to reject H0.

There is evidence that the types of schools is independent (i.e. not associated) with the language preference.

22.857

6. Determine whether to reject H0 and make decision

5.991

Another example

Cuba ini.• Darah manusia adalah diklasifikasikan sebagai A, B, AB

atau O. Tambahan, darah manusia juga boleh dikelaskan sebagai Rh positif (Rh+) atau Rh negatif (Rh-). Tinjauan 500 individu yang dipilih secara rawak mendapati keputusan seperti berikut:

• Ujikan sama ada jenis darah dan aras Rh adalah berkaitan pada aras kesignifikanan 0.05.

Aras RhJenis Darah

A B AB O

Rh+ 176 28 22 198

Rh- 30 12 4 30

H0: Jenis darah dan aras Rh adalah tidak berkaitan/tidak bersandar (independent).

H1: Jenis darah dan aras Rh adalah berkaitan/bersandar (dependent)

1. Determine the hypotheses.

Penyelesaian

2. Calculate expected frequencies for each cell category.

totalTable

)totalColumn)(totalRow()E(frequencyExpected

• Frekuensi dijangka ditunjukkan dalam kurungan

Aras RhJenis Darah

A B AB O

Rh+ 176(174.688)

28(33.92)

22(22.048)

198(193.344)

Rh- 30(31.312

12(6.08)

4(3.952)

30(34.656)

3. Specify the distribution, df, level of significance.

• Semua frekuensi dijangka E > 1, hanya 1 dpd 8 atau <20% frekuensi dijangka kurang dpd 5. Taburan χ2

• df = (r-1)(c-1) = (2-1)(4-1) = 3, and α = .05

4. Determine the critical value and rejection rule.• Daripada jadual, Nilai kritikal χ2

3,.05 = 7.815• Tolak nol jika χ2 > 7.815

5. Compute the value of the test statistic.

601.7656.34

)656.3430(

952.3

)952.34(

08.6

)08.612(

312.31

)312.3130(

344.193

)344.193198(

048.22

)048.2222(

92.33

)92.3328(

688.174

)688.174176(

2222

22222

Statistik ujian 7.601 < 7.815, maka tidak menolak H0.

Tiada bukti yang mencukupi pada aras kesignifikanan α=.05 untuk menyokong tuntutan aras Rh dan jenis darah adalah berkaitan/ bersandar.

6. Determine whether to reject H0 and make decision

Yates’ Correction

• For a 2 x 2 contingency table, Yates’ correction should be applied when calculating χ2,

where

E

)5.0EO( 22

Cuba ini.

• A driving school examined the results of 100 candidates who took their test for the first time. It was found that out of 40 men, 28 passed and out of 60 women, 34 passed. Do these results indicate, at the 5% significance level, a relationship between the gender of the candidate and the ability to pass the driving test at the first attempt?

• H0: There is no relationship between the gender of a candidate and the ability to pass at the first attempt.

• H1: There is a relationship between the gender of a candidate and the ability to pass at the first attempt.

• Contingency table: 2 x2Results driving test Total

Gender Pass Fail

Male 28 12 40

Female 34 26 60

Total 62 38 100

• Calculated expected frequencies

• All expected frequencies >5, χ2 distribution• Critical value χ2=3.841, reject null if χ2>3.841

Results driving test Total

Gender Pass Fail

Male 24.8 15.2 40

Female 37.2 22.8 60

Total 62 38 100

• Using Yates’ correction,

• χ2 <3.841, do not reject null. There is not enough evidence at the 5% sig level to indicate a relationship between gender and ability to pass driving test

O E (ӀO –Eӏ-0.52)E

28 24.8 0.293…

34 37.2 0.195…

12 15.2 0.479…

26 22.8 0.319…

ƩO=100 ƩE=100 1.289…