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# chi square

### Keywords: CHI

CHI SQUARE

FOR CATEGORICAL DATA

N O N PA R A M E T R I C T E S T

Test of
Independence

WHY USED CHI SQUARE
(2)?

• Use with categorical data – when all you have is the frequency with
which certain events have occurred.

• We measure the “goodness of fit” between our observed outcome and
the expected outcome for some variable (CHI SQUARE GOODNESS
OF FIT TEST).

• With two variables, we test in particular whether they are independent
of one another using the same basic approach (CHI SQUARE TEST OF
INDEPENDENCE) .

THE CHI SQUARE DISTRIBUTION

• Positively skewed but becomes
symmetrical with increasing degrees
of freedom.

• The values are non-negative. That
is, the values of are greater than or
equal to 0.

 2  (O  E)2
E

CHI SQUARE GOODNESS OF FIT TEST

• Used when we have distributions of frequencies across two or more
categories on one variable.

• Test determines how well a hypothesized distribution fits an obtained
distribution.

• Compares observed frequencies with theoretically expected/predicted
frequencies.

• Hypotheses:

H0: The observed data do fit the expected frequencies of the
population.

Ha: The observed data do not fit the expected frequencies for the
population.

CHI SQUARE GOODNESS OF FIT TEST

• Assumption #1: One categorical variable (i.e., the variable can be
dichotomous, nominal or ordinal).

• Assumption #2: You should have independence of observations,
which means that there is no relationship between any of the cases
(e.g., participants).

• Assumption #3: There must be at least 5 expected frequencies in
each group of your categorical variable.

• When the sample size is very small in any cell (expected value<5), Fisher’s
exact test is used as an alternative to the chi-square test.

CHI SQUARE GOODNESS OF FIT TEST

It is believed by a university that the distribution of male
and female students enrolled in the education program is
equal every year.The program coordinator is interested in
determining whether the enrollment has changed from
previous year.

# enrolled Male Female
123 185

HH0a:: The distribution of male and female students is deiqffuearle(nptM. =0.5, pF=0.5).
The distribution of male and female students is

CHI SQUARE GOODNESS OF FIT TEST

Observed frequencies (O) male female Total
Expected frequencies (E) 123 185 308

 2  (O  E)2 Critical value?
E
= .05
df = k-1 (k is number of response
categories) = 1
From table A.4

CHI SQUARE GOODNESS OF

FIT TEST male female Total
123 185 308
Observed frequencies (O) 154 154 308
Expected frequencies (E)

 2  (O  E)2 Critical value?
E
= .05
χ2 = (123 − 154)2 (185 − 154)2 df = k-1 (k is number of response
154 + 154 categories) = 1
From table A.4
(−31)2 312 Critical value =3.841
= 154 + 154

= 12.48(obtained value)

Reject Ho or Do not reject Ho ?

CHI SQUARE GOODNESS OF FIT TEST

SPSS output

A chi-square goodness-of-fit was conducted to check whether the
enrollment has changed from previous year.There were statistically
significant differences in the number of enrollment between male and
female students where χ2(1)=12.481, p<.001 .

CHI SQUARE TEST OF
INDEPENDENCE

• Used when we compare the distribution of frequencies across
categories in two or more independent samples.

• Used in a single sample when we want to know whether two
categorical variables are related (to discover if there is an association
between two categorical variables).

• Hypotheses:
H0: There is no association between the variables.
Ha: There is an association between the variables.

CHI SQUARE TEST OF INDEPENDENCE

• Assumption #1:The two variables should be measured at an ordinal
or nominal level (i.e., categorical data).

• Assumption #2:The two variables should consist of two or more
categorical, independent groups.

• Assumption #3: There must be at least 5 expected frequencies in
each group of your categorical variable.

– Expected frequency for a given cell is obtained by multiplying together

totals for the row and column which the cell is located (marginal totals)

= × .

CHI SQUARE TEST OF INDEPENDENCE

Educators are always looking for novel ways in which to
teach statistics to undergraduates as part of a non-statistics
degree course (e.g., psychology, education, law). With current
technology, it is possible to present how-to guides for
statistical programs online instead of in a book. However,
different people learn in different ways.

An educator would like to know whether gender (male/female) is associated
with the preferred type of learning medium (online vs. books).

CHI SQUARE TEST OF INDEPENDENCE

H0:Tidak terdapat perkaitan antara kaedah pengajaran dan jantina.

Ha:Terdapat perkaitan antara kaedah pengajaran dan jantina.

Male observed value Books Online Total
Female expected value 16 24 40
observed value
expected value 13 27 40
Total
29 51 80

 2  (O  E)2
E

CHI SQUARE TEST OF INDEPENDENCE

Books Online Total

Male observed value 16 24 40
Female expected value 29×40 = 14.5 51×40 = 25.5 40
80
observed value 80 80
expected value
13 27
Total 29×40 = 14.5 51×40 = 25.5

80 80

29 51

2= (16-14.5)^2 + (24-25.5)^2 + (13-14.5)^2 + (27-25.5)^2 =
14.5 25.5 14.5 25.5

Critical value=??  2  (O  E)2
df= (r-1)(c-1)=3.841 E

CHI SQUARE TEST OF INDEPENDENCE

SPSS Output

Nilai chi square=0.487 adalah lebih kecil dari nilai kritikal = 3.841. Maka Ho gagal
ditolak.Ada bukti yang mencukupi untuk gagal/tidak menyatakan terdapat perkaitan
antara kaedah pengajaran (online vs. book) dan jantina(lelaki vs perempuan).

LATIHAN

Pada tahun 2010, satu kajian telah dijalankan untuk

mendapatkan maklumat status semakan (status updates)

penggunaan Facebook dalam kalangan pelajar tingkatan 5 di

sebuah sekolah di daerah Hulu Langat Selangor.

Status Lelaki Perempuan

Setiap hari 138 164
3-5 hari/seminggu 83 97
1-2 hari/seminggu 64 84

Jalankan pengujian hipotesis dengan menggunakan ujian
inferensi yang bersesuaian dan laporkan dapatan kajian.