statistik pendidikan edu5950 sem1 2015-16 statistik inferensi: pengujian hipotesis bagi analisis...
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STATISTIK PENDIDIKANEDU5950
SEM1 2015-16
STATISTIK INFERENSI:
PENGUJIAN HIPOTESIS BAGI ANALISIS KHI-KUASA DUA
Rohani Ahmad Tarmizi - EDU5950 1
Chi – Square(x²)
Chi – Square (x²)
ANALISIS “CHI-SQUARE”(KUASA-DUA KHI)
• Ini juga merupakan analisis hubungan tetapi lebih dikenali sebagai analisis perkaitan (association)
• Analisis ini digunakan pakai bagi menentukan perkaitan antara pasangan pembolehubah yang diukur pada skala nominal atau ordinal ataupun jika salah satunya dipadankan dengan data sela dan nisbah.
• Dengan itu pembolehubah seperti – Bangsa, – Jantina, – Suka/tidak suka makanan, – Tinggi pencapaian/rendah pencapaian, – Kebimbangan tinggi/ kebimbangan sederhana/ kebimbangan rendah
• Data frekuensi dicerap dengan membilang kejadian (occurance setiap perkara). Sesuai untuk kajian tinjauan
• Daripada frekuensi yang dicerap (observed frequency) analisis “chi-square” memberi kita makluman bahawa ada/tiada perkaitan antara kedua-dua pemboleh ubah.
ANALISIS “CHI-SQUARE” (KUASA-DUA KHI)
• KATAKANLAH, penyelidik mengumpul maklumat tentang bangsa bagi responden dan juga kategori amalan pemakanan setiap responden,
• ATAU penyelidik tinjau pelajar dibeberapa buah sekolah dari segi jantina dan minta/tidak minat kepada aliran sains
• ATAU penyelidik tinjau bapa-bapa dan mengumpul maklumat tahap pendidikan (tinggi/ sederhana/ rendah) dan dikaitkan dengan kategori gaji
• Bagi ketiga-tiga contoh tersebut analisis yang sesuai dijalankan adalah analisis tak parametrik (analisis kuasa-dua khi)
• dan seterusnya dibina jadual kontingensi atau jadual“crosstabulation”.
• Daripada frekuensi yang dicerap (observed frequency) analisis “chi-square” memberi kita makluman bahawa ada/tiada perkaitan antara kedua-dua pemboleh ubah.
ANALISIS “CHI-SQUARE”(KUASA-DUA KHI)
• Terdapat dua cara/kategori – CHI-SQUARE TEST OF GOODNESS OF FIT dan TEST OF INDEPENDENCE/DEPENDENCE
• TEST GOODNESS OF FIT – menjawab persoalan “adakah terdapat perbezaan kadar bagi sesuatu perkara/kejadian/persetujuan”
• TEST OF INDEPENDENCE/ DEPENDENCE – menjawab persoalan “adakah terdapat perkaitan/kebersandaran/ hubungan antara dua perkara
ANALISIS “CHI-SQUARE”(KUASA-DUA KHI)
• Dapatan bagi analisis ini lazimnya dalam bentuk jadual frekuensi yang dipanggil jadual kontingensi atau jadual “crosstabulation”.
• Daripada frekuensi yang dicerap (observed frequency) analisis “chi-square” ini memberi kita makluman bahawa ada/tiada perkaitan yang signifikan antara kedua-dua pembolehubah yang dikaji
• Ataupun ada/tiada perbezaan frekuensi yang signifikan antara kategori-kategori yang dikaji.
•Daripada jadual tersebut kita boleh telitikan atau kajikan sama ada terdapat hubungan atau perkaitan antara kedua-dua pemboleh ubah tersebut.
•Selanjutnya analisis pengujian hipotesis perlu dijalankan ia itu untuk menguji terdapatnya perkaitan antara kedua-dua pemboleh ubah tersebut dengan signifikan.
•Pengujian hipotesis ini adalah ujian kuasa dua khi.
•Sekiranya, terdapat perkaitan yang signifikan maka langkah seterusnya adalah dengan menentukan darjah atau magnitud hubungan tersebut.
•Bagi analisis ini, data adalah dalam bentuk kekerapan dan sudah semestinya taburan skor adalah tidak normal.•Dengan itu taburan ini dipanggil taburan bebas (distribution-free).•Ujian ini juga dipanggil ujian tak parametrik oleh kerana ia tidak bertabur secara normal.•Sebagai “rule-of-thumb” penggunaan ujian parametrik digalakkan oleh kerana oleh kerana “power” atau kekuatannya, walaubagaimana pun jika data adalah dalam bentuk nominal serta juga terdapat taburan data yang tidak normal maka ujian tak parametrik diterima pakai.
•Ujian-ujian parametrik – sign test, Mann-Whitney U test, Wilcoxon matched-pairs signed ranks, Kruskal-Wallis, Chi-square.
9
Different Scales, Different Measures of Association
Scale of Both Variables
Measures of Association
Interval or Ratio Scale Pearson r
Ordinal Scale Spearman’s rho
Nominal Scale Pearson Chi-Square (χ2 ) -Indicates there is significant association between two categorical variables
Measure of association is:
Phi coefficient
Contingency coefficient
Cramer’s V coefficient
Types of Chi-Square Test
1. Goodness-of-fit To test for certain assumption regarding one categorical variable
2. Test of Independence Test on association between variables regarding contingency tables
The Chi-Square Distribution
♠ The Chi-Square distribution has only one parameter degrees of freedom
♠ The shape of the distribution curve is skewed to the right for small degrees of freedom and becomes symmetric for large degrees of freedom
♠ The entire Chi-square distribution lies to the right of the y-axis
♠ the Chi-square distribution assumes nonnegative values
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
df = 2
df = 7
df = 12
The Chi – Squares Distribution
x² Goodness - of - fit
14
Steps in Test of Hypothesis –1. State the null and alternative hypothesis.
2. Determine the appropriate sampling distribution and the
critical value.
3. Calculate the test statistics - calculation based on:
O – Observed frequency
E – Expected frequency
Formula to calculate χ ² = ( O-E) ² / E
4. Make decision.
5. Conclusion.
Goodness-of-fit
♠ Test assumption for categorical variable
♠ Only one variable
♠ calculation based on:O – Observed frequencyE – Expected frequency
♠ Formula to calculate x²
x² = Σ ( O – E )² E
1
Hyp
oth
esis
Test
1State
HO and HA
2Determine the sampling
distribution and Critical Value
3Calculate x²
4Decision
5Conclusion
Based on: O and E
x² = Σ( O – E )² E
Step in testing Hypothesis
1. State the null and alternative hypothesesHO: Statement of assumptionHA: Statement opposite of the assumption
2. Calculate the best statistic
x² value
3. Determine critical value• α• df = k-1
4. Make your decision
5. Make conclusion
Criteria Decisionx² cal > x² critical Reject HO, Accept HA
x² cal ≤ x² critical Fail to Reject HO
Criteria DecisionSig- x² < α Reject HO, Accept HA
Sig- x² ≥ α Fail to Reject HO
Manual
SPSS
Example 1:
The following table displays the age distribution for a sample summoned for traffic violations. Test the hypothesis that the proportion of people summoned for traffic violations is different for all age groups at 0.05 level of significance.
Age <20 20-29 30-39 40-49 >49
Summon 32 25 19 16 8
Assumption:p1 = p2 = p3 = p4 = p5 = 0.20
Answer:
1. Hypotheses HO: The proportion of people involve in traffic violation is the same for all age groups HA: The proportion of people involve in traffic violation is not the same for all age groups
2. Test statistic
Age O E ( O – E ) ( O – E )² E
<20 32 20 12 144 7.20
20-29 25 20 5 25 1.25
30-39 19 20 -1 1 0.05
40-49 16 20 -4 16 0.80
>49 8 20 -12 144 7.20
100 16.50
( O – E )²
3. Critical value df = k – 1 = 5 – 1 = 4
x² 4,0.05 = 9.49
4. Decision
Since x² cal (16.5) is bigger than x² critical (9.49) Reject HO, H Accept HA.
5. Conclusion: The proportion of people summoned for traffic violations is not
the same significantly for all age groups, x² (n=100) = 16.5, p < .05. This indicated that people of different age group differ significantly in frequency of traffic violation. Findings also indicated that the lower age group performed more traffic violations.
Fail to reject HO
RejectHO
α = 0.05
SPSS Chi-Square output Age group involved in traffic offences
Observed N Expected N Residual
<20 32 20.0 12.0
20-29 25 20.0 5.0
30-39 19 20.0 -1.0
40-49 16 20.0 -4.0
>49 8 20.0 -12.0
Total 100
Test Statistics
Age group involvedIn traffic offences
Chi-Square 16.500df 4Asymp. Sig. 0.002
x² is valid if less than20% of the cells withExpected values < 5
Example 2:
In the 2008 poll, adults were asked, “Do you agree with the move to increase highway speed limit to 120 km/hour?” Results revealed that 58% said yes, 31% said no and 11% said do not know. Suppose the result hold true for the 2008.A recent poll produced the following distribution in response to the same question. Test the hypothesis that the current distribution of adult belonging to the three categories is different from that for 2008 at 0.01 level of significance.
Table 2
Category Yes No Do not know
Frequency 313 146 41
Prob.
Yes 0.58
No 0.31
Do not know 0.11 variable – agreement towards the move to increase highway speed limit to 120 km/hour
Answer
1. Hypotheses HO: The current percentage distribution of adults belonged to the three categories as that for 2008 HA: The current percentage distribution of adults do not belonged to the three categories from that for 2008
2. Test statistic
Opinion O E ( O – E ) ( O – E )² E
Yes 313 290 23 529 1.824
No 146 155 -9 81 0.523
Do not know 41 55 -14 196 3.564
500 5.911
( O – E )²
3. Critical value df = k – 1 = 3 – 1 = 2
x² 2,0.01 = 9.21
4. Decision
Since x² cal (5.911) is smaller than x² critical (9.21) Fail to reject HO
5. Conclusion: The current percentage distribution of adults do belong to the
three categories as from that of 2008 at 0.01 level of significance, x² (2, n=500) = 5.911, p > .01. The current percentage is not different from the distribution in 2008. Therefore the agreement of adults in the current surveyIs similar to those in year 2008.
Fail to reject HO
RejectHO
α = 0.01
9.21
SPSS Chi-Square output Perception
Observed N Expected N Residual
Agree 313 290.0 23.0
Disagree146 155.0 -9.0
Don’t know 41 55.0 -14.0
Total 500
Test Statistics
Perception
Chi-Square 5.910df 2Asymp. Sig. 0.052
a
a. 0 cells (0.0%) have expected frequencies less than 5. The minimum expected cell frequency is 55.0.
x² Test of Independence
27
Assumptions
• Chi Square test of independence-dependence is used when
two variables are measured on a nominal scale.
• Chi-square goodness-of-fit is used for test of differences
when you have only one variable.
• It can be applied to interval or ratio data that have been
categorized into a small number of groups.
• It assumes that the observations are randomly sampled
from the population.
Assumptions
• All observations are independent (an individual
can appear only once in a table and there are no
overlapping categories).
• It does not make any assumptions about the
shape of the distribution nor about the
homogeneity of variances
29
Steps in Test of Hypothesis –1. State the null and alternative hypothesis.
2. Determine the appropriate sampling distribution and the
critical value.
3. Calculate the test statistics - calculation based on:
O – Observed frequency
E – Expected frequency
Formula to calculate χ ² = ( O-E) ² / E
4. Make decision.
5. Conclusion.
30
The Hypothesis:Whether There is an Association or Not
• Ho : The two variables are independent
• Ha : The two variables are associated or dependent
31
Calculating Test Statistics
• Contrasts observed frequencies in each cell of a contingency table with expected frequencies.
• The expected frequencies represent the number of cases that would be found in each cell if the null hypothesis were true ( i.e. the nominal variables are unrelated).
• Expected frequency of two unrelated events is product of the row and column frequency divided by total number of cases.
E or FE= Fr Fc / N
32
Calculating Test Statistics
E
EO 22 )(
33
Calculating Test Statistics
E
EO 22 )(
Observed
frequencies
Expe
cted
fre
quen
cy
Expected
frequency
34
Determine Degrees of Freedom
df = (R-1)(C-1)
Num
ber of
levels in
column
variable
Num
ber of levels in row
variable
35
Compare computed test statistic against a tabled/critical value
• The computed value of the Pearson chi- square statistic is compared with the critical value to determine if the computed value is improbable
• The critical tabled values are based on sampling distributions of the Pearson chi-square statistic
• If calculated 2 is greater than 2 table value, reject Ho
36
Example
• Suppose a researcher is interested in voting preferences on environmental control issues.
• A questionnaire was developed and sent to a random sample of 90 voters.
• The researcher also collects information about the political party membership of the sample of 90 respondents.
37
Bivariate Frequency Table or Contingency Table
AGREE (V1)/ TYPE OF VOTERS (V2)
Favor Neutral Oppose f row
Barisan 10 10 30 50
Pakatan 15 15 10 40
f column 25 25 40 n = 90
38
Bivariate Frequency Table or Contingency Table
Favor Neutral Oppose f row
Barisan 10 10 30 50
Pakatan 15 15 10 40
f column 25 25 40 n = 90
Obser
ved
frequ
encie
s
39
Bivariate Frequency Table or Contingency Table
Favor Neutral Oppose f row
Barisan 10 10 30 50
Pakatan 15 15 10 40
f column 25 25 40 n = 90
Row
frequency
40
Bivariate Frequency Table or Contingency Table
Favor Neutral Oppose f row
Barisan 10 10 30 50
Pakatan 15 15 10 40
f column 25 25 40 n = 90Column frequency
41
Determine The Hypothesis
Party Membership ( 2 levels) and Nominal Preference ( 3 levels) and Nominal
• Ho : There is no difference between B & P in their opinion on environmental control issue.
• Ha : There is difference between B & P in their opinion on environmental control issue.
• Ho: There is no association between responses to the environmental survey and the party membership in the population.
• Ha: There is association between responses to the environmental survey and the party membership in the population.
42
Calculating Test Statistics
Favor Neutral Oppose f row
Barisan fo =10
fe =13.9
fo =10
fe =13.9
fo =30
fe=22.2
50
Pakatan fo =15
fe =11.1
fo =15
fe =11.1
fo =10
fe =17.8
40
f column 25 25 40 n = 90
43
Calculating Test Statistics
Favor Neutral Oppose f row
Barisan fo =10
fe =13.9
fo =10
fe =13.9
fo =30
fe=22.2
50
Pakatan fo =15
fe =11.1
fo =15
fe =11.1
fo =10
fe =17.8
40
f column 25 25 40 n = 90
= 50*25/90
44
Calculating Test Statistics
•fo = O
•fe = E
Favor Neutral Oppose f row
Barisan fo =10
fe =13.9
fo =10
fe =13.9
fo =30
fe=22.2
50
Pakatan fo =15
fe =11.1
fo =15
fe =11.1
fo =10
fe =17.8
40
f column 25 25 40 n = 90
= 40* 25/90
45
Calculating Test Statistics
8.17
)8.1710(
11.11
)11.1115(
11.11
)11.1115(
2.22
)2.2230(
89.13
)89.1310(
89.13
)89.1310(
222
2222
= 11.03
46
Determine Degrees of Freedom
df = (R-1)(C-1) =(2-1)(3-1) = 2
3. Critical value df = 2
x² 2,0.05 = 5.99
4. Decision
Since x² cal (11.02) is bigger than x² critical (5.99) Reject HO
5.Conclusion: There is significant association between responses to the environmental survey and the party membership (Barisan or Pakatan) in
the population, x² ( 2,n=90) = 11.02, p < .05. ORThere is significant difference between the Barisan and Pakatan voters in
their opinion on environmental control issue, x² ( 2,n=90) = 11.02, p < .05.
Fail to reject HO
RejectHO
α = 0.05
48
Compare computed test statistic against a tabled/critical value
• α = 0.05• df = 2• Critical tabled value = 5.991• Test statistic, 11.03, exceeds critical value• Null hypothesis is rejected• Barisan & Pakatan differ significantly in
their opinions on gun control issues
49
Phi Coefficient• Pearson Chi-Square
provides information about the existence of relationship between 2 nominal variables, but not about the magnitude of the relationship
• Phi coefficient is the measure of the strength of the association
Symmetric Measures
-.050
.050
-.050
-.050
100
Phi
Cramer's V
Nominal byNominal
Pearson's RInterval by Interval
Spearman CorrelationOrdinal by Ordinal
N of Valid Cases
Value
Not assuming the null hypothesis.a.
Using the asymptotic standard error assuming the null hypothesis.b.
Based on normal approximation.c.
N
2
50
Cramer’s V• When the table is larger
than 2 by 2, a different index must be used to measure the strength of the relationship between the variables. One such index is Cramer’s V.
• If Cramer’s V is large, it means that there is a tendency for particular categories of the first variable to be associated with particular categories of the second variable.
Symmetric Measures
-.050
.050
-.050 .100
-.050 .100
100
Phi
Cramer's V
Nominal byNominal
Pearson's RInterval by Interval
Spearman CorrelationOrdinal by Ordinal
N of Valid Cases
ValueAsymp.
Std. Error
Not assuming the null hypothesis.a.
Using the asymptotic standard error assuming the null hypothesis.b.
Based on normal approximation.c.
)1(
2
kNV
Total Number of
cases
Smallest of number of rows or
columns
51
Contingency (C)
• When the table is larger than 2 by 3, a different index must be used to measure the strength of the relationship between the variables. One such index is C.
• If Contingency (C)is large, it means that there is a tendency for particular categories of the first variable to be associated with particular categories of the second variable.
Symmetric Measures
-.050
.050
-.050 .100
-.050 .100
100
Phi
Cramer's V
Nominal byNominal
Pearson's RInterval by Interval
Spearman CorrelationOrdinal by Ordinal
N of Valid Cases
ValueAsymp.
Std. Error
Not assuming the null hypothesis.a.
Using the asymptotic standard error assuming the null hypothesis.b.
Based on normal approximation.c.
nC
2
2
52
SPSS Output for Gun Control Example
Chi-Square Tests
11.025a 2 .004
11.365 2 .003
8.722 1 .003
90
Pearson Chi-Square
Likelihood Ratio
Linear-by-LinearAssociation
N of Valid Cases
Value dfAsymp. Sig.
(2-sided)
0 cells (.0%) have expected count less than 5. Theminimum expected count is 11.11.
a.
Test of Independence
♠ Test the null hypothesis that the two variable are not related (independent) against the alternative hypothesis that the two variable are related (dependent)♠ Contingency table – two variables♠ Calculation based on:
O – Observed frequencyE – Expected frequency
E = RT x CT GT
♠ Formula to calculate x²
x² = Σ ( O – E )² E
2
What to expect?H
yp
oth
esis
Test
1State
HO and HA
2Calculate
x² :
3Critical Value
4Decision
5Conclusion
Based on: O and E = np
x² = Σ( O – E )² E
Measure ofRelationship
• Phi• Contingency• Cramer’s V
Step in Testing Hypothesis:1. State the null and alternative hypotheses
HO: DV is independent of IVHA: DV is dependent on IV
2. Calculate the test statistic
x² Value
3. Determine critical value• α• df = (R-1)(C-1)
4. Make your decision
5. Make conclusion
Criteria Decision x² cal > x² criticalReject HO
x² cal ≤ x² critical Fail to Reject HO
Criteria DecisionSig- x² < α Reject HO
Sig- x² ≥ α Fail to Reject HO
Manual
SPSS
Example 1:
A study was conducted to test the relationship between student group (campus and PJJ) and academic performance. Data collected from a randomly selected sample follow.
1. Test the hypothesis on the relationship between the two variables at 0.01 level of significance.2. Calculate and describe an appropriate measure of association between the two variables.
Student Academic PerformanceGroup High Moderate Low
Campus 93 70 12
PJJ 87 32 6
Answer:
1. Hypotheses testing a. Hypotheses HO: Academic performance is independent of student group HA: Academic performance is dependent on student group
b. Test statistic Calculated expected value for each cell:
Student Academic Performance RowGroup High Moderate Low Totals
Campus 93 70 12 175
(105.0) (59.5) (10.5)
PJJ 87 32 6 125
(75.0) (42.5) (7.5)
Column Totals 180 102 18 300
Chi-Square:
Age O E ( O – E ) ( O – E )² E
C-H 93 105.0 -12 144 1.371
C-M 70 59.5 10.5 110.25 1.853
C-L 12 10.5 1.5 2.25 0.214
P-H 87 75.0 12 144 1.920
P-M 32 42.5 -10.5 110.25 2.594
P-L 6 7.5 -1.5 2.25 0.300
300 8.252
( O – E )²
3. Critical value df = ( R – 1 )( C – 1 ) = ( 2 – 1 )( 3 – 1 ) = 1 x 2 = 2
x² 2,0.01 = 9.21
4. Decision Since x² cal (8.252) is smaller than x² critical (9.21) Fail to reject HO
5. Conclusion There is not enough evidence from the sample to conclude that the two variables, student groups and academic performance, are dependent at 0.01 level of significance.
2. Measure of association for a 2 x 3 contingency table, both contingency and Cramer’s V coefficients are appropriate
x² x² + n
8.2528.252 + 300
0.164
√
x² n(k-1)
8.252300(2-1)
0.166
√
C =
C =
C =
√V =
V =
V =
Negligible association between student group and academic performance
Negligible association between student group and academic performance
SPSS Chi-Square output
Performance is Statistics
High Moderate Low Total
Student Campus Count 93 70 12 175
Group Expected Count 105.5 59.5 10.5 175.0
PJJ Count 87 32 6 125
Expected Count 75.0 42.5 7.5 125.0
Total Count 180 102 18 300
Expected Count 180.0 102.0 18.0 300.0
Student group * Performance in Statistics Cross tabulation
Asymp. Sig.
Value df (2-sided)
Pearson Chi-Square 8.253 2 0.016
Likelihood Ratio 8.370 2 0.015
Linear-by-Linear
Association 3.762 1 0.009
N of Valid cases 300
Chi-Square Tests
a. 0 cells (0.0%) have expected count less than 5. the minimum expected count is 7.50
Value Approx. Sig.
Nominal by Phi 0.166 0.016
Nominal Cramer’s V 0.166 0.016
Contingency Coefficient 0.164 0.016
N of valid cases 300
Symmetric Measures
a. Not assuming the null hypothesisb. Using the asymptotic standard error assuming the null hypothesis
Example 3:
Dr Irwan is interested to test the relationship between gender and program ofstudy. Data taken from a randomly selected sample follow.
1. Test the hypothesis on the relationship at 0.01 level of significance.2. Calculate and describe an appropriate measure of association between the two variables.
Gender Program of Study
Science Social Science
Male 60 110
Female 75 55
Answer:
1. Hypotheses testing a. Hypotheses HO: Academic performance is independent of student group HA: Academic performance is dependent on student group
b. Test statistic Calculated expected value for each cell:
Gender Program of Study RowScience Social Science Totals
Male 60 110 170
(76.5) (93.5)
Female 75 55 130
(58.5) (71.5)
Column Totals 135 165 300
Chi-Square:
G-POS O E ( O – E ) ( O – E )² E
M_Sc 60 76.5 -16.5 272.25 3.559
M_SS 110 93.5 16.5 272.25 2.912
F_Sc 75 58.5 16.5 272.25 4.654
F_SS 55 71.5 -16.5 272.25 3.808
300 14.933
( O – E )²
With Yate’s correction:
G-POS O E ( O – E ) (|O – E|-1/2 )² E
M_Sc 60 76.5 -16.5 256.0 3.346
M_SS 110 93.5 16.5 256.0 2.738
F_Sc 75 58.5 16.5 256.0 4.376
F_SS 55 71.5 -16.5 256.0 3.580
300 14.040
(|O – E|-1/2 )²
3. Critical value df = ( R – 1 )( C – 1 ) = ( 2 – 1 )( 2 – 1 ) = 1 x 1 = 1
x² 1,0.01 = 6.63
4. Decision Since x² cal (14.040) is bigger than x² critical (6.63) Reject HO
5. Conclusion There is not strong evidence from the sample to conclude that the two variables, gender and program of study, are dependent at 0.01 level of significance.
2. Measure of association For a 2 x 2 contingency table, phi coefficients is the most appropriate to be used
x² n
14.933 300
0.223
√
√
Ø =
Ø =
Ø =
Low association between gender and program of study
Chi-Square output
Performance is Statistics
Science Social Science Total
Gender Male Count 60 110 170
Expected Count 76.5 93.5 170.0
Female Count 75 55 130
Expected Count 58.5 72.5 130.0
Total Count 135 165 300
Expected Count 135.0 165.0 300.0
Gender * Program of Study Cross tabulation
Expected Values
Observed Values
Contingency Table, 2 x 2