The document discusses the chi-square test, which is used to determine if an observed frequency distribution differs from an expected theoretical distribution. It can be used as a test of independence to determine if two variables are associated, and as a test of goodness of fit to assess how well an expected distribution fits observed data. The steps of the chi-square test are outlined, including calculating the test statistic, determining degrees of freedom, and comparing the statistic to critical values to determine if the null hypothesis can be rejected. An example of a chi-square test of independence is shown to test if perceptions of fairness of performance evaluation methods are independent of each other.
INFERENTIAL STATISTICS: AN INTRODUCTIONJohn Labrador
For instance, we use inferential statistics to try to infer from the sample data what the population might think. Or, we use inferential statistics to make judgments of the probability that an observed difference between groups is a dependable one or one that might have happened by chance in this study.
INFERENTIAL STATISTICS: AN INTRODUCTIONJohn Labrador
For instance, we use inferential statistics to try to infer from the sample data what the population might think. Or, we use inferential statistics to make judgments of the probability that an observed difference between groups is a dependable one or one that might have happened by chance in this study.
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Most of the students may struggle with the analysis of variance (ANOVA). Here in this presentation you can clear all your doubts in analysis of variance with suitable examples.
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Chi Square Test…..
This topic comes under Biostatistics…….
This is useful for Maths students, B.Pharm Students ,M.Pharm Students who studying Biostatistics.
This Presentation Contain following...
#History and Introduction
#Conditions
#Formula
#Classification
#Types of Non-Parametric Chi Square Test
#Test of Independence
#Steps for Test of Independence
#Problem and Solution for Test of Independence
#Test of Goodness of Fit
#Problem and Solution for Test of Goodness of Fit
#Applications of Chi Square Test
Thanks for the Help and Guidance of Dr. M. S. Bhatia Sir
• Non parametric tests are distribution free methods, which do not rely on assumptions that the data are drawn from a given probability distribution. As such it is the opposite of parametric statistics
• In non- parametric tests we do not assume that a particular distribution is applicable or that a certain value is attached to a parameter of the population.
When to use non parametric test???
1) Sample distribution is unknown.
2) When the population distribution is abnormal
Non-parametric tests focus on order or ranking
1) Data is changed from scores to ranks or signs
2) A parametric test focuses on the mean difference, and equivalent non-parametric test focuses on the difference between medians.
1) Chi – square test
• First formulated by Helmert and then it was developed by Karl Pearson
• It is both parametric and non-parametric test but more of non - parametric test.
• The test involves calculation of a quantity called Chi square.
• Follows specific distribution known as Chi square distribution
• It is used to test the significance of difference between 2 proportions and can be used when there are more than 2 groups to be compared.
Applications
1) Test of proportion
2) Test of association
3) Test of goodness of fit
Criteria for applying Chi- square test
• Groups: More than 2 independent
• Data: Qualitative
• Sample size: Small or Large, random sample
• Distribution: Non-Normal (Distribution free)
• Lowest expected frequency in any cell should be greater than 5
• No group should contain less than 10 items
Example: If there are two groups, one of which has received oral hygiene instructions and the other has not received any instructions and if it is desired to test if the occurrence of new cavities is associated with the instructions.
2) Fischer Exact Test
• Used when one or more of the expected counts in a 2×2 table is small.
• Used to calculate the exact probability of finding the observed numbers by using the fischer exact probability test.
3) Mc Nemar Test
• Used to compare before and after findings in the same individual or to compare findings in a matched analysis (for dichotomous variables).
Example: comparing the attitudes of medical students toward confidence in statistics analysis before and after the intensive statistics course.
4) Sign Test
• Sign test is used to find out the statistical significance of differences in matched pair comparisons.
• Its based on + or – signs of observations in a sample and not on their numerical magnitudes.
• For each subject, subtract the 2nd score from the 1st, and write down the sign of the difference.
It can be used
a. in place of a one-sample t-test
b. in place of a paired t-test or
c. for ordered categorial data where a numerical scale is inappropriate but where it is possible to rank the observations.
5) Wilcoxon signed rank test
• Analogous to paired ‘t’ test
6) Mann Whitney Test
• similar to the student’s t test
7) Spearman’s rank correlation - similar to pearson's correlation.
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Chapter 11: Goodness-of-Fit and Contingency Tables
11.1: Goodness of Fit Notation
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The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
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Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
2. Chi-Square Test
Karl Pearson introduced a
test to distinguish whether
an observed set of
frequencies differs from a
specified frequency
distribution
The chi-square test uses
frequency data to generate a
statistic
Karl Pearson
3. A chi-square test is a statistical test
commonly used for testing independence
and goodness of fit. Testing independence
determines whether two or more
observations across two populations are
dependent on each other (that is, whether
one variable helps to estimate the other).
Testing for goodness of fit determines if
an observed frequency distribution
matches a theoretical frequency
distribution.
5. Conditions for the application of 2 test
Observations recorded and collected are
collected on random basis.
All items in the sample must be
independent.
No group should contain very few items,
say less than 10. Some statisticians take this
number as 5. But 10 is regarded as better by
most statisticians.
Total number of items should be large, say
at least 50.
6. The 2 distribution is not symmetrical and all the
values are positive. For each degrees of freedom
we have asymmetric curves.
10. 2. As a Test of Goodness of Fit
It enables us to see how well does
the assumed theoretical distribution(such
as Binomial distribution, Poisson
distribution or Normal distribution) fit to
the observed data. When the calculated
value of χ2 is less than the table value at
certain level of significance, the fit is
considered to be good one and if the
calculated value is greater than the table
value, the fit is not considered to be good.
11. EXAMPLE
As personnel director, you
want to test the perception of
fairness of three methods of
performance evaluation. Of
180 employees, 63 rated
Method 1 as fair, 45 rated
Method 2 as fair, 72 rated
Method 3 as fair. At the 0.05
level of significance, is there
a difference in perceptions?
13. H0: p1 = p2 = p3 = 1/3
H1: At least 1 is different
= 0.05
Test Statistic:
2 = 6.3
n1
= 63 n2 = 45 n3 =
Critical Value(s):
72
Decision:
Reject H0 at sign. level 0.05
Conclusion:
At least 1 proportion is different
Reject H0
= 0.05
0
5.991
2
14. 3.As a Test of Independence
χ2 test enables us to explain whether or not
two attributes are associated. Testing
independence determines whether two or more
observations across two populations are
dependent on each other (that is, whether one
variable helps to estimate the other. If the
calculated value is less than the table value at
certain level of significance for a given degree of
freedom, we conclude that null hypotheses stands
which means that two attributes are independent
or not associated. If calculated value is greater
than the table value, we reject the null
hypotheses.
15. Steps involved
Determine The Hypothesis:
Ho : The two variables are independent
Ha : The two variables are associated
Calculate Expected frequency
17. Compare computed test statistic
against a tabled/critical value
The computed value of the Pearson chisquare 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 chisquare statistic.
If calculated 2 is greater than 2 table value,
reject Ho
19. EXAMPLE
Suppose a researcher is interested in voting
preferences on gun 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.
20. BIVARIATE FREQUENCY TABLE OR
CONTINGENCY TABLE
Favor
Neutral
Oppose
f row
10
10
30
50
Republican 15
15
10
40
f column
25
40
n = 90
Democrat
25
21. BIVARIATE FREQUENCY TABLE OR
CONTINGENCY TABLE
Favor
Neutral
Oppose
f row
10
10
30
50
Republican 15
15
10
40
f column
25
40
n = 90
Democrat
25
22. Row frequency
BIVARIATE FREQUENCY TABLE OR
CONTINGENCY TABLE
Favor
Neutral
Oppose
f row
10
10
30
50
Republican 15
15
10
40
f column
25
40
n = 90
Democrat
25
22
23. BIVARIATE FREQUENCY TABLE OR
CONTINGENCY TABLE
Favor
Neutral
Oppose
f row
10
10
30
50
Republican 15
15
10
40
f column
25
40
n = 90
Democrat
Column frequency
25
24. DETERMINE THE HYPOTHESIS
•
Ho : There is no difference between D &
R in their opinion on gun control issue.
•
Ha : There is an association between
responses to the gun control survey and
the party membership in the population.
25. CALCULATING TEST STATISTICS
Favor
Neutral
Oppose
f row
fo =10
fe =13.9
Republican fo =15
fe =11.1
fo =10
fe =13.9
fo =15
fe =11.1
fo =30
fe=22.2
fo =10
fe =17.8
50
f column
25
40
n = 90
Democrat
25
40
29. COMPARE COMPUTED TEST STATISTIC
AGAINST TABLE VALUE
α
= 0.05
df = 2
Critical tabled value = 5.991
Test statistic, 11.03, exceeds critical
value
Null hypothesis is rejected
Democrats & Republicans differ
significantly in their opinions on gun
control issues
30. 2 TEST OF INDEPENDENCE
THINKING CHALLENGE
You’re a marketing research analyst. You ask a
random sample of 286 consumers if they
purchase Diet Pepsi or Diet Coke. At the 0.05
level of significance, is there evidence of a
relationship?
Diet Pepsi
Diet Coke
No
Yes
Total
No
84
32
116
Yes
48
122
170
Total
132
154
286
31. 2 TEST OF INDEPENDENCE SOLUTION*
Eij 5 in all cells
116·132
286
Diet Coke
Diet Pepsi
154·132
286
No
Yes
Obs. Exp. Obs. Exp. Total
No
84
53.5
32
62.5
116
Yes
48
78.5
122
91.5
170
Total
132
132
154
154
286
170·132
286
170·154
286
33. H0:
No Relationship
H1:
Relationship
= 0.05
df = (2 - 1)(2 - 1) = 1
Critical Value(s):
Test Statistic:
2 = 54.29
Decision:
Reject at sign. level 0 .05
Reject H0
= 0.05
0
3.841
2
Conclusion:
There is evidence of a relationship
34. 2 TEST OF INDEPENDENCE THINKING
CHALLENGE 2
There is a statistically significant relationship
between purchasing Diet Coke and Diet
Pepsi. So what do you think the relationship
is? Aren’t they competitors?
Diet Coke
No
Yes
Total
Diet Pepsi
No
Yes
84
32
48
122
132
154
Total
116
170
286
35. YOU RE-ANALYZE THE DATA
High
Income
Diet Coke
No
Yes
Total
Diet Pepsi
No
Yes
4
30
40
2
44
32
Total
34
42
76
Diet Coke
No
Yes
Total
Diet Pepsi
No
Yes
80
2
8
120
88
122
Total
82
128
210
Low
Income
Data mining example: no need for statistics here!