Day2 statistical tests

Statistical tests-when to use
which?
(Arthvidya Consulting | April 2020) 1
Nidhi Kasetwar
nidhi@arthvidya.com

Tests of hypothesis
Tests of hypothesis
Parametric tests or
standard tests of
hypotheses
Non-parametric tests
or distribution-free
test of hypotheses

What are parametric tests?
Statistical tests are normally based on a
probability distribution.
The test technique makes use of one or
more values obtained from sample data
[often called test statistic(s)] to arrive at
a probability statement about the
hypothesis.
These make use of some assertions: it
may assume that population is normally
distributed, sample drawn is a random
sample.

Parametric tests
z-test; t-test;
𝑥2-test (
Chi square),
F-test.

Non Parametric tests
One sample Sign
test
two-sample sign
test, Fisher-Irwin
test, Rank sum test
Rank correlation,
Kendall’s
coefficient of
concordance and
other tests for
dependence
Kruskal-Wallis test
analogous to
ANOVA
one sample runs
test
Chi square test

Hypothesis testing of means:
Population Sample size Variance Alternate
hypothesis
Formula Type of test
Population
normal,
population
infinite
Large or small Variance of
population
Known
Ha may be one-
sided or two-
sided
Z test
Population
normal,
population
finite
sample size may
be large or small
variance
of the
population
is known
Ha may be one-
sided or two-
sided
Z test

Population Sample size Variance Alternate
hypothesis
Formula Type of test
Population
normal,
population
infinite
sample size small variance of the
population
unknown,
H 𝛼 may be one-
sided or two-
sided
Calculation of t –test statistic t test
Population
normal,
population
finite,
sample size small variance of the
population
unknown,
Ha may be one-
sided or two-
sided:
t test

ztest- When to use?
Population •Normal, infinite or finite
Sample
size
•Large or small
Variance •Known

Example: Hypothesis testing of means contd.
A sample of 400 students is found to
have a mean height 67.47 inches.
Can it be reasonably regarded as a
sample from a large population with
mean height 67.39 inches and
standard deviation 1.30 inches?
Test at 5% level of significance.
Taking the null hypothesis that the
mean height of the population is
equal to 67.39 inches, we can write:
H0 : 𝜇H0 =67.39"
Ha : 𝜇Ha ≠ 67.39"
ҧ𝑥 = 67.47" , 𝜎p = 1.30" , n = 400.
Assuming the population to be
normal, we can work out the test
statistic z as under: ( ҧ𝑥- 𝜇H0 )/( 𝜎p
/ 𝑛) = (67.47-67.39)/(1.3/ 20)
=1.231

Example: Hypothesis testing of means
As Ha is two-sided in the given question
-- we shall be applying a two-tailed test
for determining the rejection regions at
5% level of significance
which comes to as under, using normal
curve area table:
R : | z | > 1.96
The observed value of z is 1.231 which is
in the acceptance region since R : | z | >
1.96
and thus H0 is accepted.
Conclusion:
the given sample (with mean height =
67.47") can be regarded to have been
taken from a population with mean
height 67.39" and standard deviation
1.30" at 5% level of significance.

Areas for standard normal distribution

Example: Hypothesis testing of Means contd..
We are interested in a population of 20
industrial units of the same size, all of
which are experiencing excessive labour
turnover problems.
The past records show that the mean of
the distribution of annual turnover is 320
employees, with a standard deviation of 75
employees.
A sample of 5 of these industrial units is
taken at random which gives a mean of
annual turnover as 300 employees. Is the
sample mean consistent with the
population mean? Test at 5% level.
population mean is 320 employees, we can
write:
H0 : 𝜇H0 = 320
Ha : 𝜇Ha ≠ 320
Z= = – 0.67

Example: Hypothesis testing of Means contd..
As Ha is two-sided in the given
question, we shall apply a two-tailed
test for determining the
rejection regions at 5% level of
significance which comes to as under,
using normal curve area table:
R : | z | > 1.96
The observed value of z is –0.67
which is in the acceptance region
since R : | z | > 1.96 and thus,
H0 is accepted and we may conclude
that the sample mean is consistent
with population mean i.e.,
the population mean 320 is
supported by sample results.

Example : Hypothesis testing of differences between
means:
The mean produce of wheat of a
sample of 100 fields in 200 lbs. per
acre with a standard deviation of 10
lbs. Another sample of 150 fields gives
the mean of 220 lbs with a standard
deviation of 12 lbs. Can the two
samples be considered to have been
taken from the same population whose
standard deviation is 11 lbs? Use 5 per
cent level of significance.
means of two populations do not
differ, we can write
𝐻0 ∶ 𝜇1= 𝜇2
𝐻𝑎 ∶ 𝜇1≠ 𝜇2

Example: Hypothesis testing of proportions:
The null hypothesis is that 20 per
cent of the passengers go in first
class, but management recognizes
the possibility that this percentage
could be more or less. A random
sample of 400 passengers includes
70 passengers holding first class
tickets. Can the null hypothesis be
rejected at 10 percent level of
significance?
The null hypothesis is
H0 : p = 20% or 0.20
and Ha : p ≠ 20%

Example : Hypothesis testing for difference between
proportions:
null hypothesis: that there is no
difference between the two drugs i.e.,
H0 : ෠𝑃1 = ෠𝑃2
The alternative hypothesis can be
taken as that there is a difference
between the drugs i.e.,
Ha : ෠𝑃1 ≠ ෠𝑃2
A drug research experimental unit is
testing two drugs newly developed to
reduce blood pressure levels.
The drugs are administered to two
different sets of animals. In group one,
350 of 600 animals tested respond to
drug one and in group two, 260 of 500
animals tested respond to drug two.
The research unit wants to test
whether there is a difference between
the efficacy of the said two drugs
at 5 per cent level of significance.

t test- When to use?
Population •Normal, infinite, finite
Sample
size
•small
Variance •Not Known

Two types of t-test:
Independent samples t test
used when we want to
compare the mean scores of
two different groups of people
or conditions
Paired samples t test
used when we want to
compare the mean scores for
the same group of people on
two different occasions, or
when we have matched pairs

Example: Hypothesis testing for means - t test:
The specimen of copper wires
drawn form a large lot have the
following breaking strength (in
kg. weight):
578, 572, 570, 568, 572, 578,
570, 572, 596, 544
Test (using Student’s t-
statistic)whether the mean
breaking strength of the lot
may be taken to be 578 kg.
weight (Test at 5 per cent level
of significance).
population mean is equal to hypothesised
mean of 578 kg:
H0 : 𝜇= 𝜇H0 = 578 kg
Ha : 𝜇 ≠ 𝜇H0
As the sample size is small (since n = 10) and
the population standard deviation is not
known,
we use t-test assuming normal population
and shall work out the test statistic t as
under:

Hypothesis testing for means t test contd….
S.No Xi (Xi - ҧ𝑥) (Xi - ҧ𝑥)^2
1 578 6 36
2 572 0 0
3 570 -2 4
4 568 -4 16
5 572, 0 0
6 578 6 36
7 570 -2 4
8 572 0 0
9 596 24 576
10 544 -28 784
5720= σ 𝑥𝑖 σ (Xi − ҧ𝑥)^2 =1456

Hypothesis testing for means t test
𝜎𝑠 = ෎
(𝑥𝑖− ҧ𝑥)2
𝑛 − 1
= 12.72 kg
t = (572-578)/(12.72/ 10)=-1.488
Degree of freedom = (n – 1) = (10 –
1) = 9
As Ha is two-sided, we shall
determine the rejection region
applying two-tailed test at 5 per
cent level of significance, and it
comes to as under, using table of
t-distribution for 9 d.f.:
R : | t | > 2.262
As the observed value of t (i.e. –
1.488) is in the acceptance region,
we accept H0 at 5 per cent level
and conclude that the mean
breaking strength of copper wires
lot may be taken as 578 kg.
weight.

Example : Hypothesis testing for comparing two related
samples
Memory capacity of 9 students was
tested before and after training. State at
5 per cent level of
significance whether the training was
effective from the following scores:
Stud
ent 1 2 3 4 5 6 7 8 9
Befor
e 10 15 9 3 7 12 16 17 4
After 12 17 8 5 6 11 18 20 3
• H0 : 𝜇1 = 𝜇2which is equivalent to test
H0 : ഥ𝐷 = 0
• Ha : 𝜇1 < 𝜇2 (as we want to conclude
that training has been effective)
• Where: ഥ𝐷 is the mean of differences

Assumptions of t test:
Measured at the interval or
ratio level;, using a
continuous scale
Random sampling Independence of
observations
Normal distribution
the populations from
which the samples are
taken are normally
distributed
Homogeneity of variance
Samples that are obtained
from populations should
have equal variances..

F test:
Population • Normal,
Sample size • Drawn
Observations • Independent

Used for:
The object of F-test is to test the
hypothesis whether the two samples
are from the same normal
population with equal variance
or from two normal populations with
equal variances.
Testing the homogeneity of
variances
Nowadays mostly used in Analysis
of variance

Example:
Two random samples drawn from two
normal populations are:
Sample 1: 20 16 26 27 23 22 18 24 25
19
Sample 2: 27 33 42 35 32 34 38 28 41
43 30 37
Test using variance ratio at 5 per cent
and 1 per cent level of significance
whether the two populations have the
same variances.
To test the equality of variances of two
normal populations, we make use of F-
test based on F-distribution the null
hypothesis is:
𝐻0: 𝜎 𝑃1
2
= 𝜎 𝑝2
2

F test
Wherein the F ratio is given by
F = 𝜎𝑠2
2
/ 𝜎𝑠1
2
We compare this with the table
value of F
If the calculated value of F is
greater than table value of F at a
certain level of significance
for (n1 – 1) and (n2 – 2) degrees of
freedom, we regard the F-ratio as
significant.
Degrees of freedom for greater
variance is represented as v1 and
for smaller variance as v2.

Chi square test –Conditions for application
Sample:
• Observations recorded, collected on a random
basis.
• All the items in the sample must be independent.
Number of
items
• No group should contain very few items, say less
than 10. Some statisticians take this number as 5
• The overall number of items must normally be at
least 50, howsoever small the number of groups
may be.
Constraints • The constraints must be linear.

Used :
As a non parametric test
1. test the goodness of fit;
2. test the significance of
association between two
attributes
3. test the homogeneity or the
significance of population
variance.
Test for comparing variances:

Example : Chi-square as a test for comparing variances
• Weight of 10 students is as
follows:
S.No 1 2 3 4 5 6 7 8 9 10
Weig
ht(kg) 38 40 45 53 47 43 55 48 52 49 Can we say that the variance of the
distribution of weight of all students
from which the above sample of 10
students was drawn is equal to 20 kgs?
Test this at 5 per cent and 1 per cent
level of significance.

Chi square interpretation
At 5 per cent level of significance
the table value of 𝑥2(𝐶ℎ𝑖 𝑠𝑞𝑢𝑎𝑟𝑒) =
16.92
and at 1 per cent level of significance,
it is 21.67 for 9 d.f. and both these
values are greater than the calculated
value of 𝑥2(𝐶ℎ𝑖 𝑠𝑞𝑢𝑎𝑟𝑒) which is
13.999.
Hence we accept the null hypothesis
and conclude that the variance of the
given distribution can be taken as 20
kgs at 5 percent as also at 1 per cent
level of significance.
In other words, the sample can be said
to have been taken from a population
with variance 20 kgs.

As a goodness of fit measure:
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.
.
If the calculated value of chisquare is
less than the table value at a certain
level of significance, the fit is
considered to be a good one.

Chi square: As a test of independence
Test enables us to explain whether or not
two attributes are associated.
We may be interested in knowing
whether a new medicine is effective in
controlling fever or not,
chi square test will helps us in deciding
this issue.
In such a situation, we proceed with the
null hypothesis that the two attributes
(viz., new medicine and control of fever)
are independent which means that new
medicine is not effective in controlling
fever.
On this basis we first calculate the
expected frequencies and then work out
the value of chi square.

The interpretation ….contd
If the calculated value of chi
square is less than the table value
at a certain level of significance for
given degrees of freedom-
we conclude that null hypothesis
stands which means that the two
attributes are independent or not
associated (i.e., the new medicine
is not effective in controlling the
fever).

The interpretation
But if the calculated value of
chi square is greater than its
table value
null hypothesis does not hold good
which means the two attributes are
associated and the association is not
because of some chance factor but it
exists in reality (i.e., the new medicine
is effective in controlling the fever and
as such may be prescribed).

A point to note:
Chi-square test is not a measure
of the degree of relationship or
the form of relationship
between two attributes
but is simply a technique of judging
the significance of such association
or relationship between two
attribute

Other places where chi-square can be used:
A die is thrown 132 times with
following results:
Is the die unbiased?
The table given below shows the data obtained
during outbreak of smallpox:
Test the effectiveness of vaccination in
preventing the attack from smallpox. Test your
result with the help of chisquare test at 5 per
cent level of significance.
Number
turned
up 1 2 3 4 5 6
Frequen
cy 16 20 25 14 29 28
attacked
Not
attacked Total
Vaccinated 31 469 500
Not vaccinated 185 1315 1500
Total 216 1784 2000

Anova–Conditions for application
Data type interval data of the dependent variable
normality data is normally distributed.
Homoscedasticity
different samples have the same variance, even if they
come from different populations.
no
multicollinearity
independent variables are not intercorrelated

About Anova
ANOVA
One way Anova
• take only one factor and investigate
the differences amongst its various
categories having numerous
possible values
Two way Anova
• investigate two factors at the same
time

Essence of Anova- Variation broken down into two
parts:
Variation
attributed to
chance
Variation
attributed to
specific causes

Example:
Set up an analysis of variance table for the following two way design
results:
State whether variety differences are significant at 5% level of
significance.

F ratios
The F-ratios are compared with their
corresponding table values, for given
degrees of freedom at a specified level
of significance
and if it is found that the calculated F-
ratio concerning variation between
columns is equal to or greater than its
table value, then the difference among
columns means is considered
significant.
Similarly, the F-ratio concerning
variation between rows can be
interpreted.

Judge the effectiveness in reducing
blood pressure for three different
groups of people:
• Do the drugs act differently?
• Are the different groups of people
affected differently?
• Is the interaction term significant?
• To be considered at a significant
level of 5%.
Drug
X Y Z
Group
of
people
A 14 10 11
15 9 11
B 12 7 10
11 8 11
C 10 11 8
11 11 7

ANOVA –where to use?
Used when multiple sample
cases are involved.
Examine the significant
difference for more than two
sample means at the same time.
One can draw inferences about
whether the samples have been
drawn from populations having
the same mean.

Applications of Anova:
the differences in various types of
feed prepared for a particular class of
animal
various types of drugs manufactured
for curing a specific disease may be
studied and judged to be significant
or not
a manager of a big concern can
analyze the performance of various
salesmen of his concern in order to
know whether their performances
differ significantly.

compare more than two populations such
as in comparing the yield of crop from
several varieties of seeds,
the gasoline mileage of four automobiles
Through this technique one can
explain whether various varieties of
seeds or fertilizers or soils differ
significantly so that a policy
decision could be made pertaining
to them..

Non parametric tests
Statistical tests are normally based
on a probability distribution.
The test technique makes use of one
or more values obtained from
sample data [often called test
statistic(s)] to arrive at a probability
statement about the hypothesis.
These make use of some assertions:
it may assume that population is
normally distributed, sample drawn is a
random sample.
Non parametric tests make use of no
such assumptions

Assertions in statistical tests:
An assertion directly related to
the purpose of investigation-
Hypothesis
Assertions to make a
probability statement
Model
When we apply a test (to test the
hypothesis) without a model, it is
known as distribution-free test, or
the nonparametric test.
Non-parametric tests: do not make
an assumption about the parameters
of the population and thus do not
make use of the parameters of the
distribution.

Non parametric tests:
S.No Situation analysis Test used
1. Test of a hypothesis concerning some single value for the
given data
One sample Sign test
2. Test of a hypothesis concerning no difference among two
or more sets of data
two-sample sign test, Fisher-Irwin
test, Rank sum test
3. Test of a hypothesis of a relationship between variables Rank correlation, Kendall’s
coefficient of concordance and
other tests for dependence
4. Test of a hypothesis concerning variation in the given data test analogous to ANOVA
viz., Kruskal-Wallis test.
5. Tests of randomness of a sample based on the theory of
runs
one sample runs test
6. Test of hypothesis to determine if categorical data shows
dependency or if two classifications
are independent
Chi square test

Sign Test
Sign test-
• One sample sign test
• Two sample sign test
The one sample sign test is a non-
parametric test applicable when we
sample a continuous
symmetrical population in which
case the probability of getting a
sample value less than
mean is 1/2 and the probability of
getting a sample value greater than
mean is also 1/2.

Example for Sign test
Suppose playing four rounds of golf at
the City Club 11
professionals totaled 280, 282, 290,
273, 283, 283, 275, 284, 282, 279, and
281.
Use the sign test at 5% level of
significance to test the null hypothesis
that professional golfers average 𝜇H0 =
284 for four rounds against the
alternative hypothesis 𝜇H0 < 284.
To test the null hypothesis
𝜇H0 = 284
against the alternative hypothesis
𝜇H0 < 284 at
5% (or 0.05) level of significance
We get a value of= 0.011
Since this value is less than a = 0.05, the
null hypothesis must be rejected.
In other words, we conclude that
professional golfers’ average is less than
284 for four rounds of golf.

Where to use the two sample sign test?
The sign test is used where we deal with paired data.
The following are the numbers of artifacts dug up by two archaeologists at an ancient cliff
dwelling on 30 days:
Use the sign test at 1% level of significance to test the null hypothesis that the two
archaeologists,
X and Y, are equally good at finding artifacts against the alternative hypothesis that X is
By X 1 0 2 3 1 0 2 2 3 0 1 1 4 1 2 1 3 5 2 1 3 2 4 1 3 2 0 2 4 2
By Y 0 0 1 0 2 0 0 1 1 2 0 1 2 1 1 0 2 2 6 0 2 3 0 2 1 0 1 0 1 0

Fisher Irwin test
It is employed to determine whether
one can reasonably assume that
two supposedly different treatments
are in fact different in terms of the
results they produce.
Suppose the management of a business
unit has designed
a new training programme which is now
ready and
as such it wishes to test its
performance against that of the old
training programme.

Fisher Irwing Test Methodology
Twelve newly selected workers are chosen for
an experiment through a standard selection
procedure.
This group of twelve is then divided into two
groups of six each, one group for each training
programme.
Workers are randomly assigned to the
two groups. After the training is
completed, all workers are given the
same examination and the result is as
under:

No. Passed No. failed Total
New
Training(A) 5 1 6
Old
Training(B) 3 3 6
8 4 12
How can a decision be made to check which
training is better?
We may test the hypothesis for the purpose.
The hypothesis is that the two programmes are
equally good.
Prior to testing, the significance level (or the a
value) must be specified and supposing the
management fixes 5% level for the purpose.
Calculated value= 0.27.
So we accept/ fail to reject the null hypothesis.

They do the same job-how are they different?
Chi square test
• Computationally easy
• Gives approximate p value
Fisher Exact test
• Computationally complex
• Gives exact p value
• Useful when proportions and
sample sizes are small

Mc Nemer test
After some treatment, the same number
of subjects are asked to express their
views about the given system whether
they favour it or do not favour it.
We judge the significance of any observed
change in views of the same subjects
Used when the data happen to be nominal
and relate to two related samples.
Specially useful with before-after
measurement of the same subjects.
The subjects initially are divided into equal
groups as to their favorable and
unfavorable views about, say, any system.

Application:
researcher attempts to determine if a
drug has an effect on a particular
disease.
Counts of individuals are given in a
table, with the diagnosis
(disease: present or absent) before
treatment given in the rows, and the
diagnosis after treatment in the
columns.
The test requires the same subjects to
be included in the before-and-after
measurements (matched pairs).

Wilcoxon Matched-pairs Test (or Signed Rank Test)
An experiment is conducted to judge
the effect of brand name on quality
perception.
16 subjects are recruited for the
purpose and are asked to taste and
compare two samples of product on
a set of scale items judged to be
ordinal. The following data are
obtained:

Pair Brand A Brand B
1 73 51
2 43 41
3 47 43
4 53 41
5 58 47
6 47 32
7 52 24
8 58 58
9 38 43
10 61 53
11 56 52
12 56 57
13 34 44
14 55 57
15 65 40
16 75 68
Test the hypothesis, using Wilcoxon
matched-pairs test, that there is no
difference between the
perceived quality of the two samples.
Use 5% level of significance.

Wilcoxon matched pairs test
Let us first write the null and
alternative hypotheses as under:
H0: There is no difference
between the perceived
quality of two samples.
Ha: There is difference
between the perceived
quality of the two samples.
Using Wilcoxon matched-pairs test, we
work out the value of the test statistic T
:
The table value of T at five percent level
of significance when n = 15 is 25 (using a
two-tailed test because our alternative
hypothesis is that there is difference
between the perceived quality of the
two samples).
The calculated value of T is 18.5 which is
less than the table value of 25. As such
we reject the null hypothesis and
conclude that there is difference
between the perceived quality of the
two samples.

Rank Sum tests
U test : Wilcoxon-Mann-
Whitney test:
Used to determine whether two
independent samples have been
drawn from the same population.
H-test: Kruskal – Wallis test
Analogous to Anova

How to write a research paper?
How to start
research topic?
Find information
Make your thesis
statement
Make research
paper outline
Organize your notes Literature review
The research
question(s)
Research
methodology
Writing the results,
analysis, discussion,
and conclusion
Write your first draft

Structure of a Research paper
• Title. Your title is the most important part of your paper. ...
• Abstract. The abstract is a summary of your research. ...
• Introduction. Include background information on the subject and your
objectives here.
• Research Methodology...
• Results. ...
• Discussion. ...
• Limitations. ...
• Acknowledgments.

Format of an APA research paper
• Title page
• Abstract
• An introduction to the author’s ideas
• A brief summary of the research the author conducted
• Define any abbreviations and terminology.
• Body
• Text citation and references

References:
• Kothari, C.R. (2004) Research Methodology: Methods and Techniques. 2nd Edition,
New Age International Publishers, New Delhi.
• Nargundkar, Rajendra(2002) Marketing Research: Text and Cases. Tata McGraw-Hill
Pub , New Delhi
• Malhotra, N. K. (2007). Marketing research: An applied orientation. Upper Saddle
River, NJ: Pearson/Prentice Hall.
• Ram Ahuja (2002). Research Methods 1st ed., Rawat Publications, New Delhi

Thankyou
nidhi@arthvidya.com

Day2 statistical tests

More Related Content

What's hot

Similar to Day2 statistical tests

Recently uploaded

Day2 statistical tests