ugcmodule7 new2
TRANSCRIPT
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TESTING OF HYPOTHESIS
` The greatest tragedy of research is the slaying of abeautiful hypothesis by an ugly fact.
` Statistics can be broadly classified into twocategories namely
` (i) Descriptive Statistics and`
(II) Inferential Statistics.` This unit deals with the inferential statistics,in particular, Testing of hypothesis.
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STATISTICAL INFERENCESTATISTICAL INFERENCE
Statistical Inference means making inferenceabout the population from the sample taken
from it. This is classified into two broad
categories, namely1. Estimation Theory
2. Testing of Hypothesis.
y
The overview of Statistical Inference isfurnished below.
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y
Hypothesis:A statistical hypothesis is some statementmade about the population characteristic in termsof the parameter value of parent distribution.
In testing of hypothesis we either reject oraccept the null hypothesis.
y The basic idea is take a random sample fromthe population. If the sample supports your
hypothesis then accept the hypothesis. (i.e.,the hypothesis is true), otherwise reject thehypothesis
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y Statistical testing involves verifying or
refuting statements concerning propertiesof the population, with some probability oferror, based on data from a subset of thatpopulation.
y The subset of the population, which isselected randomly, is termed a sample.
y From the sample data, a test statistic is
calculated which allows us to make adecision regarding the null hypothesis.
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Test procedure involves the following 5 stepsTest procedure involves the following 5 steps
Step 1.
Hypothesis Framing of Hypotheses
y A. Null hypothesis
y Null Hypotheses are presumptions about the statusquo in the population and are either upheld or disalloweddepending upon the result of a statistical test.
y Null Hypothesis:
yVersion 1
Null
neutral (Text Book)
y Version 2 Null nullify(RA Fisher Reject /
Discard)
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ILLUSTRATIONNull hypothesis is a neutral or null state
hypothesis
Horlicks Vs ComplanHypo 1: H > C
Hypo 2: H < CHypo 3: H = CAmong them which one is in null state?How to guarantee the null state?
Begin your hypothesis with :There is no significant difference between_____________ and ______________
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Version 2Version 2
y
According to RA FisherN
ull Hypothesis isan Hypothesis which is tested for possiblerejection under the assumption that it is true.
He insists us to select the easily reject able
hypothesis as the null hypothesisy As the consequence, Instead of simpliy
writing the inference as, null hypothesis isaccepted, it is expected to be written as since the data do not support for the rejection ofNull hypothesis, it is not rejected
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B. ALTERNATIVE HYPOTHESIS
The researcher must give an AlternativeHypothesis in the test procedure.
Whenever the Ho is rejected the H1 will beaccepted automatically.
Based on Alternative hypothesis, the test iscalled one-tailed test or two-tailed test.
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yUnless there are good reasons aboutthe hypothesis, a two tailed testshould be postulated.yOne-tailed tests are appropriate
when a theoretical perspective basedon previous research would suggest
that the difference should occurother than the one predicated.
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Step 2. FixingStep 2. Fixing lvellvel of significance basedof significance based
on Type 1 and Type 2 errorson Type 1 and Type 2 errors
ACCEPT REJECT
HO TRUE Type I error
HO FALSE Type II error
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EDUCATION:
PASS FAIL
Good Student Type I error
Poor Student Type II error
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I
ACCEPT REJECT
Good Product Type I error
Poor Quality Type II error
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Our aim is to minimize both Type Iand Type II error simultaneously.
But unfortunately they work onopposite direction. i.e., if youreduce Type I error then the Type IIerror will increase and vice versa.
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What can at best achieved is a reasonable
balance between these two errors.
In all testing of hypothesis procedures, it issimply assumed that type I error is much
more severe than type II error and so needsto be controlled.Hence what we do in statistical testingprocedure is, we are fixing the Type I errorat a specified level. Usually at 5% or 1%.But this is not mandatory.You can fix any level of your choice.Hence fixing the level of significance isthe work of the scholar.
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Step 3. Construction of Test Statistic:
The general formula for test statistic isgiven by
Example : Large Sample:
Z = ~ N[0,1]
)(StatisticErrorStandard
c)E(Statisti-Statistic:StatisticTest
=
n
x
W
Q
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SmallSample
t= ~
tn-1
n
S
x Q
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Step 4: to find the CalculatedStep 4: to find the Calculated
valuevalue
In the case of problems, we calculate
and substitute the required entities inthe test statistics from the sample and
do the simplification.Ultimately we
get a single value known as the
Calculated Value
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STEP 5. INFERENCE:
Inference can be done in any one of three
approaches. All of them will lead to the sameconclusion.
The three approaches are:
1. Critical value approach.2. P- value approach.
3. Confidence interval approach.
Note: In the SPSS output, one can see the
provisions for all these three. The researchercan select any one of the approaches. Selectp-value approach due its advantages.
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Classical approach:Classical approach: Class room teachingClass room teaching
y Under the Classical approach, with a fixedpredetermined value namely critical value or
table value, a test will produce a decision as to
whether or not to reject Ho.
y Decision rule
If the calculated value is less than the table
value , the sample does not provide any evidence
to reject the null hypothesis and hence H 0 isaccepted. Otherwise, it is rejected.
y
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RemarkRemarky But merely comparing the observed test
statistic with some critical value andconcluding as
y using a 5% test, reject Ho ory reject Ho with significance level 5% or
result is significant at 5%
(all equivalent statements)
y does not provide the recipient of the resultswith clear detailed information on thestrength for the evidence against Ho.
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P-value
y
A more informative approach is to calculate and quotethe probability value (p value ) of the observed teststatistic.
y The p-values is the lowest level at which Ho can be
rejected. The smaller the p-value, the stronger is theevidence against the null hypothesis.
y Decision Rule:
If the p value is less than the level of significance , the
null hypothesis H0 is rejected. Otherwise it is accepted.
Note the difference between these two approaches: Theresearcher may likely to have confusion but be clear.
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Let us revisit the Statistical testing procedure whichLet us revisit the Statistical testing procedure which
requires the following steps:requires the following steps:
data, statements(hypotheses) regarding the
population are formulated,
A sample, randomly selected from thepopulation,containing information
pertinent to the hypothesis, is drawn,
An acceptable significance level (alpha) is
established,
A test statistic and its observed significance
level (p value) are computed from the
sample
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SHAPE OF THE DISTRIBUTION; NORMALITY
An important aspect of the description of a
variable is the shape of its distribution,
which tells you the frequency of values from
different ranges of the variable.
Typically a researcher is interested in howwell the distribution can be approximated by
the normal distribution (see also the
Elementary Concepts section). Simple descriptive statistics ( Histogram,can
provide some information relevant to this
issue.
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TESTS OF MEANS:TESTS OF MEANS: tt TESTS:TESTS:
Although there are a variety ofparametric tests, this module concentrates on
three commonly used tests of means.
All three of these procedures assume that
the data is measured at a interval scale or
quantitative data in random samples selected
from a Normally-distributed Population.
Each of these tests employs a statistic whichfollows a t-distribution.
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One Samples t test:
The first procedure is the one-sample t-
test.
In this case the objective is to test whether the
population mean is equal to a specific value.
The null hypothesis, which is a statementof the status quo, states that the population
mean is equal to the hypothesized value.
Another way to state this null hypothesis is to
say that we are testing whether the difference
between the population mean and the
hypothesized value is equal to zero.
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H0 : = 0 or H0 : - 0 = 0
The two-sided alternate hypothesis states that this
mean difference is not zero. This procedure istermed a one-sample test since only one sample
from the population is involved.
H1 : 0 or H1 : - 0 0
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` Another procedure is the dependent-sample t-test.
` In this instance the purpose is to test whether the
population means of two different variables are
equal.` Here the null hypothesis states that the two
population means are equal, which is the same
as stating that the mean of the difference
between the two variables is equal to zero.
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Continue:Continue:
y The two-sided alternate hypothesisspecifies that this difference isunequal to zero.
y This procedure is sonamed becauseboth variables being tested comefrom the same observations and aretherefore statistically dependent.
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Assumption: differences constitute a randomAssumption: differences constitute a random
sample from a normal distributionsample from a normal distribution
Ho: Qo (=Q1 - Q2) = H
Test statistics is under Ho = 1~ no
tntS
D
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Why Paired t-test?
In testing for difference between twopopulation means, the use of independentsamples can have a major drawback.
Even if a real difference does exist, thevariability among the responses within eachsample can be large enough to make it.
The random variation within the samples
will mask the real difference between thepopulations from which they come.
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CONTINUE:
One way to control this variability external tothe issue in question is
to use a pair of responses from each subject
and then work with the differences within the
pairs.
The aim is to remove as far as possible the
subject-to-subject variation from the analysis,
and thus
to home in on any real difference between
the populations.
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` The final procedure is the independent-
samples t-test.
` The goal here is to test whether two population
means are equal.` For this test, two samples are drawn (or one
sample is divided into two mutually exclusive
groups ), and the test is performed using a
variable common to both groups.
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The null hypothesis states that the population
means of the two groups are equal.
The two-sided alternate hypothesis states that
the means are unequal.
The two samples, or sample groups, are
independent of each other because no
observation is present in both groups, hence the
name independent- samples t-test.
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Continue:
Test statistic: z =
(b) W12 , W2
2 unknown much the more
usual situation when W12
, W22
known.
2
2
2
1
2
1
21
nn
xx
WW
H
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Further, the Central Limit Theorem
justifies the use of a normal
approximation for the distribution of the
test statistic in sampling from any
reasonable populations, so therequirement that we are sampling from
normal distributions is not necessary.
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Small Samples:Small Samples:
y Under the assumptions W12=W2
2 (=W2 say), this
common variance is estimated by Sp2, and is
distributed as t with n1 + n2 2 degrees of
freedom under H0. So
y
y t =
21
21
11
nns
xx
p
H
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Continue:
y Remember that Sp2 =
y Problem
In an experiment using identical twin babies,one baby in each of 10 sets was fed food stuff, A. Whilethe other twin was fed food stuff B. the gains in weightin kg are given below
A 24 28 31 32 25 27 37 31 26 29B 19 24 32 28 28 29 31 33 2927
Identify which type of t-test has to apply.
2n
1)s-(n+s1)-(n
21
2
2
2
1
n
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Analysis of variance is one of the
powerful tool in testing of hypothesis.
If one is interested in testing equality of
several means at a time, the technique of
Analysis of Variance is used.
There are many situations where data
are classified into one, two or more ways.
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One way classification:
Let there be three treatments A,B,C each
replicated n1,n2 and n3 times.
A B C
X11 X21 X31
X12 X22 X32
. . .
. . .
. . .
X1n1 X2n2 X3n3
TotalT1 T2 T3
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ANOVA continue
y Ho :
y H1 : Atleast one treatment is different from others.
y Grand total (G) = T1 + T2 + T3 =
Correction Factor CF =
cBAQQQ !!
ji
ijx
nsobservatioofnumberis,nn
G
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Sourcesof
variation
Sum ofsquares
d.f Mean sum ofsquares
F- ratio
Treatment
ErrorTotal
SST
SSETSS
(K-1)
(N-K)(N-1)
MST = SST/K-1
MSE = SSE/N-K
F =MST/MSE
Inference : If Fcal < Ftab Accept Ho otherwisereject.
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ILLUSTRATION: YIELD IN GRAMS
Catalyst 1 Catalyst 2 Catalyst 3
100 76 108
96 80 100
92 75 96
96 84 98
92 82 100
Total 476 397 502
Based on the above data, can it be inferred that the effect of three
catalysts differ significantly with respect to yield of the chemical
product?
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y Ho: Average yield of the chemical product
under the three catalysts is same.
y H1: Average yield of the chemical product foratleast one catalyst is different from the yield
corresponding to the remaining catalysts.
y Following the general practical steps, we findthe various sums of squares, noting that here
number of levels 3 (which is equal to k ) and
n1
= 5 (number of batches tested for each
level).
y
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1. G = (100 + 96+ .. + 92) +
2. (76+80+.82) +
(108+100++100)
= 1375
2. = (1002 + 962+ .. + 922) +
(762+802+.822) +
(1082+1002++1002)= 127425
3. N = 7 n1= n
1+ n2 + n3 = 5+ 5+ 5 = 15
ji
ijx
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Anova calculations
4. CF = G2 / N = 1375-2 / 15 = 126041.67
5. TSS = 77 - CF = 127425 126041.67
= 1383.33
6. SS = 7 ( 7 / n) - CF= 4762 + 3972++ 5022) / 5 126041.67
= 1196.1
7. SS4
= TSS SS4
= 1383.33 1196.13
= 187.20
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Inference:Inference:
y Here calculated value of F is greater than the
tabulated value of F and hence the null
hypothesis is rejected. Therefore, we may
conclude that the three catalysts differ
significantly.
P blP bl
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Problem :Problem :
y The gain in weight of two random samples of
rats fed on two different diets A and B are given
below. Examine whether the difference in mean
increases in weight is significant.
D
iet A: 13 14 10 11 12 16 10 8Diet B: 7 10 12 8 10 11 9 10
Identifytheappropriatet-test.
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Problem :
Tow horses A and B were tested according to thetime (in seconds) to run a particular track withthe following results.
Horse A: 28 30 32 33 33 29 34Horse B: 29 30 30 24 27 29
Test whether the two horses have the samerunning capacity. Find the appropriate t-test.
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` A test was given to five students taken at randomfrom the fifth class of three schools of a townindividuals scores are:
`School 1 : 9 7 6 5 8
`School 2 : 7 4 5 4 5
`
School 3 : 6 5 6 7 6
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Carryout analysis of variance and stateCarryout analysis of variance and state
your conclusion.your conclusion.
Problem : Yields of three varieties of wheat
given below
Variety yield
y I 10 9 8
y II 7 7 6
y III 8 5 4
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` Parametric tests are more powerful when the
following conditions are satisfied:
` Normality
` Measurement in interval scale
` Sample size is reasonable
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Note:
If any of these conditions are
violated the parametric tests are not
appropriate. In such cases one should go for the
Nonparametric tests.
That is the focal theme of the nextmodule.