Ms Sridhar Rm

Research Methodology PART 9

Testing of hypotheses M S Sridhar Head, Library & Documentation ISRO Satellite Centre Bangalore 560017 E-mail: [email protected] & [email protected]

M S Sridhar, ISRO

Testing of Hypotheses

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Testing of hypothesis 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

Introduction • Meaning & importance of hypothesis • Hypothesis Vs theory • Problems in formulating hypothesis • Sources / Origins of hypothesis • Types of hypothesis • Characteristics of good & usable hypothesis • Statistical hypotheses & Hypothesis testing Basic concepts – Null & alternative hypotheses – Type I & type II errors – One-tailed & two tailed tests – Significance tests: significance, level and power 2 Testing of Hypotheses

Synopsis Introduction to Research & Research methodology Selection and formulation of research problem Research design and plan Experimental designs Sampling and sampling strategy or plan Measurement and scaling techniques Data collection methods and techniques Statistical techniques for processing & analysis of data Testing of hypotheses Analysis, interpretation and drawing inferences Report writing

M S Sridhar, ISRO

Testing of Hypothesis: Synopsis 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

Synopsis Introduction to Research & Research methodology Selection and formulation of research problem Research design and plan Experimental designs Sampling and sampling strategy or plan Measurement and scaling techniques Data collection methods and techniques Statistical techniques for processing & analysis of data Testing of hypotheses Analysis, interpretation and drawing inferences Report writing

M S Sridhar, ISRO

…contd.

Parametric or Standard Tests

1. 2. 3. 4. 5. 6.

7. 8.

Z- Test t- Test F- Test Chi-square Test Testing of means Differences between means Comparing two related means Testing of proportions Differences between proportions Comparing a variance to some hypothesised population variance Equality of variances of two normal populations Testing of correlation coefficients (simple, partial & multiple)

Testing of Hypotheses

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Testing of Hypothesis: Synopsis Synopsis 1. Introduction to Research & Research methodology 2. Selection and formulation of research problem 3. Research design and plan 4. Experimental designs 5. Sampling and sampling strategy or plan 6. Measurement and scaling techniques 7. Data collection methods and techniques 8. Statistical techniques for processing & analysis of data 9. Testing of hypotheses 10. Analysis, interpretation and drawing inferences 11. Report writing M S Sridhar, ISRO

…contd.

Non-parametric or Distribution Free Tests One sample tests 1. Kolmogorov-smirnov one sample test 2. Runs test for randomness 3. One sample sign test Two sample tests 4. The sign test 5. Fisher-Irwin test 6. Mc Nemer test 7. The median test 8. Chi-square test 9. Wilcoxon-Mann-Whitney U-test 10. Wilcoxon matched pair (signed rank) test More than two (k) sample tests 11 The median test 12 The Kruskal-Wallis test or H-test 13 Kendall’s coefficient of concordance

Testing of Hypotheses

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Hypothesis: Introduction • •

Meaning A Hunch, an assumption, a proposition, a guess which is tentative, provisional & explains the situation under observation but yet to be proved or disproved Selection & formulation of problem lead to clear objectives and working hypothesis (positive hypothesis) – what a researcher is looking for – A penetrating hunch with provisional explanation becomes the basis for a systematic investigation. Tentative explanations are untested but sound, plausible and reasonable Good hypothesis appear to have the requirements plausibility, consistency & chance of empirical testing Example: 1. Decline in reading habits of adults 2. Factors affecting reading habits of adults (televideo) 5 contd..

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Testing of Hypotheses

Hypothesis: Introduction

…Contd.

Importance It is very difficult, laborious & time-consuming to make adequate discriminations in the complex interplay of facts without working hypothesis 1. Gives definite point to the inquiry 2. Helps establishing direction Directs our search for order among facts & provide considerable advantage in inquiry with suggested explanation or solution

3. Prevents blind search & indiscriminate gathering of data While searching for significant & relevant facts to explain the problem, shows the essential relationship that exists between various elements within the complexity

4. Helps to delimit the field of inquiry In his search, researcher may fall back on previous experience of his own or that of others & single out those factors that are known to have explained similar situations in the past as observed in the descriptive literature or speculative philosophy

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Hypothesis Vs Theory Note: Hypothesis & theory are closely related & interdependent but they are different

•

Assumptions based on Probabilities, shrewd guesses & profound hunches

•

When facts are assembled, ordered & seen as a relationship, they constitute a theory. (Theory is not speculation but built on facts that are ordered to give meaning)

•

Facts in theory are logically analysed & relationship other than those stated in theory can be deduced (to verify their correctness) A hypo looks forward Formulation of a deduction from theory constitutes a hypo

•

Every worthwhile theory permits the formulation of additional hypo. Verified hypo becomes part of a future theoretical construction

• •

M S Sridhar, ISRO

Testing of Hypotheses

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Hypothesis

…Contd.

Problems in Formulating Hypothesis

Sources / Origins of Hypothesis

1. Absence of a clear theoretical framework 2. Lack of ability to logically utilise the theoretical framework 3. Failure to be acquainted with average research techniques Normally students tend to suggest (i) Area studies (ii) replication of pervious studies (iii) study of empirical regularities, ie., suggest type of data to be gathered (e.g.: bibliometric study)

1. General culture in the profession 2. Common sense, wisdom and practice 3. Keen observation, disciplined imagination & creative thinking 4. Literature & some formulated theoretical frame works 5. Analogies from other disciplines 6. Personal & idiosyncratic experience the way individual reacts to culture, discipline & analogies. ‘Radicals’ & ‘Marginal men’ can spring more hypotheses

Caution Researcher should not start out to prove or disprove hypothesis He should not try to defend his hypothesis when data support a conclusion exactly contrary to what he originally believed M S Sridhar, ISRO

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Types of Hypothesis by Level of Abstraction 1. Those stating the existence of empirical uniformities Known to everybody and there may not be any hypothesis to test at all ! Simple level hypothesis that seeks empirical generalisation play important role in the growth of discipline e.g.: Scientists are more library oriented than engineers

2. Those concerned with complex ideal types Aim at testing the existence of logically derived relationships between empirical uniformities e.g.: Bradford law & ‘Half life time’ of journals in a discipline

3. Those concerned with the relation of analytic variables More abstract than ideal types Most sophisticated and the most flexible mode of formulation of hypothesis Allow better measurement of the relation between the variables eg.: Use of information by a user may show empirical regularities with many factors like (i) user’s initiative/ drive, motivation, etc (ii) accessibility, ease of use and perceived utility of source of information (iii) the need M S Sridhar, ISRO

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Characteristics of Good & Usable Hypothesis 1. Precise and simple but not obvious 2. Conceptually clear – Clear operational definition of concepts 3. Should have empirical referents Example: ‘Librarians are useless creatures’ – wrong, no referent 4. Specific & limited in scope (narrower hypothesis is more testable) i.e., Practicable - capable of being tested & amenable to test within a reasonable time Significant - explain the facts that give rise to the need 5. Consistent with most known facts 6. Should state relationship between variables (if it is relational hypothesis) 7. Should be related to available techniques except when it is to stimulate development of new techniques 8. Related to a body of theory Should help to refute, qualify or support any existing theories M S Sridhar, ISRO

Testing of Hypotheses

Examples: 1. High performance causes high communication among engineers 2. The effect of informal social relationship is much stronger than the formal organization structure on interpersonal communication 3. The informal communication among peers and colleagues is much more effective and cordial than between superior and subordinate 4. The communication stardom and technological gate keepership are intrinsic to individuals and not contingent to a particular organisation 5. Co-authorship relations result in lasting and highly effective informal communication among co-authors 10

Statistical Hypotheses 9 While attempting to make decisions some necessary assumptions or guesses about the populations or statements about the probability distribution of the populations made are called statistical hypothesis. 9 These assumptions are to be proved or disproved 9 I.o.w, a predictive statement usually put in the form of a null hypothesis and alternate hypothesis 9 Capable of being tested by scientific methods, that relates an independent variable to some dependent variable

Hypothesis Testing ¾ Researcher bets in advance of his experiment that the results will agree with his theory and cannot be accounted for by the chance variation involved in sampling ¾ Procedures which enable researcher to decide whether to accept or reject hypothesis or whether observed samples differ significantly from expected results M S Sridhar, ISRO

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Hypotheses testing Steps

Procedure for hypothesis testing

1. Plan & conduct experiment so that if the results are not explained by the chance variation, theory is confirmed 2. Collect data 3. Set null hypo i.e.. assume that results are due to chance alone 4. Use a theoretical sampling distribution 5. Obtain probability of sample data as if it is chance variation 6. If probability at 5 is less than some predetermined small percentage (say 1% or 5%) reject the null hypothesis and accept the alternate hypothesis

1. Making a clear formal statement indicating use of one-tailed test or twotailed test 2. Select significance level based on: (i) the magnitude of the difference between sample means (ii) the size of samples (iii) the variability of measurements within samples (iv) whether the hypothesis is directional or non-directional 3. Deciding the appropriate sampling distribution to be used ( normal or tdistribution) 4. Selecting a random sample and computing an appropriate value (draw a sample to furnish empirical data) 5. Calculation of the probability that sample result would diverge as widely as expected in null hypothesis 6. Comparing the probability with given significance level (alpha or alpha by two)

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Flow Diagram for Hypothesis Testing State Ho as well as Ha Specify the level of Significant ( or the α value ) Decide the correct sampling distribution

Sample a random sample(s) and workout an appropriate value for sample data

Flow Diagram for Hypothesis Testing

…Contd.

Calculate the probability that sample result would diverge as widely as it has from expectations, if Ho were true Is this probability equal or smaller than α value in case of one tailed test and α/2 in case of two-tailed test No

Yes Reject Ho

Thereby run the risk of committing Type I error

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Accept Ho

Thereby run some risk of committing Type II error

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Basic Concepts 1. Tests of hypothesis (or tests of significance or rules of decision): Procedure which enable deciding to accept or reject hypothesis or to determine whether observed samples differ significance from expected results Decision rule is a sort of basis according to which hypothesis is accepted

2. Null (Ho) & Alternative (Ha) hypotheses: Ho = While computing two methods assuming that both are equally good Ha = A set of alternative to H0 or rejecting the H0 (what one wishes to prove) Alternative hypothesis

To be read as follows

Ha : µ =/ µ

(The alternative hypothesis is that the population mean is not equal to 100, I.e., it may be more or less than 100)

Ha : µ > µ

(The alternative hypothesis is that the population mean is greater than 100)

Ha : µ < µ

(The alternative hypothesis is that the population mean is less than 100)

M S Sridhar, ISRO

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Basic Concepts

…Contd.

3. Type I and type II errors: ‰ Error is determined in advance as level of significance for a given sample size ‰ If we try to reduce type I error, the probability of committing type II error increases ‰ Both type errors cannot be reduced simultaneously ‰ Decision maker has to strike a balance / trade off examining the costs & penalties of both type errors

M S Sridhar, ISRO

Decision

Decision

Accept H0

Reject H0

H0 (true)

Correct decision

Type I error (α error)

H0 (false)

Type II error (β error)

Correct decision

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4. One-tailed (sided) and Two-tailed (sided) tests

…contd.

Mathematically we can state: Acceptance Region A : Z > - 1.645 Region of acceptance or non-significance Rejection Region R : Z ≤ - 1.645 Critical region or region of significance STANDARD VARIATE SCORES (Z SCORES) --------------------------------------------------------------------------------------------------------α 0.10 0.05 0.01 0.005 0.002 --------------------------------------------------------------------------------------------------------One tailed ± 1.28 ± 1.645 ± 2.33 ± 2.58 ± 2.88 Two tailed ± 1.645 ± 1.96 ± 2.58 ± 2.81 ± 3.08 -------------------------------------------------------------------------------------------------------NOTE: Accepting H0 on the basis of sample information does not mean one constitute the proof that H0 is true. It only means that there is not statistical evidence to reject it. TABLE – Normal Distribution Z Prob. Z Prob. Z Prob. 3.0 .999 0.8 .788 -1.4 .081 2.8 .997 0.6 .726 -1.6 .055 2.6 .995 0.4 .655 -1.8 .036 2.4 .992 0.2 .579 -2.0 .023 2.2 .986 0.0 .500 -2.2 .014 2.0 .977 -.2 .421 -2.4 .008 1.8 .964 -.4 .345 -2.6 .005 1.6 .945 -.6 .274 -2.8 .003 1.4 .919 -.8 .212 -3.0 .001 1.2 .885 -1 .159 1.0 .841 -1.2 .115

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4. One-tailed (sided) and Two-tailed (sided) tests

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Testing of Hypotheses

…contd.

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4. One-Tailed / sided and Two Tailed / Sided Tests:

M S Sridhar, ISRO

Testing of Hypotheses

…Contd.

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4. One-tailed (sided) and Two-tailed (sided) tests

M S Sridhar, ISRO

Testing of Hypotheses

…contd.

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Basic Concepts

… contd.

5. The significance of “significances tests”: ¾ The accuracy of a given sample result alone is not enough ¾ Significance attached to the result is necessary, I.e., Importance, validity, etc. ¾ Statistical significance refers to real difference and not due to chance alone 6. The level of significance (α): ¾ Some percentage (usually 5%) chosen with great care, thought & reason so that how will be rejected when the sampling result (observed evidence) has a probability of < 0.05 of occurring if Ho is true ¾ Researcher is willing to take as much as a 5% risk of rejecting Ho ¾ Significance level is the maximum value of the probability of rejecting Ho when it is true ¾ It is usually determined in advance, I.e., the probability of type I error (α) is assigned in advance and hence nothing can be done about it contd. M S Sridhar, ISRO

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Basic Concepts

… contd.

7. Power of a test (1- β): ¾ Value (1-β) indicates how well the test is working, i.e., value nearer to 1 means working well (test is rejecting Ho when it is not true) and value nearer to 0 means poorly working (not rejecting Ho when it is not true) ¾ It indicates how well given test will enable us to minimise the probability of type II error (β), i.e., avoid making wrong decisions. Hypothesis testing cannot be foolproof. Sometimes test does not reject a Ho which is false (type II error). We would like β to be as small as possible or (1- β) to be as large as possible. ¾ Power function (H) : plotting values of 1-β for each possible values of the population parameter for which Ho is not true we get ‘Power curve’ associated with test. The function defining this curve is power function ¾ Operating Characteristic Function (L) L = 1 - H : Shows conditional probability of accepting Ho for all values of population parameters for a given sample size, whether or not the decision happens to be a correct one ¾ OC curve - graphs showing the probabilities of type II error (β) under various hypotheses M S Sridhar, ISRO

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Parametric or Standard Tests ¾ Require measurements equivalent to at least an interval scale

¾ Assume certain properties of parent population like i) observations are from a normal population ii) large random sample iii) population parameters like mean, variance, etc. must hold good ¾ Situations where above assumptions are not possible, non-parametric tests are used; As there is no model, these tests are also called distribution-free tests Four important parametric tests based on the assumption of normality X - μ 1. Z – Test Z = ------------- Finite pop multiplier 1/√ (N – n)/(N – 1) σs / √n Based on the normal probability distribution and even binomial or L. D. Str in case of large samples. For testing mean, variance, two individual samples, median, mode, correlation, coefficients etc., X - μ 2. t – Test t = -----------σs / √n ¾ It is based on t-distribution and only incase of small samples ¾ Used for testing difference between means of two samples, coefficient of simple & partial correlations, etc. M S Sridhar, ISRO

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Parametric or Standard Tests 3. F – Test Based on F – Distribution σ2 s 1 F = ------------σ2 s2 Used in the context of ANOVA and for the testing the significance of multiple correlation coefficients, comparing the variance of two independent samples, etc.

M S Sridhar, ISRO

… contd.

4. χ2 - Test Based on Chi-square distribution χ2

( Oij - Eij )2 = ∑ ---------------Eij

Used for comparing a sample variance to a theoretical population variance χ2

σ 2s = ---------- ( n – 1 ) σ 2p

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SOME IMPORTANT PARAMETRIC TESTS 1.

Testing of means Z and t tests & Sandler’s a statistic,

2.

Differences between means

X1 - X2 Z = --------------------------√σ2p1 / n1 + √ σ2p2 / n2

3.

∑Di2 A = ----------- Where Di = ( Xi - µH0 ) (∑Di ) 2

X1 - X2 t = -----------------------------------------------------------∑(X1i – X1)2 + ∑(X2i – X2)2 -------------------------------------- (√ 1/n1 + √ 1/n2) n1 + n2 + n3

Comparing two related means ¯D – D ∑ Di t = ------------- ; df = n – 1; ¯ D = ----------; σdif σdif /√n n

=√

∑Di2 - √(¯ D)2i n/n - 1

∑Di2 A = -----------(∑Di)2

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Parametric tests

… contd.

4. Testing of proportions Z = ^p - p / √pq/n 5. Differences between proportions ^p1 - ^p2 Z = ------------------------------------------√ [^ p1 ^ q1 /n1 + ^ p2 ^ q2 /n2] 6. Comparing a variance to some hypothesised population variance σs2 χ2 = --------- (n–1) σp2 σs12 7. Equity of variances of two normal populations F = ---------σs22 8. Testing of correlation coefficients For simple correlation coefficients: t = rrx√ n-2 / 1 - rrx2 (df = n-2) For partial correlation coefficients: t = rp √ n-k / 1 - rp2 (df = n-k) n = number of paired observation; k = number of variable involved R2 / k - 1 For multiple correlation coefficients: F = ---------------------(1 - R )/ n - k (k - 1) = df for variance in numerator; (n - k) = df for variance in denominator M S Sridhar, ISRO

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Non-parametric or Distribution-free Tests ‰ When do not depend on any assumptions about properties/ parameters of the parent population ‰ Most non-parametric tests assume only nominal or ordinal data ‰ Non-parametric tests require more observations than parametric tests to achieve the same size of the type I and type II errors ‰ Some important applications are (I) Concerning single value for the given data (ii) Differences among two or more sets of data (iii) Relations between variables (iv) Variation in the given data (v) Randomness of a sample (vi) Association or dependency of categorical data (vii) Comparing theoretical population with actual data in categories

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Non-parametric

or Distribution-free Tests

1. Hypothesis Testing of Means Example : Given below (X1) are annual addition of books to 10 sample branch libraries of public library department of a State. Test the hypothesis that the mean annual addition of books to branch libraries is 578 at 5% significance level

-----------------------------------------------------------------------------------------------------------S. no. X1 (X1 - X) (X1 - X)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 ----------------------------------------------------------------------------------------------------------n = 10 ∑ Xi = 5720 ∑(Xi - X)2=1456 ----------------------------------------------------------------------------------------------------------M S Sridhar, ISRO

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1. Hypothesis Testing of Means

… contd.

Table showing computations for A-Statistic ------------------------------------------------------------------------------------------------------------------S. no. X1 Hypothesised mean Di = (Xi - µµ0) Di2 µµ0 = 578

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

1 578 578 0 0 2 572 578 -6 36 3 570 578 -8 64 4 568 578 - 10 100 5 572 578 - 6 36 6 578 578 0 0 7 570 578 -8 64 8 572 578 -6 36 9 596 578 18 324 10 544 578 - 34 1156 ------------------------------------------------------------------------------------------------------------------n = 10 ∑ Di = - 60 ∑ Di 2 = 1816 -------------------------------------------------------------------------------------------------------------------

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1. Hypothesis Testing of Means H0 = µ = µµ0 = 578 Ha: µ ≠ µµ0 (Two tailed test) ∑ Xi ¯X = --------- = 572 n σs = √1456/9 = 12.7 2 572 – 578 t = --------------- = - 1. 488 12.72/ √ 10 Table value of t for df = n-1 = 9 is 2.62 and hence H0 is accepted

M S Sridhar, ISRO

… contd.

Therefore, ∑Di 2 1816 A= ------------ = ------------ = 0.5044 (-60)2 (∑Di)2 Null hypothesis H0 : µµ0 = 578 Alternative hypothesis Ha : µµ = 578 0 H0 = µ = µµ0 ≠ 578 Table value of A-static for df = (n-1)=9 at α = 0.05 is 0.276 and hence H0 is accepted

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1. Hypothesis Testing of Means Example: A college library which had an average daily issue of 500 books recorded the following daily issue data for 12 days during examination. Does it mean that daily issue has considerably increased ? Test at 5% significance level

H0 : µ = 500 Ha = µ >500 (One tailed test) To find ¯X and σs we make the following computation

-----------------------------------------------------------------------------------------------------------

S.No. Xi (Xi- ¯X) (Xi- ¯X) ---------------------------------------------------------------------------------------------------------1 550 2 4 2 570 22 484 3 490 -58 3364 4 615 67 4489 5 505 -43 1849 6 580 32 1024 7 570 22 484 8 460 -88 7744 9 600 52 2704 10 580 32 1024 11 530 -18 324 12 526 -22 484 ---------------------------------------------------------------N = 12 ∑Xi=6576 ∑(Xi- ¯X)2 = 23978 ----------------------------------------------------------------

M S Sridhar, ISRO

Testing of Hypotheses

… contd.

6576 ¯X = --------- = 548 12 σs = √∑((Xi-X)2/(n-1) = 46.68 548 - 500 T = --------------- = 3.558 46.68/√12 Table value of t at α = 0.5 for df = 12-1 = 11 is 1.786 Hence H0 is rejected

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2. Hypothesis Testing for differences between means Example: Given below are the time taken in minutes by 7 untrained users (X1i) and 5 trained users (X2i) for executing a query on an online database. Is there any evidence at 5% significance level that the training has reduced the time taken for executing a query user 10 and 8 as assumed means for X1i and X2i respectively.

H0 : µ1 = µ2

;

Ha: µ1< µ2

(X1i – A1) 28 ¯X1 = A1 + ------------------- = 10 + ---------- = 14 7 n1 (X2i – A2) 15 8 + --------- = 11 ¯X2 = A2 + ------------------ = n2 5 2 2 ∑ (X1i – A1) - [∑((X1i – A1)] / n1 2 σs1 = ---------------------------------------------------- = 3.667 n1-1

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2. Hypothesis Testing for difference between means … contd. --------------------------------------------------------------------------------------------------------------------Sample one Sample two --------------------------------------------------------------------------------------------------------------------S.No. X1i X1i –A1 X1i –A1 S.No. X2i X1i –A1 X1i –A1 (A1 = 10) (A2= 8) --------------------------------------------------------------------------------------------------------------------1 12 2 4 1 8 0 0 2 15 5 25 2 10 2 4 3 11 1 1 3 14 6 36 4 16 6 36 4 10 2 4 5 14 4 16 5 3 5 25 6 14 4 16 7 16 6 36 ---------------------------------------------------------------------------------------------------------------------n1 =7 ; ∑(Xii-A1) = 28; ∑(Xii-A1)2 = 134 n2 =5; ∑(X2i-A2) = 28 ; ∑(X2i-A2)2 = 134 ----------------------------------------------------------------------------------------------------------------------

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2. Hypothesis Testing for difference between means … contd. ∑(X2-A2)2 –[∑(X2i-A2)]2/n2 σs22 = ---------------------------------------- = 6 n2 - 1 T

¯ X1 - ¯ X2 = -------------------------------------------------------------√(n2 – 1) σs12 +(n2 – 1) σs22 X √1/ n1 + √ 1/ n2 ------------------------------------n1+n2-2

14 - 11 = --------------------------------------------------------- = 2.381 √(7 – 1)(3.667)+(5-1)(6) X √1/7 + √ 1/5 -------------------------------7+5-2 Table value of t for 10 df at α =0.5 for one tail test is 1.812. Hence H0 is rejected

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3. Testing Hypothesis for Comparing Two Related sample Memory capacity of 9 students was tested before & after training. Given below are their scores before & after training. Using paired t-test & A-test check whether the training was effective at 5% significance level H0 : µ1 = µ2 or Ha = D = 0 Ha: µ1< µ2 (One tailed test) df = n – 1 = 9 – 1 = 8 ----------------------------------------------------------------------------------------------------------Student 1 2 3 4 5 6 7 8 9 ----------------------------------------------------------------------------------------------------------Before(Xi) 10 15 9 3 7 12 16 17 4 After (Yi) 12 17 8 5 6 11 18 20 3 ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Student Score Score Difference Difference squared training training Di = Xi – Yi Xi Yi Di2 ------------------------------------------------------------------------------------------------1 10 12 -2 4 2 15 17 -2 4 3 9 8 1 1 4 3 5 -2 4 5 17 6 1 1 6 12 11 1 1 7 16 18 -2 4 8 17 20 -3 9 9 4 3 1 1 ----------------------------------------------------------------------------------------------∑Di2= 29 n=9 ∑Di = -7 35 -----------------------------------------------------------------------------------------------M S Sridhar, ISRO Testing of Hypotheses

3. Testing Hypothesis for Comparing Two Related sample …Contd. ∑Di -7 Mean Difference, ¯D = ---------- = --------- = - 0.778 n 9 = √∑Di2 –(¯D)2n / n – 1 = √29-(-0.778)2 (9)/ 9 – 1 = 1.715 ¯D – 0 - 0.778 -0 t = ---------- = --------------= - 1.361 σdif/ √n 1.715/ √9

S.D of difference

σdif

Tabulated value for 8 df at ∝ = 0.5 Is –1.860 and hence H0 is accepted, ie., training is not effective 29 ∑Di2 A = ------------- = --------- = 0.592 (√∑Di)2 (-7)2 Tabulated value is 0.366 & hence H0 is accepted M S Sridhar, ISRO

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3. Testing Hypothesis for Comparing Two Related sample …Contd. Example 2: Sales data of an item in six shops before and after a special

promotional campaign are given below. Judge the success of the campaign at 5% significance level using paired t-test and A-test -----------------------------------------------------------------------------------------------Shops A B C D E F -----------------------------------------------------------------------------------------------Before the promotional 53 28 31 48 50 42 campaign Xi After the campaign Yi 58 29 30 55 56 45 --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Shops Sales before Sales after Difference Difference campaign(Yi) (Di = Xi - Yi ) Squared (Di 2) campaign (Xi) ----------------------------------------------------------------------------------------------------A 53 58 -5 25 B 28 29 -1 1 C 31 30 1 1 D 48 55 -7 49 E 50 56 -6 3 F 42 45 -3 9 ----------------------------------------------------------------------------------------------------∑ Di2 = 121 n=6 ∑ Di = -21 ----------------------------------------------------------------------------------------------------M S Sridhar, ISRO

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3. Testing Hypothesis for Comparing Two Related sample ∑Di ¯D = ----------- = n σdiff

- 21 ------6

…Contd.

= - 3.5

√ ∑ Di 2 - (¯D)2(n) = ---------------------------- = n-1

√ 121-(-3.5)2 X 6 ---------------------------6-1

=

3.08

4. Testing the Equality of Variances of Two Normal Populations Example: Given below are two random samples (X1i & X2i ) drawn from two normal populations. Test using variance ratio at 5% and 2% level of significance whether the two populations have the same variance H0 = σp12 = σp22 ∑X1i 220 ¯X1 = -------- = ------- = 22 10 n1

M S Sridhar, ISRO

420 ∑X2i ¯X2 = -------- = ------- = 35 n2 12

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4. Testing the Equality of Variances of Two Normal Populations …contd. --------------------------------------------------------------------------------------------------Sample 1 Sample 2 X1i (X1i - ¯X1) (X1i - ¯X1)2 X2i (S2i - ¯X2) (X2i - ¯X2)2 --------------------------------------------------------------------------------------------------20 -2 4 27 -8 64 16 -6 36 33 -2 4 26 4 16 42 7 49 27 5 25 35 0 0 23 1 1 32 -3 9 22 0 0 34 -1 1 18 -4 16 38 3 9 24 2 4 28 -7 49 25 3 9 41 6 36 19 -3 9 43 8 64 30 -5 25 37 2 4 ----------------------------------------------------------------------------------------------------∑ X1i =220 ∑ (X1i - ¯X1)2 = 120 ∑ X2i=420 ∑(X2i- ¯X2)2 = 314 ----------------------------------------------------------------------------------------------------n1 = 10 n2 = 12 ------------------------------------------------------------------------------------------------------

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4. Testing the Equality of Variances of Two Normal Populations …Contd. ∑ (Xii- ¯X1)2 σ2s1 = -----------------------n1 - 1 ∑ (X2i - ¯X2)2 σ2s2 = -----------------------n2 - 1

=

120 --------------10 - 1

=

314 ------------12 - 1

28.55 σs22 F = -------- = ----------- = 2.14 σs12 13.33 df:

V1 = n2 - 1 = 12 - 1 = 11 V2 = n1 - 1 = 10 - 1 = 9

= 13.33

= 28.55

(Since σs22 > σs12 )

(Since V1 > V2)

Table values of F for V1 = 11 & V2 = 9 at 5% and 1% significance levels respectively are 3.11 and 5.20. Hence accept H0 at both levels of significance ie., samples have been drawn from two populations having the same variances. M S Sridhar, ISRO

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5. Chi-Square Test Has wide applications in research: 1. To test the homogeneity or the significance of population variance (I.e., test for comparing variances) (Example in the next slide) σs2 H0 = σs2 = σp2 χ2 = -------- (n-1) σp2

2. To test independence or significance of association between attributes 3. To test the goodness of fit (Oij - Eij)2 χ2 = ∑ -------------Eij (2 and 3 already discussed with examples under statistical techniques and two sample tests as a non-parametric tests)

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5. Chi-Square Test Example: Weight of 10 students is as follows: 38 40 45 53 47 43 55 48 52 & 49 (in kilograms). 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% and 1% level of significance ¯X =

Σ Xc ------ = n

470 -------- = 47 kgs 10

(Xi - ¯X)2 280 σs2 = ∑ ------------ = ----------- = 31.11 n-1 100-1

[H0: σp2 = σs2]

31.11 σ s2 2 χ = -------- (n-1) = --------- (10-1) = 13.999 20 σ p2 Table value of χ2 for df = 9 at α = 0.01 is 21.67 and at α = 0.05 is 16.92 As both values are greater than calculated value H0 is accepted. In other words, sample can be said to have been taken from a population with variance 20 kgs M S Sridhar, ISRO

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Non-parametric or distribution-free tests Two kinds of assertions in statistical tests: 1. Assertion directly related to the purpose of investigation, i.e., hypothesis to be tested 2. Assertion to make a probability statement Set of all assertions is called the model Testing a hypothesis without a model is non-parametric test. I.o.w., tests which do not make basic assumptions about and without having the knowledge of the distribution of the population parameters Characteristics 1. Do not depend on any assumptions about properties / parameters of the parent population, I.e., do not suppose any particular distribution & consequential assumptions (Parametric tests like ‘t’ & ‘F’ tests make assumption about homogeneity of the variances) & No such assumptions or less restricting assumptions 2. When measurements are not so accurate, non-parametric tests come very handy 3. Most non-parametric tests assume only nominal or ordinal data I.e., more suitable (than parametric tests) for nominal & ordinal (or rated data) 4. Involves few arithmetic computations 5. Usually less efficient & powerful than parametric tests as they are based on no assumption 6. Greater risk of accepting a false hypothesis and committing type II error; Non-parametric tests require more observations than parametric tests to achieve the same size of type I and type II errors 7. Null hypothesis is somewhat loosely defined & hence rejection of null hypothesis may lead to less precise conclusion than parametric tests 8. It is a trade off between loss in sharpness of estimating intervals and gain in the ability of using less information & to calculate faster M S Sridhar, ISRO

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Non-parametric or distribution-free tests

…Contd.

Some important applications are (I) concerning single value for the given data (II) difference among 2 or more sets of data (III) relations between variables (IV) variation in the given data (V) randomness of a sample (VI) association or dependency of categorical data (VII) comparing theoretical population with actual data in categories Typical situation 1. Data not likely to be normally distributed 2. Nominal data from responses to questionnaire 3. Partially filled questions, i.e., to handle incomplete / missing data. I.o.w., to make necessary adjustments to extract maximum information from average data 4. Reasonably good results from even very small sample but need more observations than parametric tests to achieve the same size of type I and type II errors M S Sridhar, ISRO

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One Sample Tests

They are tests for goodness of fit ¾ Significant difference between the observed & expected frequency ¾ Reasonableness to believe that the sample has been drawn from a specified population ¾ Reasonableness to accept that the sample is a random sample from some known population 1. KOLMOGOROV-SMIRNOV ONE SAMPLE TEST : Degree of agreement between the distribution of observed values and some specified theoretical distribution on ordinal scale assumes that the variable’s distribution is continuous Example: A sample of 160 users are questioned about how do they rate a library & replies recorded

___________________________________________________________________ RATING ___________________________________________________________________ Excellent Very good Good Poor ___________________________________________________________________ No. of Users 30 45 60 25 Sn(x) = Observation Cumulative Distribution 0.1875 0.4688 0.8438 1.0000 F0(x) = Theoretical cumulative Distribution 0.25 0.50 0.75 1.0 ___________________________________________________________________ | F0(x) - Sn(x) | 0.0625 0.0312 0.0938 0 -------------------------------------------------------------------------------------------------------------___________________________________________________________________ M S Sridhar, ISRO Testing of Hypotheses

45

One Sample Tests:

…contd.

1. Kolmogorov- Smirnov One Sample Test …contd. D = MAX |F0(x) - Sn(x)| = 0.0938 Critical value of ‘D’ at 5% significance for large ‘n’ is 1.36 1.36 = --------- = --------- = 0.1075 √n √160 Hence null hypothesis is not rejected H0 = There is no significant preferred rating of library at 5% significance level 2. RUNS TEST FOR RANDOMNESS To test the randomness of a sample based on the order or sequence in which the individual observations occurred Run is a sequence of identical symbols or element s which are followed and preceded by different types of symbols or elements or no symbols on either side The total number of runs is an indication of whether or not the sample is random; Too few or too many runs indicate lack of non randomness M S Sridhar, ISRO

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2. RUNS TEST FOR RANDOMNESS

… contd.

Example: The arrival pattern of 60 users/visitors to a lib counter by their gender is as follows MMWWWMWWMMWWWWMMMWWMMWMMMWWWMM WWMMWWMWWMMWWMWWMMWW H0: The gender wise arrival pattern of users to library is random H1: The gender wise arrival pattern of users to library is not random No. of males arrived, n1 = 23 Total, n = n1 + n2 =50 No. of women arrived, n2 = 27 Test Statistics (n>20) Run, r = 24 √ 2(n1)( n2)(2n1n2 -n1- n2) 2(n1)( n2) Mean, E(r)= -------------- + 1 S. D. ( r ) = --------------------------------√ (n1+ n2)2(n1+n2-1) n 1 + n2 |r – E(r)| /z/ = -------------- If n≤ 20, table can be used to find critical value of S.D.(r) r at α = 0.05 (2)(23)(27) E(r ) = ----------------- + 1 = 25.8 23 + 27

S.D (r ) = 3.48

|24 - 25.81| /z/ = ------------------ = 0.52 3.48 At α = 0.05, critical value of Z=1.96, hence null hypo accepted M S Sridhar, ISRO

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3. ONE SAMPLE SIGN TEST ¾ ¾ ¾ ¾

When sample is taken from a continuous symmetrical population The probability that the sample value < mean and > mean is 1/2 For small sample binomial probability table is used For large sample [n(1-p) >5] normal distribution (Z) is used Example: The number of books issued in a library for 11 days is as follows. Use the sign test at α = 0.05 to test the null hypothesis that the average issue. µµ0 = 284 as against the alternative hypothesis is µh0 < 284. 280, 282, 290, 273, 283, 283, 275, 284, 279 and 281 Replace sample values > µ0 by + and < µ0 by - (=µ0 are ignored) + - Examine whether one +sign observed in 10 trials support H0, the probability of one or fewer successes with n=10 & p= ½ is 10C , p1q9 + 10C p0q10 = 10(½)1(½)9 + (1)(½)0(½)10 1 0 = 0.010 + 0.001 = 0.011 (Value can be had from table also) Since the value is less than α = 0.05, H0 is rejected

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3. ONE SAMPLE SIGN TEST

…contd.

Example 2: Daily arrival of number of issues of Journals in a library is as follows. Use one sample sign test to test the Ho that the average arrival μ=23 against Ha that μ < 23 17, 15, 20, 29, 19, 18, 22, 25, 27, 9, 24, 20, 17, 6, 24, 14, 15, 23, 24, 26, 19, 23, 28, 19, 16, 22, 24, 17, 20, 13, 19, 23, 24, 17, 20, 13, 19, 10, 23, 18, 31, 13, 20, 171, 24, 14 There are 12 +ve and 30 -ve signs X = 12, i.e., Number of +ve or -ve signs whichever is lower n=46, p=½, q= ½

H 0: p = ½

H1: p<½

X - np 12 - (46)(1/2) Z = ------------ = ----------------------- = - 3.2437 ⇒ |Z | = 3.2437 √npq √(46) (½)(½) As tabulated critical value of Z at α=0.05 is 1.645 H0 is rejected M S Sridhar, ISRO

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Two Sample Tests To evaluate the effectiveness or responses if two treatments or methods or stimuli administered either on two independent samples or the same sample studied twice (i.e., before and after) 4. The sign test : Used when measurements of observations are on a qualitative basis H0 p(XA>XB) = p(XA<XB) = ½ Example: Time taken by 8 scientists to carry out online search of a database before and after imparting training are as follows: Before training (in min) 9 7 3 16 12 12 5 6 After training (in min) 5 3 4 11 7 5 5 1 Use sign test to decide the effectiveness of training at α = 0.10 Signs are + + - + + + + There are 6 + and 1 Probability of 6 or more successes in 7 trials with p= 1/2 is 0.063 (see binomial probability distribution table) This is less than α = 0.10 hence Ho is rejected or the training is effective Note: For large sample (i.e., both n.p & n.q are > 5) normal distribution can be used M S Sridhar, ISRO

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5. Fisher-Irwin Test ¾ To test that 2 different treatments are different in terms of the results they produce, i.e., there is no difference among 2 sets of data ¾ Applicable for situations where observations of each item could be classified to one of the 2 mutually exclusive categories Example: No. Passed No. Failed Total New Training (A) 5 1 6 Old Training (B) 3 3 6 ------------------------------------------------------Total 8 4 12 H0: Two programmes are equally good Probability of Group A doing as well or better = Probability (5 passing & 1 failing) + Probability (6 passing & 0 failing) 8C X 4C 8C X 4C 224 28 5 1 6 0 = --------------- + ------------------ = -------- + -------- = 0.24 + 0.03 = 0.27 12C 12C 924 924 6 6 Alternatively, probability of Group B doing as well or worse = Probability (3 passing & 3 failing) + Probability (2 passing & 4 failing) 8C X 4C 8C X 4C 3 3 2 4 = --------------- + ------------------ = 0.27 12C 12C 6 6 Comparing this probability at α = 0.05 we find that H0 is valid M S Sridhar, ISRO

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Note: 1. Elementary Probability theory ‘n’ different object taken ‘r’ at a time. nCr nCr C(n,r), Cn,r, n r nC = nC Ex: 8C5 = 8C3 r n-r nP n(n-1)……..(n-r+1) n! r nC = --------------------------- = ------------- = -------r r(r-1) …… 3.2.1 r!(n-r)! r!

8X7X6 336 8C = ----------------- = --------- = 56 3 3X2X1 6 Note 2. Probability table can also be consulted for given n and y

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6. Mc Nemer Test Useful for testing nominal data of two related samples and before – after measurements of the same subjects with a view to judge the significance for any observed change after treatment AFTER TREATMENT Unfavorable Favorable BEFORE TREATMENT

Favorable Unfavorable

A C

(200) (400)

B D

(300) (100)

H0: There is no change in people’s attitude before and after the treatment H0: P(A) = P(D) , i.e., Probability (Favourable before + Unfavourable after) = Probability (Unfavourable before + Favourable after) (|A – D| - 1)2 (|200-100| - 1)2 (99)2 χ2 = ------------------ = -------------------- = --------- = 32.67 with df = 1 A+D 200 + 100 300 Table value of χ2 for df = 1 at α = 0.05 is 3.84. Hence null hypothesis is rejected

7. The Median Test To test whether two independent samples belong to the same population (or even different population) with same or different sizes but same median Combining both, sample median is determined and a 2x2 table is formed by assorting items above the median and below the median Example: PRECIS and POPSI were adopted for indexing 8 sets of micro documents and given below are their effectiveness. Test the hypothesis that there is no difference between these two scores at α = 0.05 Set No. 1 2 3 4 5 6 7 8 A. PRECIS 49 32 44 48 51 34 30 42 B. POPSI 40 45 50 43 37 47 55 57 Combined series lead to median as 44.5 By grouping the elements above & below median PRECIS POPSI Total Above median 3 (a) 5 (b) 8 (n1) Below median 5 (c) 3 (d) 8 (n2)

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7. The Median Test

…contd.

Since the number of elements (n1 + n2) is small, we can find the probability n1 n2 8 8 a b 3 5 P = ---------------- = ------------ = 0.244 n1 + n2 16 a + b 8 Since this is greater than α = 0.005, H0 is accepted ie., effectiveness of both PRECIS and POPSI has same median. If number of elements is large use Chi-square test with the formula. N [ |ad – bc| - N/2]2 χ2 = ---------------------------------- where N = n1 + n2, df = 1 (a+b)(c+d)(a+c)(b+d)

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8. Chi-Square Test • • •

An important non-parametric test for significance of association as well as for testing hypothesis regarding (i) goodness of fit and (ii) homogeneity or significance of population variance When responses are classified into two mutually exclusive classes like favor - not favor, like - dislike, etc. To find whether differences exist between observed and expected data χ2 = Σ (Oij - Eij)2 / Eij Where, Oij = Observed frequency of the cell in ith row & jth column Eij = Expected frequency of the cell in ith row & jth column Expected frequency of any cell =

total for the row x total for the column of that cell of that cell Grand total

df = (c-1) (r-1) If the calculated value of χ2 is equal or more than that of tabulated for the given df the association is significant M S Sridhar, ISRO

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8. Chi-Square Test

…contd.

Note: 1. χ2 is not a measure of degree of relationship 2. Assumes random observations 3. Items in the sample are independent 4. Constraints are liner, no cell contains less than five as frequency value and over all no. of items must be reasonably large (Yate’s correction can be applied to a 2x2 table if cells frequencies are smaller than five); Use Kolmogorov - Smirnov Test 5. PHI Coefficient, φ = √ χ2 / N , as a non-parametric measure of coefficient of correlation helps to estimate the magnitude of association; 6. Cramer’s V-measure, V = φ2 / √min. (r-1), (c-1) 7. Coefficient of Contingency, C = √ χ2 / χ2 + N , also known as coefficient of mean square contingency, is a non-parametric measure of relationship useful where contingency tables are higher order than 2x2 and combining classes is not possible for Yule’s coefficient of association M S Sridhar, ISRO

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8. Chi-Square Test

… contd.

Example 1: An opinion poll conducted by a library among its users about the duration for which books should be issued showed the following result: 15 days, 21 days & 30 days are respectively preferred by 18, 31 & 20 respondents. Test the data whether deference in observed data at 0.5 significance level Duration of issue of books 15 days 21 days 30 days Observed preference 18 31 20 Expected preference 23 23 23

χ2 =

(Oi - Ei)2

(18 - 23)2

(31 - 23)2

(20 - 23)2

∑ ---------- = ------------- + -------------- + --------------- = 4.261 Eij 23 23 23

df = x-1 = 3 - 1 = 2 Tabulated value of χ2 at 2 df and α = 0.05 is 5.991. Hence the H0 is accepted or there is no significant difference

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8.

Chi-Square Test

…contd.

Example 2: given below is the data regarding reference queries received by a library. Is there a significant association between gender of user and type of query ? LR SR Total query query Male users 17 18 35 Female users 3 12 15 Total 20 30 50 Expectation of E11 = 20X35 / 50 = 14 Expected Frequencies Home work: A library has incurred the L S Total following expenditure for two different years. Is the pattern of M 14 21 35 expenditure changed significantly W 6 9 15 between the years ? (α = 0.5) Total20 30 50 Cells Oij Eij (Oij - Eij) (Oij- Eij )2 / Eij Year Expenditure in lakhs of Rupees 1,1 17 14 3 9/14 = 0.64 1,2 18 21 -3 9/21 = 0.43 Journals Books Others Total 2,1 3 6 -3 9/6 = 1.50 1990-91 55 23 14 92 2,2 12 9 3 9/9 = 1.00 1994-95 81 33 21 135 2 Total (∑) χ = 3.57 Total 136 56 35 227 df =(C-1) (r-1) = (2-1) (2-1) = 1 Table value of χ2 for 1 df at 5 % significance is 3.841. Hence association is not significant M S Sridhar, ISRO

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9. Wilcoxon-Mann-Whitney U-Test ¾ Most powerful non-parametric test to determine whether two independent samples have been drawn from the same population. Used as alternative to ttest both for qualitative and quantitative data ¾ Both the samples are pooled together and elements arranged in ascending order to find U ¾ For large sample, i.e., n2 larger than 20 U is approximated to follow normal distribution n1 - n2 U1 + U2 Mean, µu = ------------- ( or ------------) 2 2 Standard Deviation, σu = √ n1n2 (n1+n2+1)/12 U - µu U - n1n2 / 2 Z = ------------ = ------------------------------σu √ n1n2 (n1+n2+1)/12

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9. Wilcoxon-Mann-Whitney U-Test

…Contd

Example 1: S1: 11, 14, 16 and S2: 8, 10, 12, 15 When neither of the samples (n1 & n2) is greater than 8, U is the number of times that a score in the group with n2 elements precedes a score in the group with n1 elements (if n1
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9. Wilcoxon-Mann-Whitney U-Test

…contd.

Example 2: Sample half-life of books in the areas of physics and chemistry are as follows: Physics : 6.9 11.2 14.0 13.2 9.1 13.9 16.1 9.3 2.4 6.4 18.0 11.5 Chemistry : 15.5 11.1 16.0 15.8 18.2 13.7 18.3 9.0 17.2 17.8 13.0 15.1 Test the hypothesis that there is no difference in the ‘half-life’ of two sets of books Combined rank: 2.4 6.4 6.9 9.0 9.1 9.3 11.1 11.2 11.5 13.0 13.2 13.7 13.9 14.0 15.1 15.5 15.8 16.0 16.1 17.2 17.8 18.0 18.2 18.3 R1 =113, R2 = 187, U1 = 12(12)+12(13) / 2 - 113 = 109, U2 = 35 ie. ,R1 = 1 + 2 + 3 + 5 + 6 + 8 + 9 + 11 + 13 + 14 + 19 + 22 = 113 R2 = 4 + 7 + 10 + 12 + 15 + 16 + 17 + 18 + 20 + 21 + 23 + 24 = 187 For two tailed test at α = 0.05, n1=12, n2=12, U=37 which is less than worked out U. Hence H0 is rejected, I.e., there is a difference between ‘half life’ of books of physics & chemistry samples

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9. Wilcoxon-Mann-Whitney U-Test

…contd.

Example 3 : Following are the scores obtained by two groups of students in an examination. Test whether the two groups belong the same population Group A: 51, 68, 90, 81, 30, 46, 99, 98, 11, 06, 19, 43 Group B: 95, 82, 65, 85, 65, 81, 50, 60, 15, 05, 35, 52 Score : 05 06 11 15 19 30 32 35 43 46 50 51 60 65 65 68 81 81 82 85 90 95 98 99 Group : B A A B A A B B A A B A B B B A A B B B A B A A Rank : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

R1 = 148.5

R2 = 1515.5 n1(n1+1) U = n1n2 + ---------------- - R1 2

=

15(16) 15 X 15 + ----------- - 148.5 = 196.5 2

196.5 - (15)(15) / 2 86 Z = ------------------------------- = ------------- = 3.84 √(15)(15)(31)/12 24.109 At 5% significance, the critical value of Z is 1.96 Hence H0 is rejected, I.e., two groups of students do not belong to the same population M S Sridhar, ISRO

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10. Wilcoxon Matched Pair or Signed Rank Test ¾ Used in the context of two-related samples where we can determine both direction and magnitude of difference. Examples: wife & husband, subjects studied before & after experiment, comparing output of two machines, etc. ¾ As it attaches greater weight to pair which shows a larger difference it is more powerful test than sign test ¾ Null hypothesis (Ho ) is that there is no difference in the two groups with respect to characteristics under study Steps : • Find the differences di between each pair of values • Assign rank to the differences from smallest to largest without regard to sign • Find sum of the ranks of +ve &-ve separately • T is the smaller of the two sums • For small sample use table values of ‘T’ where ‘n’ is the number of pairs (excluding those with di = 0) • For large sample (n>25), Z test is used with T - μT μT = n(n + 1) / 4, σT = √ n(n+1)(2n+1) / 24 and Z = -----------σT M S Sridhar, ISRO

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10. Wilcoxon Matched Pair or Signed Rank Test

…contd.

Example: 16 subjects have evaluated sample from two brands judged on an ordinal scale test the H0 that there is no difference between the perceived quality of the two samples at α = 0.05 P a ir 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

B ra n d A 73 43 47 53 58 47 52 58 38 61 56 56 34 55 65 75

B ra n d B 51 41 43 41 47 32 24 58 43 53 52 57 44 57 40 68

D iffe r e n c e di 22 2 4 12 11 15 28 0 -5 8 4 -1 -1 0 -2 25 7

R ank of I d iI 13 2 .5 4 .5 11 10 12 15 6 8 4 .5 1 9 2 .5 14 7 T o ta l

R a n k w ith s ig n s + 13 2 .5 4 .5 11 10 12 15 1 -6 8 4 .5 -1 -9 -2 .5 14 7 1 0 1 .5 -1 8 .5

T = 1 8 .5 T a b le v a lu e o f T a t α = 0 .5 w h e n n = 1 5 is 2 5 (fo r 2 ta ile d te s t) H e n c e H 0 is r e je c te d . ⇒ P e r c e iv e d q u a lity o f th e tw o s a m p le s a r e n o t s a m e .

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K Sample (i.e., more than two sample) Tests 11. Median Test : Extension of the median test for two samples; Elements are pooled to find combined median and tabulated into 2 X k matrix with respect to combined median n1 n2 n3 For small sample, a X b X c P = -----------------------------n1 + n2 + n a + b + For large sample, Chi-square test is used. 12 The Kruskal-Wallis Test or H Test: Similar to U test; H0, ‘K’ individual random samples come from identical universes; does not require approximation of normal distribution as H follows Chi-square distribution; use Chi-square table. Steps: Samples are pooled together and ranked with lowest score rank as 1 Sum of the ranks R1, R2, R3 etc are worked out n = n1 + n2 +….+ nk 12 H = ----------n(n+1) M S Sridhar, ISRO

∑k

I=1

R2i -------- - 3(n + 1) ni Testing of Hypotheses

df = k-1

66

K Sample Tests (12. The Kruskal-Wallis Test or H Test: …contd. Example: Time taken (in minutes) by three professional staff (A, B& C) in a library to answer four short range reference queries are given below. Test the null hypothesis that there is no significant (α =0.5) difference in time taken by them Time taken 3.8 4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5.0

Rank Staff Data 1 C A 4.8 4.9 4.3 4.0 2 A B 4.2 4.8 5.0 4.6 3 C C 4.4 4.5 3.8 4.1 4 B 5 A n1 = n2 = n3 = 4, k = 3; n = 12 6 C 7 C R1 = 27, R2 = 34, R3 = 17 8 B 9 A 12 272 342 172 10 B H = ------------- X ------ + ------ + ----- - 3(12 + 1) 11 A 12(12+1) 3 3 3 12 B = 543.5/ 13 - 39 = 2.8

Tabulated value of χ2 for df = 2 at α = 0.005 is 5.991. Hence H0 is accepted low. There is no significant difference in time taken by three professor staff to answer short range reference queries M S Sridhar, ISRO

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13. Kendall’s Coefficient of Concordance (W) ¾ Important non-parametric measure of association (W) among k sets of ranking of N objects / individuals ¾ A standard method of ordering objects according to consensus ¾ Value of W varies from 0 (maximum disagreement among k judges) to 1 (perfect agreement among k judges) ¾ W does not take -ve value and is an index of divergence of actual agreement Steps 1. All objects(N) ranked by k judges is put in K X N Matrix 2. For each object, find the sum of ranks (Rj) 3. Determine Rj and S = Σ (Rj - ¯Rj)2 4. Workout S W = ------------------------1/12 k2 (N3 - N) In case of tied rank use average method Σ (t3 - t) If ties are numerous, correction factor T = ---------------12 t = No. of observations in a group tied for a given rank S W = ---------------------------------1/12 k2 (N3 - N) - KΣT 68

M S Sridhar, ISRO

Testing of Hypotheses

13. Kendall’s Coefficient of Concordance (W)

…contd.

Steps continued 5. N ≤ 7 Use standard table for critical values of S for given N, k and significance level N > 7 Use χ2 value for judging W’s significance with df = N-1 for given level of significance χ2 = k (N - 1) W

Note: 1. If W is significant, it only means that judges have essentially applied the same standard in ranking and not that orderings observed are correct i.e., W provides the bests estimate of the ‘true’ rankings by the order of the various sums of ranks, Rj (best estimate is related to the lowest value observed amongst Rj) 2. We can also be determined by averaging the rsj for all possible pairs of ranking (i.e., W is linearly related to rs ) (kw - 1) Average of rs = -------------(k - 1)

(All possible pairs

kC

k(k -1) 2 = -----------) 2

It is very tedious if k is large M S Sridhar, ISRO

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13. Kendall’s Coefficient of Concordance

…contd.

Example: Seven core journals in physics are ranked by four professionals in the following way. Test the significance of agreement in ranking assigned by professors at 5% level. Also point out the best estimate of the true ranking. Journals Professor J1 J2 J3 J4 J5 J6 J7 A 1 3 2 5 7 4 6 B 2 4 1 3 7 5 6 k = 4 C 3 4 1 2 7 6 5 D 1 2 5 4 6 3 7 N = 7 Rj 7 13 9 14 27 18 24 ( Rj - Rj )2 81 9 49 4 121 4 64 ΣRj = 112

⇒

Rj = 16

S = Σ ( Rj - Rj )2 = 332

Since N ≤ ? Comparing the value of S with the critical value in table for k=4 & N = 7 at 5% significance level, i.e., 217.0 H0 is rejected. Hence professors are applying essentially the same standard in ranking the journals, I.e. W is significant. S 332 W = ------------------------ = ------------------------- = 0.741 1/12 k2 (N3 - N) 1/12 (42)(73 - 7) The lowest value observed amongst Rj is 7. As such the best estimate of true rankings is in the case of journal J1 low. In other words, professors on the whole place the journal J1 as first. M S Sridhar, ISRO

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References 1.

Anderson, Jonathan, et. al. Thesis and assignment writing. New Delhi: Wiley, 1970. 2. Best, Joel. Damned lies and statistics. California: University of California Press, 2001. 3. Best, Joel. More damned lies and statistics; how numbers confuse public issues. Berkeley: University of California Press, 2004 4. Body, Harper W Jr. et.al. Marketing research: text and cases. Delhi: All India Traveler Bookseller, 1985. 5. Booth, Wayne C, et. al. The craft of research. 2 ed. Chicago: The University of Chicago Press, 2003. 6. Chandran, J S. Statistics fdor business and economics. New Delhi: Vikas, 1998. 7. Chicago guide to preparing electronic manuscripts: For authors and publishers. Chicago: The University of Chicago Press, 1987. 8. Cohen, Louis and Manion, Lawrence. Research methods in education. London: Routledge, 1980. 9. Goode, William J and Hatt, Paul K. Methods on social research. London; Mc Graw Hill, 1981. 10. Gopal, M.H. An introduction to research procedures in social sciences. Bombay: Asia Publishing House, 1970. 11. Koosis, Donald J. Business statistics. New York: John Wiley,1972. M S Sridhar, ISRO

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References 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.

…Contd.

Kothari, C.R. Research methodology: methods and techniques. 2 ed., New Delhi: Vishwaprakashan, 1990. Miller, Jane E. The Chicago guide to writing about numbers. Chicago: the University of Chicago Press, 2004. Rodger, Leslie W. Statistics for marketing. London: Mc-Graw Hill, 1984. Salvatoe, Dominick. Theory and problems of statistics and econometrics (Schaum’s outline series). New York: McGraw-Hill, 1982. Spiegel, Murray R. Schauim’s outline of theory and problems of statistics in SI units. Singapore: Mc Graw Hill , 1981. Simpson, I. S. How to interpret statistical data: a guide for librarians and information scientists. London: Library Association, 1990. Slater, Margaret ed. Research method in library and information studies. London: Library Association, 1990. Turabian, Kate L. A manual for writers of term papers, theses, and dissertations. 6 ed. Chicago: The University of Chicago, 1996. Young, Pauline V. Scientific social surveys and research. New Delhi: Prentice-Hall of India Ltd., 1984. Walizer, Michael H and Wienir, Paul L. Research methods and analysis: searching for relationships. New York: Harper & Row, 1978. Williams, Joseph M. Style: towards clarity and grace. Chicago: The University of Chicago Press, 1995.

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About the Author Dr. M. S. Sridhar is a post graduate in Mathematics and Business Management and a Doctorate in Library and Information Science. He is in the profession for last 36 years. Since 1978, he is heading the Library and Documentation Division of ISRO Satellite Centre, Bangalore. Earlier he has worked in the libraries of National Aeronautical Laboratory (Bangalore), Indian Institute of Management (Bangalore) and University of Mysore. Dr. Sridhar has published 4 books, 81 research articles, 22 conferences papers, written 19 course materials for BLIS and MLIS, made over 25 seminar presentations and contributed 5 chapters to books. E-mail: [email protected], [email protected], [email protected] ; Phone: 91-80-25084451; Fax: 91-80-25084476.

M S Sridhar, ISRO

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