BUSINESS STATICS

11.3.1 Null Hypothesis and Alternative Hypothesis Hypothesis • In the context of statistical analysis, while comparing any two methods, the NOTES following concepts or assumptions are taken into consideration: o Null Hypothesis: While comparing two different methods in terms of Sclf:fnstmctiomt! Mltcrial 291 their superiority, wherein the assumption is that both the methods are equally good is called null hypothesis. It is also known as statistical hypothesis and is symbolized as Ho. o Alternative Hypothesis: While comparing two different methods, regarding their superiority, wherein, stating a particular method to be good or bad as compared to the other one is called alternate hypothesis. It is symbolized as ~. 11.3.2 Comparison of Null Hypothesis with Alternative Hypothesis • Following are the points of comparison between null hypothesis and alternate hypothesis: o Null hypothesis is always specific while altemate hypothesis gives an approximate value. o The rejection of null hypothesis involves grcal nsk, which is not in the case of alternative hypothesis. Null hypothesis is more frequently used in statistics than alternate hypothesis because it is specific and is not based on probabilities. The hypothesis to be tested is called the Null Hypothesis and is denoted by /:{).This is to be tested against other possible states of nature called alternative hypothesis. The alternative is usually denoted by ~- The null hypothesis implies that there is no difference between the statistic and the population parameter. To test whether there is no difference between the sample mean X and the population Jl, we write the null hypothesis. fl.J: X=~ The alternative hypothesis would be, ~: x ~~ This means X > ~or X < ~- This is called a two-tailed hypothesis. The alternative hypothesis~: X>~~ is right tailed. The alternative hypothesis ~: X < ~ is left tailed. These are one sided or one-tailed alternatives. NOte 1: The alternative hypothesis H1 implies all such values of the parameter, which are not specified by the null hypothesis H0. NOte 2: Testing a statistical hypothesis is a rule, which leads to a decision to accept or reject a hypothesis. A one-tailed test requires rejection of the null hypothesis when the sample statistic is greater than the population value or less than the population value at a certain level of significance. 1. We may want to test if the sample mean X exceeds the population mean p. Then the ·null hypothesis is, fl.J: x > ~

Hypothesis 2. In the other case the null hypothesis could be, NOTES fiJ: X< 1! Each of these two situations leads to a one-tailed test and has to be dealt with in the same manner as the two-tailed test. Here the critical rejection is on one side only, right for X > 11 and left for X < ll· Both the figures 11.1 and 11.2 here show a five per cent level oftest of significance. For example, a minister in a certain government has an average life of 11 months without being involved in a scam. A new party claims to provide ministers with an average life of more than 11 months without scam. We would like to test if, on the average the new ministers last longer than 11 months. We may write the null hypothesis fiJ: X = 11 and alternative l'oypothesis ~: X > 11. Fig. 11.1 &: X > 11 Check Your Progress Fig. 11.2 &: X < 11 I. Define the term hypothesis. 11.4 CRITICAL REGION 2. Wnat do you understand by The Critical Region (CR), or Rejection Region (RR), is a set of values for testing statistic 'type I' and 'type II' errors? for which the null hypothesis is rejected in a hypothesis test. It means, the sample space 3. Explain null and alternative for the test statistic is partitioned into two regions; one region as the critical region will lead us to reject the null hypothesis &, the other not. So, if the observed value of the test hypotheses. statistic is a member of the critical region, we conclude that 'reject &'; if it is not a member of the critical region then we conclude that 'do not reject&'. 292 Self-Instructional Mlterial We shall consider test problems arising out of Type I Error. The level of significance of a test is the maximum probability with which we are willing to take a risk of Type I Error. If we take a 5% significance level (p= 0.05) we are 95% confident (p= 0.95) that a right decision has been made. A 1% significance level (p = 0.01) makes us 99% confident (p = 0.99) about the correctness of the decision. The critical region is the area of the sampling distribution in which the test statistic must fall for the null hypothesis to be rejected. We can say that the critical region corresponds to the range of values of the statistic, which according to the test requires the hypothesis to be rejected.

• Two-tailed and One-tailed Tests: A two-tailed test rejects the null hypothesis Hypothesis if the sample mean is either more or less than the hypothesized value of the NOTES mean of the population. It is considered to be apt when null hypothesis is of some specific value whereas alternate hypothesis is not equal to the value of null hypothesis. In a two-tailed curve there are two rejection regions, also called critical regions. Acceptance and rejection regions in case of a two-tailed Rejection test I region (With 5% significance level) I I Acceptance region (Accept liJ I I :Rejection I. :regwn if the sample mean Xfalls in I this region) I I I I t: I \\~ :J 0.475 of 0.475 of area area I {Both taken together equals 0.95 or 95% of area} Reject 1{, if the sample mean ( X) falls in either of these two regions • Conditions for the Occurrence of One-tailed Test: When the population mean is either lower or higher than some hypothesised value, one-tailed test is considered to be appropriate where the rejection is only on the left tail of the curve. This is known as left-tailed test. Se/t:Jnstmctiona/ Mitcnilf 293

Hypothesis I 1~--------------, Acceptance region (Accept LR_e..:.je_c_ti_o_n_re-=g:..._.ion______.~...- 1-fJ ifthe sample mean falls NOTES in this region) 0.45 of area {Both taken together equals 0.95 or 95% of area} z =-1.645 Reject 1-fJ ifthe sample mean ( X) falls in these regions For example, what will happen if the acceptance region is made larger? a will decrease. It will be more easily possible to accept 1-fJ when HJ is false (Type II error), i.e., it will lower the probability by making a Type I error, but raise that ofp, Type II error. Noll:: a, p are probabilities of making an error; 1 - a, 1- Pare probabilities of making correct decisions. Acceptance region 1-a Critical Critical region region Example 11.1: Can we say a+ p= l? Solution: No. Each is concerned with a different type of error. But both are not independent of each other. 294 Self-Instructional Mlterial

11.5 PENALTY Hypothesis Usually type II error is considered the worse of the two though it is mainly the NOTES circumstances of a case that decide the answer to this question. If type I error means accepting the hypothesis that a guilty person is innocent and if type II error means accepting the hypothesis that an innocent person is guilty then type II error would be dangerous. The penalties and costs associated with an error determine the balance or trade offbetween Type I and Type II errors. Usually type I error is shown as the shaded area, say 5% of a normal curve which is supposed to represent the data. If the sample statistic, say the sample mean, falls in the shaded area the hypothesis is rejected at 5 per cent level of significance. 11.6 STANDARD ERROR The concept of Standard Error (SE) of statistic is used to test the precision of a sample and provides the confidence limits for the corresponding population parameter. The statistic may be the sample arithmetic mean X, the sample proportion p etc. The SE of any such statistic is the standard deviation of the sampling distribution of the statistic. Given below is SE in common use. - JcDr SE(X)= SE(p)=~ SE of difference between two means X 1. X 2 or two proportions Pi. fJJ. sample sizes 11), J1l can be stated as, SE(P.- JJ1)= Where, n is the number of observations X is the sample mean Jl is the population mean s is the sample standard deviation cr is the population standard deviation pis the sample proportion, q = I - p Pis the population proportion, Q= I- P 11.7 DECISION RULE In this concept of hypothesis, you will formulate a rule provided both null hypothesis and alternative hypothesis are given. Formulating a decision means either accepting null hypothesis and rejecting alternate hypothesis or rejecting null hypothesis and accepting alternative hypothesis. It can be easily understood with the help of an example, wherein Self-Instructional 1\\lllterial 295

Hypothesis you test 10 items and formulate a decision on the basis of the rule that states, a null hypothesis will be accepted if out of those 10 items, either none is defective or only 1 is NOTES defective otherwise alternate hypothesis will be accepted. Suppose u is distributed normally with mean 0 and standard deviation 1, Briefly we write u- f\\(0, 1). If the expected value of u is written B... u) then the standardized normal variate is, cS:O.O = ~l) cS:O.O = ~l) .n-1 c~.o ()~.I=~ O='S'. ()~.1-='S'. z=-u--,E;_(u;_) SE(u) The total area under the normal curve is 1 (corresponding to 100 per cent). The area between z= -1.96 and z= + 1.96 is 0.95. This is the region of acceptance with 95 per cent confidence. We write .A-1.96 5 z 51.96) = 0.95 i}Q.l O=s. iW.l- The size of the critical region is 0.05 (shaded area 0.025 on the left and 0.025 on the right). This is a two-tailed test. If 121 remains between the range ± 1.96, we are in the hypothesis acceptance region. The two values - 1.96 and 1.96 are the 5 per cent critical values. Ifl21 > 1.96 we are in the critical region, i.e., the region ofrejection of the hypothesis. 11.7.1 Large Sample Tests Two-Tailed Decision Rule At 5 per cent level of significance, Accept the null hypothesis if 121 51.96 Using Normal Tables . (1) Reject the null hypothesis if 121 > 1.96 (2) At 1 per cent level of significance, (3) Accept ifl21 52.58 Reject ifl21 > 2.58 At 4.55 per cent level of significance, Accept ifl21 52 Reject ifl21 > 2 296 Self-Instructional Mlterial

At 0.27 per cent level of significance, Hypothesis Accept if 121 .5' 3 (4) NOTES Reject ifl21 > 3 For other levels of significance, normal tables should be used. The values of 121 .5' 2 and 121 .5' 3 given in equations (3) and (4) are popularly used. One-Tailed Decision Rule Right Tailed Test & : 11 > 11 o, Accept if z< 2.33 at 1 per cent Accept if z< 1.645 at 5 per cent Left Tailed Test & : 11 < 11 o, Accept if z<- 2.33 at 1 per cent Accept if z< -1.645 at 5 per cent Example 11.2: A dice is rolled 49152 times and the number of times it shows 4, 5 or 6 is 25149. Test the hypothesis that the dice is unbiased. & : P= 1/2 (The probability of getting 4, 5 or 6) H.: p :;C l/2 ~Solution: This is a two-tailed test with SE = 25149 z=l:;;l= ~~2 =5.4>3 -x- ___L_f_ 49152 The dice is biased. We reject the null hypothesis at 0.27 per cent level of confidence. Example 11.3: A coin is tossed 200 times. It shows heads 116 times. Can you say the coin is biased? z=. np-nP = 116-100 = 2.26 > 1.96 r::n;, Solution: vnPQ 200x!x! 22 The difference is significant at 5 per cent level. But since z< 2.58, it is not significant at l per cent level. We can also proceed thus, p-P Z =~P-Q-n- Test for a Sample Proportion p Check Your Progress To test the null hypothesis that the population proposition has a specified value PJ, i.e., 4. Define critical region of lfl:p=PJ hypothesis. For large n, if & is true then, 5. State the concept of the standard error. z= p- A is approximately normal. SE(p) Self-Instructional Mlterial 297 The critical region for z; depending on the desired level of significance, can be found out.

Hypothesis We can also write z= np- n\"n = n~pn-PnQ\"n = _~pP-_Q.ln1nJ_ SE(np) NOTES Example 11.4: A dice was rolled 400 times. It showed 6 coming up 80 times. Can you 298 Self-Instructional Mlterial say the dice is unbiased? Given, p= 80/400 = 115 P= 116 Solution: z= l/5 -l/6 =~1.77 <1.96 -1 x5- x1- J51120 6 6 400 The result suggests that dice is unbiased. NOli:: If the requirements are given it is possible to determine the sample size using the formula n= ~ 12 pqe'- when e is the error permitted. Example 11.5: It is desired to find the percentage of citizens in a town wanting iodized salt. How large a sample will be needed with 95 per cent confidence that the estimate is within 1 per cent ofthe true percentage given? Given, (a) A prior estimate of25 per cent want the scheme (b) No estimate ofthose wanting it Solution: (a) Use the formula n= ~12 JXie'- when e= 1% error= 0.01 D= 1.962 X 0.25 X 0.75 = 7200 (0.01)2 ~~(b) If no estimate ofpis available, taken= = 4 x1(.906.021)2 = 9600 4e 11.8 SUMMARY In this unit, you have learned about the hypothesis, type I and type II errors, null and alternative hypotheses, critical region, penalty, standard error and decision rule. A hypothesis is an approximate assumption that is tested to find its logical or empirical consequence. It should be clear and accurate. Null and alternative hypotheses help to verify the testability of an assumption. It refers to a provisional idea whose merit needs evaluation. In decision-making process, a hypothesis has to be verified and then accepted or rejected as it plays significant role in different areas such as marketing, industry and management. The Critical Region (CR), or Rejection Region (RR), is a set of values for testing statistic for which the null hypothesis is rejected in a hypothesis test. The penalties and costs associated with an error determine the balance or trade off between Type I and Type II errors. It is often referred as a convenient mathematical approach for simplifying cumbersome calculation. Testing a statistical hypothesis on the basis of a sample enables us to decide whether the hypothesis should be accepted or rejected. These concepts help in understanding various situations where hypothesis is applicable. 11.9 ANSWERS TO 'CHECK YOUR PROGRESS' 1. Hypothesis is an assumption that is tested to find its logical or empirical consequence. 2. Type I Error: In this type of error, you may reject a null hypothesis when it is true. It means rejection of a hypothesis, which should have been accepted. It is denoted by a (alpha) and is also known alpha error.

Type II Error: In this type of error, you are supposed to accept a null hypothesis Hypothesis when it is not true. It means accepting a hypothesis, which should have been NOTES rejected. It is denoted by fJ (beta) and is also known as beta error. Sell~Instructional Mlterial 299 3. Null Hypothesis: While comparing two different methods in terms of their superiority, wherein the assumption is that both the methods are equally good is called null hypothesis. It is also known as statistical hypothesis and is symbolized as 1-fJ. Alternate Hypothesis: While comparing two different methods, regarding their superiority, wherein, stating a particular method to be good or bad as compared to the other one is called alternate hypothesis. It is symbolized as~. 4. The Critical Region (CR), or Rejection Region (RR), is a set of values for testing statistic for which the null hypothesis is rejected in a hypothesis test. 5. The concept of Standard Error (SE) of statistic is used to test the precision of a sample and provides the confidence limits for the corresponding population parameter. 11.10 QUESTIONS AND EXERCISES Short-Answer Questions 1. What is a hypothesis? 2. Why is a hypothesis tested? Explain. 3. Explain the characteristics of hypothesis. 4. Explain the importance of statistical decision-making. 5. What are type I and type II errors? 6. Differentiate between null and alternative hypotheses. 7. Describe critical region with the help of an example. 8. What are the conditions for the occurrence of one-tailed test? 9. What is penalty? 10. How is standard error calculated? 11. What is decision rule? 12. Differentiate between one-tailed and two-tailed decision rules. Long-Answer Questions 1. The normal sales of a commodity are likely to be 10,000 units. As a result of a campaign the sales go up. If the standard error is known show that sales have increased decisively. 2. Is a particular significance level important? 3. A dice is rolled 9000 times. Of these 3220 times it shows 3 or 4. Test the hypothesis that the dice is unbiased. 4. A coin is tossed 200 times. It shows heads 116 times. Can you say the coin is biased? 5. A dice was rolled 400 times. It showed a 6 coming up 80 times. Can you say the dice is unbiased? Given, p= 80/400=115, P=l/6 6. A company supplied 500 units of an item. The number of defectives were found to be 42 as against the company's conviction that 6 per cent items couh.: be defective. Examine the tenability of the company claim. 7. A dice is rolled 49152 times. Of these 25149 times it shows 4, 5 and 6. Test the hypothesis that the dice is unbiased.

Hypothesis 8. A company supplied 700 units of an item. The number of defectives were found to be 89 as against the company's conviction that 8 per cent items could be NOTES defective. Examine the tenability of the company claim 9. To test the goodness of a coin, it is tossed 5 times. It is considered a bad coin if more than 4 heads show up. (a) What is the probability of Type I error? (b) If the probability of a head is 0.2, what is the probability of Type IT error? 10. In a sample of 500 people, 280 are tea takers and the rest coffee takers. Are tea and coffee equally popular? 11. A minister in a certain government has an average life of 11 months without being involved in a scam A new party claims to provide ministers with an average life of more than 11 months without scam We would like to test if, on the average the new ministers last longer than 11 months. We may write the null hypothesis H0: p = 11 and alternative hypothesis H1: p > 11. 12. A drug is known to benefit one half of its users. A new drug is claimed to do better. We have Hl :p= 0.5 1ft: p> 0.5 Test the hypothesis. 13. The normal sales of a commodity are likely to be 10,000 units. As a result of a campaign the sales go up. If the standard error is known, calculate that if sales have increased decisively. In an examination, a certain batch of candidates scores, (a) unusually high marks, (b) usually low marks. Is each of these a proper case for a one tailed test? 14. A company supplied 500 units of an item. The number of defectives was found to be 42 as against the company's conviction that 6 per cent items could be defective. Examine the tenability of the company claim 15. It is desired to find the percentage of citizens in a town wanting iodized salt. How large a sample will be needed with 95 per cent confidence that the estimate is within 2 per cent of the true percentage given? Given (a) A prior estimate of 50 per cent want the scheme. (b) No estimate of those wanting it. 11.11 FURTHER READING Kothari, C.R. 1984. Quantitative Teclmiques, 3rd Edition. New Delhi: Vikas Publishing House Pvt. Ltd. Chandan, J.S. 1998. Statistics For Business and Economics. New Delhi: Vikas Publishing House Pvt. Ltd: Chandan, J.S., Jagjit Singh and K.K. Khanna. 1995. Business Statistics, 2nd Edition. New Delhi: Vikas Publishing House Pvt. Ltd. Levin, Richard I. 1991. Statistics for !tfanagenrnt New Jersey: Prentice-Hall. Meyer, Paul L. 1970. Introductory Probability and Statistical Applicatiom. Massachusetts: Addison-Wesley Publishing Co. 300 Self-Instructional Mlterial

UNIT 12 TESTING OF HYPOTHESIS Testing ofHypothesis and AND DECISION-MAKING Decision-MI.ldng Structure NOTES 12.0 Introduction Self-Instructional Ml.terial 301 12.1 Unit Objectives 12.2 A Note on Statistical Decision-Making 12.3 Test for a Sample Mean X 12.4 Test for Equality of Two Proportions 12.5 Large Sample Test for Equality of Two Means ~, ~ 12.6 Small Sample Tests of Significance 12.7 Paired Observations 12.8 To Test the Significance of an Observed Correlation Coefficient r 12.9 z2 Test oflndependence 12.10 12.9.1 Contingency Tables Test for a given Population Variance 12.11 F-Test 12.12 Summary 12.13 Answers to 'Check Your Progress' 12.14 Questions and Exercises 12.15 Further Reading 12.0 INTRODUCTION In this unit, you will learn about the hypothesis testing and decision-making. Statistical decisions have to be made in the presence of uncertainty. The null hypothesis is tested about the population mean which has a specified value J.l. An F-test is any statistical test in which the null hypothesis is true, and the test statistic has an F-distribution. 12.1 UNIT OBJECTIVES After going through this unit, you will be able to: • Explain statistical decision-making • Explain tests of equality of two proportions • Define the concept of standard errors of statistics • Do large sample tests for equality of two means • Do small sample tests of significance • Analyse paired observations • Define observed correlation coefficient • Define F-test 12.2 A NOTE ON STATISTICAL DECISION-MAKING Statistical decisions have to be made in the presence of uncertainty. In testing of hypothesis, the choice is between H0 and H1. In estimation, there are several choices available. The design of experiments requires one to choose between the nature and extent of observations. All this has to be done in the presence of uncertainty. A decision function /J...x) assigns to every possible outcome a unique action. This may result in loss, positive or negative, depending on an unknown parameter w.

Testing ofHypothesis and So the loss function is 4 w, Ilj, which depends on the outcome xis a random variable. Its Decision-Miking expected value is called the Risk Function. NOTES 12.3 TESTS FOR A SAMPLE MEAN X We have to test the null hypothesis that the population mean has a specified value f..L, i.e., H0: X= f..L. For large n, if H> is true then, I _ttlz= X- is approximately nominal. The theoretical region for z depending on the SE(X) desired level of significance can be calculated. For example, a factory produces items, each weighing 5 kg with variance 4. Can a random sample of size 900 with mean weight 4.45 kg be justified as having been taken from this factory? n=900 X=4.45 f..L=5 u=J4=2 z=l!z~l =I;Jnl=14::3~5~ =825 We have z > 3. The null hypothesis is rejected. The sample may not be regarded as originally from the factory at 0.27% level of significance (corresponding to 99.73% acceptance region). 12.4 TEST FOR EQUALITY OF TWO PROPORTIONS If Pt. P2 are proportions of some characteristic of two samples of sizes n1, 11]. drawn from populations with proportions /'t, P2 then we have HJ: P1 = P2 vs lit:P1 ;tt P2 Case (a): If HJ is true then let P1 = P2 = p Where pcan be found from the data: p= ~I?+~~ ~+~ q=l- p pis the mean of the two proportions. Check Your Progress z= I?-~ ,Pisapproximatelynorrnal(O,l) l. What is statistical decision- SE(J?-~) making? We write.z -1'(0, 1) 302 Self-Instructional Miterial The usual rules for rejection or acceptance are applicable here. Case (b): If it is assumed that the proportion under question is not the same in the two populations from which the samples are drawn and that ~. P2 are the true proportions, we write,

SE(~-~)= ( ~'1 + P~qz) Testing ofHypothesis and We can also write the confidence interval for P 1- P2. ~cision-Making For two independent samples of sizes n~o n.z selected from two binomial populations, the NOTES 100 ( 1 - a) % confidence limits for P1 - P2 are, Self-Instructional Material 303 (~-~)±.?u/2 (l~~'1 +~~Ch)1 The 90%, confidence limits would be [with a =0.1, 100 (1- a)= 0.90] (~-~)±1.645 ( ~'1 + P~qz) Example 12.1: Out of 5000 interviewees, 2400 are in favour of a proposal. and out of another set of2000 interviewees 1200 are in favour. Is the difference significant? Solution: 2400 ~ = 5000 = 0.48 P. = 1200 = 0.6 2 2000 l1J = 5000 n.z= 2000 SE= ( 0.48 x 52 + 0·6 x 0.4) = 0.013 (using Case (b)) 5000 2000 z=~~-~~= 0.12 =9.2>3 SE 0.013 The difference is highly significant at 0.27% level. 12.5 LARGE SAMPLE TEST FOR EQUALITY OF TWO MEANS .Xi, Xi Suppose two samples of sizes 11J, n.z are drawn from populations having means /JJ. p2 and standard deviations CT~o CT2 X;To test the equality of means ~, we write, Ho :Jli =J.12 ~=lli*Il2 If we assume li.J is true then, ~ X; ,z= - approximately normally distributed with mean 0, and S.D. = 1. 0\"2 0\"2 _ I +___1_ .q ~ We write z- N(O, 1) As usual, if Iz I> 2 we reject li.J at 4.55% level of significance and so on. Example 12.2: Two groups of sizes 121 and 81 are subjected to tests. Their means are found to be 84 and 81 and standard deviations 10 and 12. Test for the significance of difference between the groups. Solution: ~=84 X;=81

Testing ofHypothesis and z= --r.=:\\=)=-==.x=;== , z= 84 - 81 = 1.86 < 1.96 Dccision-Aflking --,==== NOTES -100+1-44 _0\"12 +_0\"22 304 Self-Instructional Aflterial llj ~ 121 81 The difference is not significant at the 5% level of significance. 12.6 SMALL SAMPLE TESTS OF SIGNIFICANCE The sampling distribution of many statistics for large samples is approximately normal. For small samples with n < 30, the normal distribution, as shown above, can be used only if the sample is from a normal population with known a. If a is not known we can use student's t distribution instead of the normal. We then replace a by sample standard deviation s with some modification as given below. Let x~. x2...., Xn be a random sample of size n drawn from a normal population with mean JL and S.D. a. Then, t= X-J.l si.Jn-1 · Here tfollows the student's tdistribution with n- 1 degrees of freedom !VOtJ:: For small samples of n < 30 the term .Jn-1 in SE = si.Jn-1 corrects the bias resulting from the use of sample standard deviation as an estimator of a Also, i_= n-l or S= S~n-l Sn n Procedure: SmaU Samples Ho:To test the null hypothesis J.1 = J.lo against the alternative hypothesis ~: J.l =t= J.lo Calculate I~ = X-_tL and compare it with the table value with n- 1 degrees of freedom SE(X) (d.f.) at level of significance a per cent If this value> table value, reject IiJ If this value< table value, accept IiJ (Significance level idea same as for large samples) We can also find the 95% (or any other) confidence limits for p. For the two-tailed test (use the same rules as for large samples; substitute tfor z) the 95% confidence limits are, X±fo_si.Jn-1 a=0.025 Rejection Region. At d'lo level for two-tailed test if I tl > fa.J2 reject. For one-tailed test, (right) if t> fa reject (left) if t> -fa reject At 5 per cent level the three cases are, If I t I> ~.o2s reject two-tailed

If t> ~.o5 reject one-tailed right Testing ofHypothesis and Decision-Making If t ~ ~.o5 reject one-tailed left NOTES For proportions, the same procedure is to be followed. Example 12.3: A firm produces tubes of diameter 2 em. A sample of I0 tubes is found to have a diameter of2.01 em and variance 0.004. Is the difference significant? Given ~.05,9= 2.26 Solution: .X-~ 2.01-2 = 0.01 =0.48 t =-s1-.J-;:::n=-=1 '-'0.004/10-1 0.021 It ISince < 2.26 the difference is not significant at 5% level. 12.7 PAIRED OBSERVATIONS ~Test for the Difference of Means Let (x\" y1), (x2, ~.), •••,(xll> yJ be the pairs of values for the same subjects, e.g., sales data before(~ and after an advertisement campaign (.0 Performance of candidates before (~and after training (.0 We have to test the significance of the difference between x, yvalues. For each pair (xi. .n) find 4 = Xi- .Yi Ho: J.li = ).lz, i.e., no difference before and after and H1: p 1 :;& p 2 We find the mean d of d values and use the statistics: t--d- Where, S= J'i)d- d)' - SI.Jfl n-l Example 12.4: Eleven students were given a test and their marks noted. After training, their marks in a second test were noted. Do the marks indicate any benefit from training? Solution: Student 1 2 3 4 s 6 7 8 9 10 11 X 23 20 19 21 18 20 18 17 23 16 19 y 24 19 22 18 20 22 20 20 23 20 17 4 -1 1 -3 3 -2 -2 -2 -3 0 -4 2 Self-Instructional Material 305

Testing ofHypothesis and d= -11=-1 Decision-Mlking 11 NOTES ~=2.24 306 Self-Instructional MJterial df=ll-1=10 t= IJI Jll=l.S 2.24/Jli 2.24 Table value t10. .os = 2.23 The difference is not significant. 12.8 TO TEST THE SIGNIFICANCE OF AN OBSERVED CORRELATION COEFFICIENT R To know the significance of correlation coefficient based on sample data, following formulae are applied: ErI;ryxVFor st.mp1e correIat1.0n coeffitct.ent: t=1_ null hypothesis is either accepted or rejected based on the value of t rpln-Fo• pa<tial correlation coefficient: I= ~ 1-rP null hypothesis is either accepted or rejected based on the value oft ~(k-l)For multiple correlation coefficient: F= (1- )/(n- k) Where R is any multiple coefficients, k being the number of variables involved and n being the number of paired observations. 12.9 X.2 TEST OF INDEPENDENCE In the test of independence, the row and column variables are independent of each other and this is the null hypothesis. The following are properties of the test for independence: • The data are the observed frequencies. • The data is arranged into a contingency table. • The degrees of freedom are the degrees of freedom for the row variable times the degrees of freedom for the column variable. It is not one less than the sample size, it is the product of the two degrees of freedom. • It is always a right tail test. • It has a chi-square distribution. • The expected value is computed by taking the row total times the column total and dividing by the grand total. • The value of the test statistic doesn't change if the order of the rows or columns are switched. • The value of the test statistic doesn't change if the rows and columns are interchanged (transpose of the matrix).

12.9.1 Contingency Tables Testing ofHypothesis and Decision-Making Suppose the frequencies in the data are classified according to attribute A into r classes (rows) and according to attribute Binto cclasses (columns) as follows: NOTES ...Class B. ~ Be Total A ~I 912 ... ~e <A> .-4 lhi lh2 ... lhc (Az) ... ... ... ... ... ... A,. Or! Q.2 ... Q.c (A,.) ...Table N (B.) (~) (Be) The totals of row and column frequencies are (1\\-), (/J;). To test if there is any relation between A, Bwe set up the null hypothesis of independence between A, B. The expected frequency in any cell is calculated by using the formula: £.. = (1\\)(Bj) yN z (Q.--£..)2 Use 2 = :E 1J 1J with d.f. = (r- l)(c- 1) ~j For example, find from the following test if there is any association between sex and education. Given X~.oo.os = 5.99, X~.oo.OI = 9.2 Observed Frequencies Boys School I College University Total 10 15 25 50 Girls 25 10 15 50 Total 35 25 40 100 Expected Frequencies Boys Schnol College University Total Girls Total 17.5 12.5 20 50 Where, 17.5 12.5 20 50 35 25 40 100 35 X 50= 17.5 100 Self..lnstructional Material 307

Testing ofHypothesis and 25x50 =12.5 Decision-Mlking 100 NOTES 40x50 =20 100 df= (2 - I)(3 - 1) = 2 z2 =L(O- E)2 1E=9.9 This is greater than the table value. It is not true that education does not depend on sex, i.e., the two are not independent. 12.10 TEST FOR A GIVEN POPULATION VARIANCE In the test for given population variance, the variance is the square of the standard deviation, whatever you say about a variance can be, for all practical purposes, extended to a population standard deviation. =To test the hypothesis that a sample x1, x2, •• .Xn ofsize n has a specified variance cr2 cr~ Null hypothesis Ho :cl- =c?o or u =uo Hj: cl- > c?o Test statistic z2 = nc$?o2 :L<x-x)2 = c?o iIf is greater than the table value we reject the null hypothesis. 12.11 F-TEST An F-test is any statistical test in which if the null hypothesis is true, the test statistic has an F-distribution. A great variety of hypotheses in applied statistics are tested by F-tests. Among these are given below: • The hypothesis that the means of multiple normally distributed populations, all having the same standard deviation, are equal. This is perhaps the most well- known of hypothesis tested by means of an F-test, and the simplest problem in the analysis ofvariance (ANOVA). • The hypothesis that the standard deviations of two normally distributed populations are equal, and thus that they are of comparable origin. If there are two independent random samples from normal populations we have to test the hypothesis that the population variances crf ,cr~ ar~ the same Then, ca. \"' ( -- )2 estimate of u 2 = £...J Xj 1Xj I q ' ~- 308 Self-Instructional Mlterial

522 = ~::Cx2- X2)2 estimate of o-22 Testing ofHypothesis and Decision-Making , 11z -I NOTES To carry out the test of significance, find See the Ftables for n1- I, 11z- I degrees of freedom (n1- I corresponds to the numerator of Fwith the greater variance, 11z- I for the denominator) at 5% level of significance. If the observed F is less than the table value we assume the two populations have a common variance. At 1% level of significance, the same procedure is to be followed. Example 12.5: One sample of 10 bulbs gives a S.D. of9 hours oflife and another sample of II bulbs gives a S.D. of I0 hours of life. Can you say the variances are different at I% level of significance? (Note: Here S1 = 9, ~ =IO) Solution: Sf =: ___!i_ ~2 = _J_Q_ X (9)2 = 90 .q -1 10-1 Si Si=: _!!l_ = _l_l_ X (10)2 =: jj 0 11z -I 11-1 F=S2i =1-10= 1.22< mble value 5( 90 (Table value F9.w.oo.l = 4.94) We accept the null hypothesis. The population variances may not be different. 12.12 SUMMARY In this unit, you have learned about the hypothesis testing and statistical decision-making. Statistical decisions are made in the presence of uncertainty. In testing of hypothesis, the choice is between H0 and H1. The design of experiment requires choice between the nature and extent of observations. All this has to be done in the presence of uncertainty. A decision function D.,x) is assigned to every possible outcome. You can test the null hypothesis that the population mean has a specified value fl· The test of independence is done when the row and column variables are independent of each other. An F-test is any statistical test in which if the null hypothesis is true, the test statistic has an F-distribution. 12.13 ANSWERS TO 'CHECK YOUR PROGRESS' I. Statistical decisions have to be made during uncertain situations. In testing of Check Your Progress hypothesis, the choice is between 1k and lft. In estimation, there are several 2. Defme properties of the test for independence. choices available. The design of experiments requires one to choose between the nature and extent of observations. 3. What is test for given population variance? 2. The following are properties of the test for independence: 4. What is F-test? • The data are the observed frequencies. Self-Instructional Material 309

'R:sting ofHypothesis and • The data is arranged into a contingency table. Decision-Mlking • The degrees of freedom are the degrees of freedom for the row variable times NOTES the degrees of freedom for the column variable. It is not one less than the sample size, it is the product of the two degrees of freedom. 310 Self-Instructional Mlterial • It is always a right tail test. • It has a chi-square distribution. • The expected value is computed by taking the row total times the column total and dividing by the grand total. • The value of the test statistic doesn't change if the order of the rows or columns are switched. • The value of the test statistic doesn't change if the rows and columns are interchanged (transpose of the matrix). 3. In the test for given population variance, the variance is the square of the standard deviation, whatever you say about a variance can be, for all practical purposes, extended to a population standard deviation. 4. An F-test is any statistical test in which if the null hypothesis is true, the test statistic has an F-distribution. A great variety of hypotheses in applied statistics are tested by F-tests. 12.14 QUESTIONS AND EXERCISES Short-Answer Questions l. Describe the significance of decision making in statistics. 2. How will you test that the population mean has a specified value 1.1.? 3. How will you test the large samples for equality of two means? 4. Describe small sample tests of significance. 5. Explain the properties ofchi-square test of independence. 6. Describe the role ofF-test in statistics. Long-Answer Questions I. The average score of two groups A and B were found to be 25 and 22 with standard deviation 4 and 5.5 respectively. Test for the equality of the two group scores. Given n1 = 11J. = 400. 2. 800 ore pieces from a mine were found to contain an average of74.5 gm of gold. From a nearby mine 1600 pieces had 75 gm gold. Test the equality of the averages from the two mines, each having a standard deviation of2.4. 3. 10 persons randomly selected arc found to have heights 63, 63, 66, 67, 68, 69, 70, 71. 71. 71 inches. Discuss the suggestion that the mean height in the population is 66 inches. 4. 360 persons out of 600 are found to suffer from pollution induced bronchitis in one city. In another, 400 out of 500 are found to suffer from bronchitis. Is there any significant difference in the incidence of bronchitis? 5. In two large populations there are 30% and 25% smokers. Is this difference likely to be hidden in samples of 1200 and 900 from the populations? 6. A factory produces items, each weighing 5 kg with variance 4. Can a random sample of size 900 with mean weight 4.45 kg be justified as having been taken from this factory?

7. A sample of 400 persons has a mean height of 171.38 em Can the sample be Testing ofHypothesis and regarded as having been drawn from a population with mean height of 171.17 frcision-Mlking em and standard deviation 3.3? NOTES 8. Out of 5000 interviewees, 2400 are in favour of a proposal, out of another set of 2000 interviewees 1200 are in favour. Is the difference significant? Self.Instructional Mlterial 311 9. The number of defectives produced by a machine is 20 in a batch of 400. Another machine produced 10 defectives in a batch of300. (a) Is there any difference? (b) If the second machine is the same as the first machine after overhauling, would you say the machine has improved. 10. Of 2000 men aged 25 working in a factory, 400 die before they are 50. Out of 1000 men of age 25 in another factory 175 die before they are 50. Test if there is a difference in the proportion of deaths. 11. 450 men are found to be smokers in a sample of 600 men. In another sample of 900 men, 450 men are smokers. Are the two groups different in the matter of smoking? 12. Two groups of sizes 121 and 81 are subjected to tests. Their means are found to be 84 and 81 and standard deviations 10 and 12. Test for the significance of difference between the groups. 13. The average score of two groups A, B were found to be 25 and 21 with standard deviation 4 and 5.5 respectively. Test for the equality of the two group scores. Given 11! = 112 = 400. 14. 800 ore prices from a mine were found to contain an average of74.5 gm of gold. From a nearby mine 1600 pieces had 75 gm gold. Test the equality of the averages from the two mines, each having a standard deviation of2.4. 15. 5 students attend a class and get respectively 42, 39, 48, 60, 41 marks. 7 other students attending another class get marks 38, 42, 56, 64, 68, 69, 62. Is there any significance difference between the classes? 16. 11 students were given a test and their marks noted. After training, their marks in a second test were noted. Do the marks indicate any benefit from training? Student 2 3 4 5 6 7 8 9 10 11 X 23 20 19 21 18 20 18 17 23 16 19 y 24 19 22 18 20 22 20 20 23 20 17 d; -1 -3 3 -2 -2 -2 -3 0 -4 z 17. There are 100 slips in an urn supposed to has all digits in equal numbers. A check gives the following result. Is the result consistent with the hypothesis that the urn contains all digits in equal number, i.e., 10? Problem Worked Problem Digit 0 (Observed fi:equencies) E (Expected frequencies) (~Ej!E 09 10 1110 11 10 e 1110

71:sting ofHypothesis and 2 12 10 4/10 Decision-MJkiiii! 3 12 10 4/10 NOTES 4 10 10 0 58 10 4/10 67 10 9/10 7 10 10 0 89 10 l/10 9 12 10 4/10 28/10= 2.8 18. In a town with 8000 graduates there are 800 female graduates. Out of 1500 graduate employees, 120 are females. Is there any distinction in employment on sex basis? Given x~.o.os = 3.84 19. Check whether the following samples from normal populations have the same variance, Sample 1: 0 5 11 14 16 22 25 25 . 27 Sample 2: 1 6 7 25 18 3 25 26 28 31 12.15 FURTHER READING Kothari, C.R. 1984. Quantitative Teclmiques, 3rd Edition. New Delhi: Vikas Publishing House Pvt. Ltd. Chandan, J.S. 1998. Statistics For Business and Ec0Il0111ics. New Delhi: Vikas Publishing House Pvt. Ltd Chandan, J.S., Jagjit Singh and K.K. Khanna. 1995. Business Statistics, 2nd Edition. New Delhi: Vikas Publishing House Pvt. Ltd Levin, Richard I. 1991. Statistics for Mmagement New Jersey: Prentice-Hall. Meyer, Paul L. 1970. Introductory Probability and Statistical Applications. Massachusetts: Addison-Wesley Publishing Co. 312 Self-Instructional Mlterial

NJD-Pararrrtric Tests UNIT 13 NON-PARAMETRIC TESTS Structure NOTES 13.0 Introduction Self-Instructional l'vfaterial 313 13.1 Unit Objectives 13.2 Chi-square Test and Goodness of Fit 13.3 Other Non-parametric Tests of Significance 13.4 Solved Problems 13.5 Sunnnary 13.6 Answers To 'Check Your Progress' 13.7 Questions and Exercises 13.8 Further Reading 13.0 INTRODUCTION Parametric tests ofsignificance are based on the premise/assumption that population of data from which a sample (or samples) is (are) drawn is normally distributed. These statistical tests allow considerable latitude and deviations from normality. The central limit theorem, for instance, allows the normality assumption to be bypassed for samples sufficiently large. Ifthe distribution from which a sample is drawn is badly skewed or is otherwise grossly non-normal, however, for smaller samples, these statistical tests will not yield meaningful results. A second assumption upon which these tests rest is that meaningful sample statistics such as the mean and standard deviation can be derived from the sample(s) and used to estimate the corresponding population parameters. Data which are nominal in nature (such as 'increase, decrease, no change') or ordinal (ranked) do not yield such meaningful results when parametric tests are employed. Non-parametric tests (or distribution-free tests) ofsignificance come to our rescue when: (i) the nature ofthe population distribution from which the samples were drawn is not known or is known to be non-normal. (ii) the variables are expressed in the nominal form i.e., classified into categories and represented by frequency counts. (iii) the variables are expressed in ordinal form i.e:, ranked in order and expressed as 1, 2, 3, ... etc. (iv) the samples drawn are very small. (v) the data collected (say, from mailed questionnaires) is partial or incomplete. Thus, non-parametric tests are useful to test hypotheses about data which are non-normal or for which meaningful sample statistics cannot be calculated. Since these tests do not depend on the shape ofthe distribution, they are called distribution-free tests. These tests do not depend upon the population parameters, such as the mean and the variance and hence are called non-parametric tests.

Business Statistic:s-H 13.1 UNIT OBJECTIVES NOTES After going through this unit, you will be able to: • Define non-parametric tests ofsignificance • Explain the Chi-Square Test and Goodness ofFit • Describe other non-parametric tests ofsignificance 13.2 CHI-SQUARE TEST AND GOODNESS OF FIT From the observation (of data), different statistics are constructed to estimate the population parameters. In general (but not always) the sampling distribution ofthese statistics depends on the parameters and form ofthe parent population. The difference between distnbutions have been previouslystudied through constants like mean, standard deviation etc. which are the estimates ofthe parameters, but generally these do not give all the features of these distributions. This caused the.necessity to have some index which can measure the degrees of difference between the actual frequencies of the various groups and can thus compare all necessary features ofthem. An index ofthis type is \"Karl Pearson's J} (chi-square)\" which is used to measure the deviations of observed frequencies in an experiment from the expected frequencies obtained from some hypothetical universe. In the present chapter, we are going to study a distribution called x.2-distribution which enables us to compare a whole set ofsample values with a corresponding set ofhypothetical values. X2 distribution was discovered in 1875 by Helmert and was again discovered independentlyin 1900 by Karl Pearson who applied it as a test ofgoodness offit. If t; and (, denote the observed and corresponding expected frequencies of a class interval (or cell), then chi-square is defined by the relation: t·'X.2 ~ ~ ~t;)'} where the summation extends to the whole set ofclass-intervals. Another form ofx.2 is obtained as follows:- = ~{rf;. +f.-2r.} eo e

= L[1)+N-2N Nm-Paramctric Tests = =( :. L J; L ~ N the total frequency) NOTES 1)-x2= L[ N :t can be proved that ifx1, Xi• ···, xn be in nindependent normal variates each having zero mean and (if +ximit variance, then the sum ofthe squares ofthese nvariates i.e., + ... + ~) is a statistic ;alled X2 with 1rJ degrees of freedom or a stochastic variate having c2 distribution with 1rJ degrees )ffreedom (The no. of independent variates is called the no. of degrees of freedom). Degrees of Freedom and constraints: Let the individuals ofa sample be grouped into 'Ii classes or cells but instead ofthese Jeing independent, let those be subject to 'v independent linear constraints, then the no. Jf degrees of freedom n is defmed by the relation v= (n- c). i.e., ( degrees ) = (no. of) (no. of linear) .,, offreedom groups - constraints NJte Each independent linear constraint reduces the no. of degrees of freedom by one. fhe chisquare distribution For large sample size, the sampling (probability) distribution of x2 can be closely 1pproximated by a continuous curve known as the chi-square distribution. The probability function ofX2 distribution is given by ..14\\c2) -- e x( 2)(v2 -l) ·ex2/2 +-+-+ ..where e= 1 1 1 ad J·Df 1! 2! 2.71828 u = no. ofdegrees of freedom c = a constant depending only on u. fhe chi-square distribution has only one parameter u, the no. of degrees of freedom. fhis is similar to the case ofthe t-distribution. Hence f(x2) is a family ofdistributions, Jne for each value of u (i) X2 distribution is a continuous probability distribution which has the value zero at its lower limit and extends to infinity in the positive direction. Negative value ofx2 is not possible (since the differences between the observed and expected frequencies are always squared). Self-Instructional Mlterial 315

Business Statistics-I/ 2 F(x ) NOTES (ii) The exact shape of distribution depends upon the no. of degrees offreedom v. For different values of v, we shall have different shapes ofdistribution. In general, when vis small, the shape ofthe curve is skewed to the right and as vgets larger, the distribution becomes more and more symmetrical and can be approximated by the normal distribution. ~ (iii) The mean ofthe x2distribution is given by the degrees offreedom i.e, E(x2) = v and variance is twice the degrees of freedom, i.e., l{x2) = 2 v. (iv) As vgets larger, x2 approaches the normal distribution with mean vand standard deviation J2V. In practice, it has been determined that the quantity ~2X2 provides a better approximation to normality than x2 itselffor values of 30 or more. The distribution of ~2x2 has a mean equal to .J(2 v-1) and a standard deviation equal to 1. (v) The sum of independent X2 variates is also a X2 variate. Therefore, if xf is a X2 variate with l! degrees offreedom and x~ is another x2 variate with l2 degrees offreedom, then their sum (xf + x~ ) is also a c2 variate with ( l! + ll) degrees of freedom. This property is known as the additive property xof 2• Chi-square Test The x2 test is one ofthe simplest and most widelyused non-parametric tests in statistical work. It makes no assumptions about the population being sampled. The quantity X2 describes the magnitude ofdiscrepancy between theory and observation, i.e., with the help ofx2test we can know whether agiven discrepancy between theory and observation can be attributed to chance or whether it results from the inadequacyofthe theory to fit the observed facts. If x2 is zero, it means that the observed and expected frequencies completely coincide. The greater the value ofX2, the greater would be the discrepancy c' ~ t\" ~f,)'}between observed and expected frequencies. The formula for computing chi-square is l: where ~ = observed frequency and t; = theoretical or expected frequency. The calculated value ofX2is compared with the table value ofX2for given degrees offreedom at specified level ofsignificance. Ifthe calculated value ofx2 is greater than the table value, the difference between theory and observation is considered to be significant, i.e., it could not have arisen due to fluctuations ofsimple sampling. On the other hand, if the calculated value of x2 is less than the table value, the difference 316 Self-InstructionalMaterial

•etween theory and observation is not considered significant i.e., it could have arisen NJn-Paranrtric Tests lue to fluctuations ofsampling. NOTES The no. of degrees of freedom is described as the no. of observations that are ree to vary after certain restrictions have been imposed on the data. For a uniform Self-Instructio111lll'tilterial 317 listribution we place one restriction on the expected distribution- the total ofsample 1bservations. In a contingency table, the degrees offreedom are calculated in a slightly different nanner. The marginal total or frequencies place the limit on our choice ofselecting cell requencies. The cell frequencies ofall columns but one i.e., (c- 1) and ofall rows but me ie, (r- 1) can be assigned arbitrarily and so the no. ofdegrees offreedom for all cell requencies is (c- 1) (r-1) where 'c stands for the no. of columns and 't for the no. >frows. :onditions for the application of X2-test :;-or the application ofchi-square test, the following conditions must be satisfied: (i) N(i.e., the total frequency) must be sufficiently large, not less than 50 i.e., the sample must contain at least 50 observations. (ii) The smallest call frequency should be 5 or more i.e., there should not be less than 5 observations in any one cell. For less than 5 observations, the value ofX2 shall be over-estimated and result in too many rejections ofthe null hypothesis. In such cases, the respective cells are combined into a single cell having a combined frequency of5 or more. (iii) The sample observations (experimental data) must be independent ofone another. (iv) The constraints imposed on the cell frequencies must be linear. (v) The sample data must be drawn at random from the target population. (vi) The data should be expressed in original units for convenience of comparison (and not in percentage or ratio form). Applications of X2-distribution Some ofits important applications out ofmany in statistics, are as follows: (i) It is used to test the independence ofattributes (ii) It is used to test the goodness of fit (iii) It is used to test whether the estimates ofthe population variance are homogeneous or not. (iv) It is used to test whether the hypothetical value ofthe population variance is s2 or not. Cautions while Using X2-test It is very popularly used in practice. However, it is unfortunate to fmd that the no. of misuses of x2 has become surprisingly large. The test must be used with greater care keeping in mind the assumptions on which it is based. Some sources of error in the application of this test revealed by a survey of all papers published in the Journal of Experiment Psychology are: (i) Small theoretical frequencies.

Business Statistics-/I (ii) Neglect offrequencies ofnon-occurrence NOTES (fu) Indeterminate theoretical frequencies (iv) Incorrect or questionable categorizing (v) Failure to equalize the sum of the observed frequencies and the sum of the theoretical frequencies. (vi) Use ofnon-frequency data. It should also be noted that x2 test is not the only non-parametric test. There are many other non-parametric tests that can be used in business decisions. Use of the Chi-Square l'able To facilitate its many applications, the chi-square distribution has been extensively tabulated. the table of areas found in the appendix gives values of x2 for various probabilities and various degrees offreedom. The value ofa is given in the column headings, the degrees of:freedom vare given in the rows and the body ofthe table gives the x2 values. As depicted in the following figure, the value ofx2 in the appendix table are given for various combinations ofu and (1- a). Some important applications ofx2test are discussed below: (i) Sampling Distribution of the Sample Variance The sampling distribution ofthe sample variance i- is particularly important in problems crwhere one is concerned about the variability in a random sample. Since must always be positive, the distribution ofi- cannot be a normal distribution. The distribution ofi- is a unimodal distribution which is skewed to the right, is a chi-squared distribution. When the parent population is normal, with variance s2 and if random samples ofsize nwith sample variance i- is drawn can be shown to be related as s_2·lr-i -_ -x2c-l- v (n -1) c2 = (n-1); cl- follows x2 distribution with u = (n- 1) 318 Self-InstructionalMaterial

(ii) Confidence Interval for Variance: N:m-Param:tric Jests Confidence m· tervall1'-0r vari·ance a2 1· s based on the samp1m· g d\"1stn\"butl·on of (n -I)i NOTES a2 which follows x2 distribution with v= (n-l).A 100 (1-a) per cent confidence interval for a2 is constructed by first obtaining an interval about (n- 1) ;./a2. Two values of x2 are selected from the table (given in appendix) such that a/2 is to the left of the smaller value and a/2 to the right ofthe larger values. Since the chi-square distribution is not symmetrical, -x~12 does not give the approximate value of the left side of the distribution. The point that does give the correct probability is that of x2 cutting off (1- a/2) of the right tail. Therefore, a 100 (1 - a)% confidence interval for (n- 1) s2/a2 is given by 2 (n-l)s2 2 Xa.f2 < a 2 < Xi-a./2 So1vm. g these m. equa11. t1. es 1l'.0r a-7, we get (n- I )s2 2 (n-I)s2 which is the <2 2 <a X(l-a./2) Xa./2 required 100 ( 1 - a)% confidence interval for a2. (iii) Tests of Hypothesis Concerning Variance: In testing hypothesis about the variance of a normally distributed population, the null a =a; a;hypothesis is H0: 2 where is some specified value ofthe population variance. . We know that x2 = (n -I)s2 a2 where ;. is computed from a random sample of size n. If x2 < X~I-a./2) and x2 > x~;2 , i.e., when the computed value of x2 lies in the rejection region, we reject the null hypothesis, otherwise, we accept the null hypothesis. This is shown in the diagram given below: 2 F(x) 2 Self-Instructional 1\\lhterial 319 <x) (vi) Test of Independence A very useful and important application ofx2-test is to investigate (or test) the compatibility ofobserved and expected frequencies to test the independence oftwo attributes when the data is arranged in two-way tables which are known as contingency tables. On the hypothesis of independence, the theoretical frequencies are calculated and then x2 is

Business Statistics-0 evaluated. The hypothesis is rejected ifx2-test shows that the divergence between the NOTES observation and expectation is significant. 320 Self-InstructionalMaterial Contingency Table Let the given data be classified into pclasses A,_, ~. ···, A,according to attribute xand finretoquqencclyasosefsthe~,cel~l .bel·o·,nBgm9 gactcoobrdoitnhgthteocalattsrsiebsutfe\\ y. Let lij devote the observed (i = 1, 2, ···, p) and B.iU= 1, 2, ···, q). . Let the total ofall the frequencies belonging to the class f\\ be denoted by (I\\) and similarly let (B) denote the total ofall the frequencies belonging to the class B_t Then the given data can be set into a table ofrrows and s columns in the following manner: aasses Bl B2 ... Bi ... Bq-1 Bq Totals AI ljl lj2 ... ljj ... lj, q-1 ljq (1\\) ... ... ~... Iii li2 ... lij ... /i,q-1 /iq (I'\\) ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ~ J;l 10. ... fjj ... .t;, q-1 fiq (I\\) ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... 4-I ~1, I ~1,2 ... ~l,j ... ~l,q-1 fF_I,q (Ap_l) ... ... ~q-1 fPq (Ap) ~ fpt ~ fg Totals (~) (~) ... (B) ... (Bq_l) (Bq) N Calculation of vfor Contingency Table The theoretical frequencies in acontingencytable are calculatedbyimposing the limitations that the row totals, column totals and the grand total remain constant (i.e., unchanged). Therefore, ifthere be prows and qcolumns, then each ofprow totals and qcolumn-- totals gives rise to one constraint and so we have (p + q) constraints. But, the sum ofthe border rows and the sum ofthe border columns must each be equal to the grand total and so one constraint is diminished and so there are only (p + q- 1) constraints. So, the no. of degrees offreedom is given as follows: u = (n- c) = [pq- (p + q- 1)] = [pq- p- q + 1] = [p(q-1)- (q-1)] = {p-1) (q- 1) Calculations of X1 to test the Hypothesis of Independence In order to test whether any relationship between the attributes Aand Bexists, we take the null hypothesis ofindependence ofA and Band on the basis ofthis hypothesis we calculate the expected frequencies in all the cells. Since we are to test the independence only, expected frequencies are calculated by keeping the class totals as constant (i.e., fixed). This means imposition oflinear constraints that the class totals ofthe observed and expected frequencies agree.

Let e.. denote the expected frequency ofthe cell common to A and B. Since the N:Jn·Pararretric 7.Csts lj 1 1 NOTES attributes Aand Bare independent, so in any class the proportions ofmembers belonging to J\\, A.z, ···, ~are the same and equal to the proportions in the total. :. we have p (B) ~)AJ N i=l =These give e.. (Aj) X (Bj) IJ N- - - - ' - - - - = ' - i.e., the expected frequency in any cell is equal to the product of the class totals of the two classes to which this cell belongs divided by the total no. of observations. e .. = (Ai) x (Bj) IJ N Coefficient of Contingency C The coefficient ofcontingency is given by =Ifthe null hypothesis is correct, then lij erso that x? = 0 and C= 0. Also, it is evident that C can never be equafto unity. The maximum value of C depends on the no. of cells and its maximum value is 0.707 for 2 x 2 table. (v) Test of Goodness of Fit Chi-square test is mainly applied to test the compatibility (i.e., capability ofcoexisting) of observed and theoretical frequencies. Ifit is used in this way, it is called a test of'goodness offit'. This expression within inverted commas is used in two ways. In the first place, it may mean the 'fit' ofthe observed data to the hypothetical data. In the second place, it is used to fit a curve to the given data (without any reference to a hypothesis) and the value ofchi-square is found by taking into account the given values and the corresponding values on the fitted curve. For example we can test how will a normal curve fit the given Self-Instructional !odlterial 321

Business Statistics-II data, calculations are similar in both the cases and would be clear from the examples that follow. NOTES We calculate pto find the goodness offit (i.e., how good the fit is). The low value Check Your Progress ofpis generally taken as descnbing a poor fit and very high value ofpgives an excellent fit provided we assume the closeness of the curve to the data (i.e., if we use the 1. Who discovered the Chi- 'goodness of fit' in the second sense as given above). But, while testing a certain Square Test? hypothesis, a very high value ofpdoes not necessarily mean that the hypothesis is true and similar is the case when the value ofpis very low. 2. Name two sources of error in the application of (vi) Test of Homogeneity the Chi-Square test. It is frequently of interest to explore the proposition that several populations are 322 Self..InstructionalMtterial homogeneous w.r.t. some characteristic ofinterest. For example we may be interested in knowing whether some raw material available from several retailers is homogeneous or not. Another way ofstating the problem is to say that we are interested in testing the null hypothesis that several populations are homogeneous w.r.t. the proportion ofsubject falling into several categories or some other criterion ofclassification. Arandom sample is drawn from each ofthe population and the number in each sample falling into each category is determined. The sample data is displayed in a contingencytable. The analytical procedure is same as that discussed for test ofindependence. The main difference is that in tests of independence we are concerned with the problem whether the two attributes are independent or not while in tests ofhomogeneity we are concerned whether the different samples come from the same population. Another difference is that test of independence involves a single sample but test of homogeneity involves two or more samples, one from each population. When there are two populations involved, and when the characteristics ofinterest consist of two categories, the test of homogeneity is the same as testing hypothesis about the difference between two populations proportions which was discussed in the section A 1 of this chapter. 13.3 OTHER NON-PARAMETRIC TESTS OF SIGNIFICANCE Sometimes, the rigid requirements ofparametric tests are not satisfied by the data. This is particularly so in case ofnominal scaled or ordinal scaled data. Under such a situation, non-parametric tests (also known as distribution-free tests) can be used for testing the significance ofthe results obtained from the statistical analysis ofsuch data. The non- parametric tests are useful under the following conditions: (i) When the nature ofthe population distribution from which the samples were drawn is not known or is known to be non-normal. (ii) The variables are expressed in: * Nominal form, i.e., classified into categories and represented by frequency counts, or * Ordinal form i.e., ranked in order and expressed as 1, 2, 3, ... etc.

Most non-parametric tests, thus, are applied to nominal or ordinal scaled data. Nm-Paranrtric 'li:sts Non-parametric tests do not involve detailed mathematical analysis. Typically, NOTES these tests require the use ofmedian rather than mean, sign differences (i.e., plus and minus), or rank order rather than ratios, as determining a mean and use ofratios, etc., Self-Instructional Mlterial 323 require cardinal or ratio-scaled data, whereas determining median, sign difference or rank order requires only ordinal or nominal scaled data. However, it must be understood that when basic assumptions about the parametric tests are valid, the non-parametric tests are less powerful than the parametric tests. Thus, there is a greater risk ofaccepting a false hypothesis and thus committing a Type- II error. Secondly, in non-parametric methods, the null hypothesis is somewhat loosely defmed, thereby when the null hypothesis is rejected, the non-parametric tests yield less precise conclusions compared to the parametric tests. In case ofcomparison between two samples, the specific form ofthe difference between two distributions is not clearly specified. Remarks 1. We should also point out that there are a number of disadvantages associated with non-parametric tests. Primarily, they do not utilize all the information provided by the sample, and thus a non-parametric test will be less efficient than the corresponding parametric- procedure when both methods are applicable. Consequently, to achieve the same power, a non-parametric test will require a larger sample size than will the corresponding parametricJest. 2. It should be pointed out that even under the standard normal theory assumptions, the efficiencies ofthe non-parametric tests are remarkably close to those ofthe corresponding parametric procedures. On the other hand, serious departures from normality will render the non-parametric method much more efficient than the parametric procedure. 3. Parametric tests can be used even when there are slight departures from normality, particularly when the sample size is large. However, when the sample size is small, non-parametric tests are preferred. This is more so when the population distribution is grossly non-normal (badly skewed) or is not known. Various Non-Parametric Tests of Significance (1) Sign Test The sign test is used to test hypothesis on a population median. In the case ofmany of the non-parametric procedures, the mean is replaced by the median as the pertinent location parameter under test. Given a radom variable X the population median 'jl (corresponding to the sample median) is defined such that P(X>ii)::;0.5 and P( X< ii) ::; 0. 5 . In the continuous case, P( X> ii) = P( X< ii) = 0. 5 . Of course, if the distribution is symmetric, the population mean and median are equal. In testing the null hypothesis H0 that ii =ii 0 against an appropriate alternative hypothesis, on the basis of a random sample of size n, we replace each sample value exceeding iio with a plus sign and each sample value less than iio with a minus sign. Ifthe null hypothesis is true and the population is symmetric, the sum ofthe no. ofplus

Business Statistics-II signs should be (approximately) equal to the sum ofthe no. ofminus signs. When one sign appears more frequently than it should, based on chance alone, we reject the NOTES hypothesis that the population median l:l is equal to ilo. In theory, the sign test is applicable only in situations where ilo cannot equal the value of any of the observations. Although there is a zero probability of obtaining a sample observation exactly equal to ilo when the population is continuous, nevertheless, in practice a sample value equal to ilo will often occur from a lack of precision in recording the data. When sample values equal to ilo are observed, they are excluded from the analysis and the sample size is correspondingly reduced. The appropriate test statistic for the sign test is the binomial random variable .X representing the no. of plus signs in our random sample. If the null hypothesis that jl =ilo is true, the probability that a sample value results in either a plus sign or a minus sign is equal to 21. Therefore, to test the null hypothesis that il = jl0 , we are actually i·testing the null hypothesis that the no. of plus signs is a value of a random variable having the binomial distribution with the parameter P = P-values for both one-sided and two-sided alternatives can then be calculated using this binomial distribution. For example, in testing H0 : il = ilo, against~: il <jj. 0 , we shall reject H0 infavour of~ only ifthe proportion ofplus signs is sufficiently less than.!_, that is, when the value of xofour random variable is small. Hence, ifthe computld P-value, P = P (X;S; x when p =i) is less than or equal to some preselected significance level a, we reject ~in favour of~- For example, when n= 15 and x= 3, we find from table C.8 (iv) inAppendix-C that P(P= X;S;3whenp=i) so that the null hypothesis jl = ilo can certainly be rejected at the 0.05 level of significance but not at the O.Ollevel. To test the hypothesis H0 : il = ilo ~: il > ilo we reject H0 in favour of ~ only if the proportion of plus signs is sufficiently igreater than i.e., when x is large. Hence, if the computed p-value, i.e., -1:P= X~ xwhen p= ~) is less thana, we reject H0 in favour of~- Finally, to test the hypothesis 324 Self-Instructional Mlterial

~: fl *flo, NJn-Param::tric 'Jests we reject H0 in favour of~ when the proportiu.! ofplus signs is significantly less NOTES i ·than or greater than This, of course, is equivalent to x being sufficiently small or Self-Instructional Material 325 x<;sufficiently large. Therefore, if and the computed P-value, a, x>;P= 2P ( x:::;; xwhen p=~) is less than or equal to or if and the computed i)P-value, P= 2P (X;:: x when P = is less than or equal to a, we reject H0 in favour of~. Whenever n > 10, binomial probabilities with P =~ can be approximated from the normal curve, since np = nq > 5. Suppose, for example, that we wish to test the hypothesis H0 : fl = flo Ha··-1.1 <-llo at a = 0.05 level of significance for a random sample of size n = 20 that yields x= 6 plus signs. Using the normal-curve approximation with ji = np = 20 x 0.05 = 10 and cr = ~npq=·h0x0.5x0.5=2.236, we find that z= (6.5-10)=-1.57 2.236 Therefore, P = P(X:::;; 6)- P(Z<- 1.57) = 0.0582, which leads to the non-rejection ofthe null hypothesis. (2) Signed-Rank Test As is evident, the sign test utilizes only the plus and minus signs of the differences between the observations and llo in the one-sample case, or the plus and minus signs ofthe differences between the pairs of observations in the paired-sample case, but it does not take into account the magnitudes of these differences. A test utilizing both direction and magnitude, proposed in 1945 by Frank Wilcoxon, is now commonly referred to as the Wilcoxon Signed-rank test. The analyst can extract more information from the data in a non-parametric fashion ifit is reasonable to invoke an additional restriction on the distribution from which the data were taken. The Wilcoxon signed-rank test applies in the case of a symmetric continuous distribution. Under this condition, we can test the null hypothesis 1.1 = llo· We first subtract llo from each sample value, discarding all differences equal to zero. The remaining differences are then ranked without regard to sign. A rank of 1 is assigned to the smallest absolute difference (i.e., without considering sign), a rank of2 to the next smallest and so on. When the absolute value of two or more differences is the same, assign to each the average ofthe ranks that would have been assigned ifthe differences were distinguishable. For example ifthe 4th and 5th differences are equal in absolute value, each would be assigned a rank of4.5. If the hypothesis 1.1 = llo is true, the total

Business Statistics-II ofthe ranks corresponding to the positive differences should nearly equal the total of NOTES the ranks corresponding to the negative differences. Let us represent these totals by 326 Self-InstructionalMlterial w wand W\" respectively. We designate the smaller of these and w- by w. wIn selecting repeated samples, we would expect and W\" and therefore wto vary. w,Thus, we may think of w-and was values ofthe corresponding random variables W, W and W.The null hypothesis f.1. = f.l.o can be rejected in favour ofthe alternative wf.1. < f.1.0 only if is small and W\" is large. Likewise, the ahemative f.1. > f.1.0 can be accepted wonly if is large and W\" is small. For a two-sided alternative, we may reject H0 in wfavour of~ ifeither or W\" and hence if wis sufficiently small Therefore, no matter what the alternative hypothesis may be, we reject the null hypothesis when the value ofthe appropriate statistic W, W, or WIS sufficiently small. Two Samples with Paired Observations To test the null hypothesis that we are sampling two continuous symmetric populations with f.1.1= 1-12 for the paired-sample case, we rank the differences ofthe paired observations without regard to sign and proceed as with the single-sample case. The various test procedures for both the single-sample and paired-sample cases are summarized as follows: s·12ned-Rank11est 1b test~ lersus ~ Compute f.l.=f.l.o f.1.<f.1.o w+ f.l.t = f.l.2 f.1.>f.1.o f.l.* f.l.o w- f.l.t<~ w f.l.t>~ w+ f.l.t ;C f.l.2 w- w It is not difficult to show that whenever n < 5 and the level of significance does not exceed 0.05 for a one-tailed test or 0.10 for a two-tailed test, all possible values of wf-, w-, or wwilllead to the acceptance ofthe null hypothesis. However, when 5 :s; n :s; 30, approximate critical values of Wand W for levels ofsignificance equal to 0.01, 0.025 and 0.05 for a one-tailed test and critical values of Wfor levels of significance equal to 0.02, 0.05 and 0.10 for a two-tailed test can be obtained from the standard w,tables. The null hypothesis is rejected ifthe computed value of W\", or wis less than or equal to the appropriate tabled value. For example, when n= 12, from the said table, it can be seen that a value of w :s; 17 is required for the one-sided alternative f.1. < f.l.o to be significant at the 0.05 level. Remark When n~ 15, the sampling distribution of W (or W) approaches the normal distribution with mean \" = n(n + 1) and variance 2= n(n + 1)(ln + 1) ' ...-w 4 ' crw+ 24 Therefore, when n exceeds the largest value in the said table the statistic Z= ( W'\"cr;w) can be used to determine the critical region for our test. This is what is known as Normal Approximation for large samples.

(3) Rank-Sum Test Mn-Paranrtric Tests When one is interested in testing equality ofmeans oftwo continuous distributions that NOTES are obviously non-normal, and samples are independent (i.e., there is no pairing of observations), the Wllcoxonrank:-sum test (or Wilcoxon two-sample test) is an appropriate alternative to the two-sample t-test. We shall test the null hypothesis that H0 : lli = ~ against some suitable alternative. First we select a random sample from each of the populations. Let n1 be the no. of observations in the smaller sample, and ~ be the no. of observations in the larger sample. When the samples are of equal size, n1 and ~ may be randomly assigned. Arrange the (n1 +~)observations ofthe combined sample in ascending order and substitute a rank of 1, 2, ..., (n1 +~)for each observation. In the case ofties (identical observations), we replace the observations by the mean ofthe ranks that the observations would have ifthey were distinguishable. For example, ifthe sixth and seventh observations are identical, we would assign a rank of6.5 to each ofthe two observations and so on. The sum ofthe ranks corresponding to the n1 observations in the smaller sample is denoted by \"'I. Similarly, the value \"2 represents the sum ofthe 11:2. ranks corresponding to the larger sample. The total (\"'I + \"2) depends only on the no. of observations in the two samples and is in no way affected by the results of the experiment. Hence, if n1 = 3 and~ =4, then(\"'t + \"2) =(1 + 2 + ...+7) = 28, regardless ofthenumerical values of the observations. In general, ( \"'I + \"2) -- (nl + n2 )(n1 + n2 + 1) , the an'thmetI'C sum ofthe m· tegers 1, 2, ..., (n1 2 + ~). Once we have determined \"'I it may be easier to fmd \"2 by the formula _[(n n\"2- 1 + n2 )(n1 + 2 + 1) ] 2 -WI In choosing repeated samples ofsize n1 and~· we would expect \"'I and therefore \"2• to vary. Thus, we may think of \"'I and \"2 as values of the random variables \") and ~ respectively. The null hypothesis J.11 = J.12 will be rejected in favour of the alternative J.lJ < J.12 only if \"'I is small and \"2 is large. Likewise, the alternative J.11 > J.12 can be accepted only if \"1 is large and \"2 is small. For a two-tailed test, we may reject H0 in favour of~ if \"1 is small and \"2 is large or if \"'I is large and \"2 is small. In other words, the alternative J.11 < J.12 is accepted if \"'I is sufficiently smal~ the alternative lli > J.12 is accepted if \"2 is sufficiently small; and the alternative J.11 ::;:. J.12 is accepted if the minimum of \"'I and \"2 is sufficiently small. In actual practice, we usually base our decision on the value +1)] _[ +1)]u-_ I [ w,1 - n1(n1 or u- w:2 - n2 (n2 2 2 I Uz,of the related statistic ~ or or on the value u of the statistic U, the minimum of Uz.~ and These statistics simplify the construction oftables ofcritical values, since q_both and ~have symmetric sampling distributions and assume values in the interval from 0 to n1~ such that ( u1 + ~) = n1~· From the formulas for u1 and ~·we see that u1 will be small when \"'I is small and ~ will be small when \"2 is small. Consequently, the null hypothesis will be rejected lJ2whenever the appropriate statistic ~, or Uassumes a value less than or equal to the desired critical value given in the pertinent table. The various test procedures are summarized in the following table. Self-Instructional Material 327

Business Statistics-II Rank-Sum Test NOTES 1b test~ lersus ~ Co11J'Ufe J.ll < 1-'2 J.li 328 Self-Instructionall'Jaterial J.li = 1-'2 J.ll > 1-'2 1-'2 J.ll :# J.12 J.1 The table segment in the example-4 gives critical values of Vi and q for levels ofsignificance equal to 0.025 for a one-tailed test, and critical values of Ufor levels ofsignificance equal to 0.05 for a two-tailed test. Ifthe observed value of u1, ~or u is less than or equal to the tabled critical value, the null hypothesis is rejected at the level ofsignificance indicated by the table. Suppose, for example, that we wish to test the null hypothesis that J.lt = J.12 against the one-sided alternative that J.lt < ~at the 0.05level ofsignificance for random samples of size n1 = 3 and ~ = 5 that yield the value ~ = 8. It follows that ul = [s-3;4] =2 Our one-tailed test is based on the statistic Vi. From the said table, we reject the null hypothesis ofequal means when u1 ~ 1. Since u1 = 2 falls in the acceptance region, the null hypothesis cannot be rejected. (4) Kruskai-Wallis Test As can be seen from sectionA.2, the technique ofanalysis ofvariance is prominent as an analytical technique for testing equality of k 3 2 population means. However, the applicabilityofANOVAor the Ftest is based on the premise ofnormality. In this section, we investigate a non-parametric alternative to ANOVA. The Kruskal-Wallis test, also called the Kruskal-Wallis .B.test, is a generalization ofthe Rank-Sum test to the case ofk> 2 samples. It is used to test the null hypothesis 1{, that kindependent samples are from identical populations. Introduced in 1952 by W.H. Kruskal and W.A. Wallis, the test is a non-parametric procedure for testing the equality ofmeans in the one-factor ANOVA when the experimenter wishes to avoid the assumption that the samples were selected froin normal populations. Let ni(i= 1, 2, ..., Jq be the no. ofobservations in the Ah sample. First, we combine all ksamples and arrange then= (n1+ ~ + ···· +nJ observations in ascending order, substituting the appropriate rank from 1, 2, ···, n for each observation. In the case of ties (identicalobservations), we follow the usual procedure ofreplacing the observations by the means ofthe ranks that the observations would have iftheywere distinguishable. The sum of the ranks corresponding to the ni observations in the Jlli sample is denoted by the random variable ~·Now let us consider the statistic ±H= [ 12 Rl -3 (n+1)]. n(n+1) i=l ni which is approximated very well by a Chi-squared distribution with (k- 1) degrees of freedom when~ true and ifeach sample consists ofat least 5 observations. Note that the statistic H assumes the value h, where L -h= [ 12 k d -3 (n+1)] 1 n(n + 1) i=l ni

where ~ assumes the value r1, ~ assumes the value r 2 and so on. The fact that h Nm-Pararretric Jests is large when the independent samples come from populations that are not identical NOTES allows us to establish the following decision criterion for testing H0 • NJte To test the null hypothesis H0 that kindependent samples are from identical populations, compute ~f:;:h = [ 12 rf -3 (n + 1)] n(n + 1) ni If hfalls in the critical region H> :x.~ with u = (k-1) degrees of freedom, reject ~at the a-level of significance; otherwise accept H0• (5) Runs Test for Randomness Applying the many statistical concepts discussed throughout this chapter, it was always assumed that our sample data had been collected by some randomization procedure. The runs test, based on the order in which the sample observations are obtained, is a useful technique for testing the null hypothesis ~ that the observations have indeed been drawn at random. To illustrate the runs test, let us suppose that 12 people are polled to find out ifthey use a certain product. One would seriously question the assumed randomness ofthe sample if all 12 people were of the same sex. We shall designate a male and female by the symbols Mand F, respectively and record the outcomes according to their sex in the order in which they occur. A typical sequence for the experiment might be. MM FFF M FF MMMM where we have grouped subsequences ofsimilar symbols. Such groupings are called runs. Thus, a run is a subsequence ofone or more identical symbols representing a common property of the data. Regardless ofwhether our sample measurements represent qualitative or quantitative, data, the runs test divides the data into two mutually exclusive categories : male or female, defective or non-defective, heads or tails, above or below the median, and so forth. Consequently, a sequence will always be limited to two distinct symbols. Let n1 be the no. of symbols associated with the category that occurs the least and .11z be the no. ofsymbols that belong to the other category. Then the sample size n= (n1 + .llz). For the n = 12 symbols in our poll we have five runs, with the first containing two JIJs, the second containing three Fs and so on. Ifthe no. ofruns is larger or smaller than what we would expect by chance, the hypothesis that the sample was drawn at random should be rejected. Certainly, a sample resulting in only two runs, M M M M M M M F F F F F, or the reverse, is most unlikely to occur from a random selection process. Such a result indicates that the first 7 people interviewed were all males followed by 5 females. Likewise, ifthe sample resulted in the maximum no. of 12 runs, as in the alternating sequence, M F M F M F M F M F M F, we would again be suspicious of the order in which the individuals were selected for the poll. The runs test for randomness is based on the random variable V, the total no. ofruns that occur in the complete sequence ofour experiment. In the following table segment, values of P( V~ u* when H0 is true) are given for u* = 2, 3, 4, 5 runs and values of n1 = 5 and .11z from 5 to 10. The P-values for both one-tailed and two-tailed tests can be obtained using these tabled values. Self-Instructional Material 329

Business Statistics-D ~VS V' when H 0 is true) in the Runs Test NOTES (Dt,ni) 2 \"* 4 5 0.167 330 Self-InstructionalMlterial (5,5) 0.008 3 0.110 0357 (5,6) 0.004 0.076 0262 (5, 7) 0.003 0.040 0.054 0.197 (5,8) 0.002 0.024 0.039 0.152 (5,9) 0.001 O.ot5 0.029 0.119 (5, 10) 0.001 0.010 0.095 0.007 0.005 * For an exhaustive list ofvalues, one should consult the textbooks on Nonparametric Statistical Methods cited in the bibliography. In the poll taken above, we exhibit a total of 5 Fs and 7 ]'d). Hence, with n1 = 5, ~ = 7 and u = 5, we note from the said table for a two-tailed test that the P-value is P= 2P(Vs 5 when H0 is true)= 2(0.197) = 0.394 > 0.05. That is, the value of v= 5 is reasonable at the 0.05 level ofsignificance when~ is true, and therefore we have insufficient evidence to reject the hypothesis ofrandomness in our sample. When the no. ofruns is large, for example if v= 11 and n1 = 5 and ~ = 7, then P-value in a two-tailed test is P= 2P(V~ 11 when ~is true) = 2 [1 - P( Vs 10) when H0 is true] = 2 (1- 0.992) = 0.016 < 0.05, which leads us to reject the hypothesis that the sample values occurred at random. The runs test can also be used to detect departures in randomness of a sequence ofquantitative measurements over time, caused by trends or periodicities. Replacing each measurement in the order in which they are collected by a plus symbol ifit falls above the median, by a minus symbol if it falls below the median, and omitting all measurements that are exactly equal to the median, we generate a sequence of plus and minus symbols that are tested for randomness. Remarks 1. When n1 and~ increase in size, the sampling distribution of Vapproaches the normal 1]distribution with means, f.1v = [ 2n1n2 + Dt +n2 and van.ance, ay2 = [2n1n2 (2n2 -n 1 -n2 )] (n1 +n2)2 (n1 + n2 -1) Consequently, when n1 and n.z are both greater than 10, one could use the statistic ::v)Z= ( V to establish the critical region for the runs test. 2 The runs test, although less powerful, can also be used as an alternative to the Wilcoxon two-sample test to test the claim that two random samples come from populations having the same distributions and therefore equal means. If the populations are symmetric, rejection of the claim of equal distributions is equivalent to accepting the alternative hypothesis that the means are not equal. In performing the test, we first combine the observations from both samples and arrange them in ascending order. Now assign the letter A to each observation taken from one of the populations and the letter Bto each observation from the second population, thereby generating a sequence consisting of

the symbols A and B Ifobservations from one population are tied with observations from NJn-Paranr:tric 1i:sts the other population, the sequence ofA and Bsymbols generated will not be unique and consequently the no. ofruns is unlikely to be unique. Procedures for breaking ties usually NOTES result in additional tedious computations, and for this reason one might prefer to apply Check Your Progress the Wilcoxon rank-sum test whenever these situations occur. 3. Name three non- 13.4 SOLVED PROBLEMS parametric tests of significance. Problem 13.1 A Personnel Manager is interested in trying to determine whether absenteeism is greater on one day ofthe week than on another. His records for the past 4. Who devised the Signed- year show this sample distribution: Rank Test? Day of ~nday Tbesday 'Rednesday Thursday Friday Self-Instructional Material 331 the !leek 57 54 48 75 66 NJ. of Absentees Test whether the absence is uniformly distributed over the week. Let us take the (null) hypothesis that absenteeism is uniformly distributed over the week. On the basis ofthis hypothesis, we should expect (66 +57+ 54+ 48 + 75)/5 = 300/5 = 60 absentees on each day ofthe week. -(; .t; (fo-ri Fe 66 ro 57 ro 0.00 54 ro 0.15 48 ro 75 ro o.ro 2.40 3.75 l: (fo- f\"e)2 7.50 Fe x?=!: ( f 0- f}2 =7.5 e 4 u = (n- 1) = (5 - 1) = 4 for u = 4, x20.05 = 9.49 The calculated value ofx2 is less than the table value. Hence, the (null) hypothesis is accepted. Problem 13.2 Records taken ofthe no. ofmale and female births in 800 families having 4 children are as follows:

Business Statistics-II NJ.of NJ.of NJ.of rmle births female births fa.nilies NOTES 0 4 32 1 3 178 2 2 290 3 1 236 4 0 64 Total=800 Test whether the data are consistent with the hypothesis that the binomial law holds, and that the chance ofa male birth is equal to that of a female birth, namely q = p = +·You may use the table given below: D:grees offreedom 12 345 5% value ofchi-square 3.84 5.90 7.82 9.49 11.07 ++rThe theoretical frequencies ofO, 1, 2, 3, 4 male births are the successive terms ofthe binomial expansion N(p + q)n i.e., of 800 ( + ·m' HJ'G)'[(H GH)'ie.• of soo +4 + 4c, +4c,(i)'(H +4c4 ·M] i.e.,of soox(~r x[1+4+6+4+1] i.e., 50, 200, 300, 200, 50 :. we have OI:XJerved 32 178 290 236 64 300 200 50 frequencies (fJ Theoretical 50 200 frequencies (t;) Hence 332 Self-InsiructionalMaterial = {(32- 50)2 + (178- 200)2 + (290- 300)2 + (236- 200)2 + (64- 50)2 } = 19 63 50 200 300 200 50 . Also, the no. of degrees offreedom, u = (n- 1) =(5-1)= 4 From tables, x?o.os = 9.49 for u = 4 The calculated value ofx? (19.63) is much greater than the table value ofx? (9.49) and so the hypothesis that the binomial law holds' does not hold for the given data.

Problem 13.3 A book has 700 pages. The no. ofpages with various nwnbers ofmisprints N:m-Paranrtric Tests is recorded below. At 5% significance level are the misprints distributed according to Poisson law? NOTES NJ. ofmisprints (x) 01234 5 Self-Instructional Mlterial 333 1X NJ. ofpa.ges with 616 70 10 2 1 misprints By applying Poisson law, we find the expected values as follows: Xf fx 0 616 0 1 70 70 2 10 20 32 6 41 4 51 5 N=700 'f.fx= 105 x or m= 'f.fx - =1005 . 1 5 -= N 700 P(O) =e-m= e-D· 15 = 0.8607 (NJte: e 1 2.718) N/(0) = 700 X 0.8607 =602.5 N/(1) = Nl(0 x m= 602.5 x 0.15 = 90.4 ~N/(2) = N/(1) x = 90.4 x ( 0·; 5 ) = 6.8 ~N/(3) = Nl(2>x = 6.8 x ( 0 ·; 5 ) = 0.4 N/(4) = N/(3) X ; = 0.4 X 0~ 5 i 0 m 0.15 5N/(5) = N/(4) X =0X -5- = 0.. Applying x2- test, we obtain the following result. ~ ~ (~-fr/ ( f0 ~r; J 616 602.5 182.25 0.302 70 90.4 416.16 4.604 10 6.8 10.24 1.506 2 0.4 12.96 32.400 1 0.0 1 0.0 - - - - t;i~ (fo- = 38.812 t;, u = (3 - 1) = 2 (Since the last two frequencies did not figure in calculations at all and the 3rd and 4th frequencies are clubbed together as the cell value 2 falls below 5).

Business Statistics-ll For u = 2, x~.os =5.99. NOTES As the calculated value ofx2 (i.e., 38.812) is greater than the table value (ie, 5.99), 334 Self-InstructionalJl.flterial we reject the null hypothesis that 'the Poisson law holds for the give data'. Hence, at 5% level of significance, the misprints are not distributed according to Poisson law. Problem 13.4 An automobile company gives you the following information about age groups and the liking for particular model ofcar which is plans to introduce. Age groups Persons Below20 20-39 40-59 / 60 Total 20 280 Who liked the car 140 00 40 00 .220 Who disliked 50 30 the car ro 100 500 Total 200 130 70 On the basis ofthis data, can it be concluded that the model appeal is independent oftheagegroups. (Given v=3, x~.05 =7.815) Let the null hypothesis be that the model appeal is independent of the age group. Applying x2 test: f = 280x200 = 112 e11 500 l(. Row total x Column totall l.e., ) Grand total f = 280 X 130 = 72.8 112 500 fen = 280 x 70 = 39.2 and so on. 500 Thus, the table of expected frequencies will be obtained as follows: 112 72.8 39.2 56 280 88 572 30.8 44 220 200 130 70 100 500 ~ ~ (~-rtf (~-ftf!J; 112.0 784.00 7.000 140 88.0 784.00 8.910 72.8 51.84 0.712 ro 51.84 0.906 572 0.64 0.016 00 392 0.64 0.021 50 30.8 1296.00 23.143 40 56.0 1296.00 29.454 30 44.0 20 l~ {fo~fe)2 } = 70.162 00

Thus, x2 = 70.162 N:m-Paranrtric Jests u = (r- 1)(c- 1) = (2 - 1) (4 - 1) = 3 NOTES For u = 3, x~.05 = 7.815 Self-Instructional Mltcrial 335 The calculated value is much greater than the table value. Hence, the null hypothesis is rejected. We therefore, conclude that the model appeal is not independent ofthe age groups. Problem 13.5ln experiments on Pea breeding, Mendel obtained the following frequency of seeds: Round and yellow .. 315 Wrinkled and yellow .. 101 Round and green .. 108 Wrinkled and green .. 32 Theory predicts that the frequencies should be in proportion 9 : 3 : 3 : 1. Examine the correspondence between theory and experiment. The total frequency= 315 + 101 + 108 + 32 = 556 It is given that the theoretical frequencies are in proportions 9 : 3 : 3 : 1. The sum of these proportions = 9 + 3 + 3 + 1 = 16 .·. The corresponding theoretical frequencies are 9 3 -3 X 556 and - I X 55 6 - X 55 6 - X 55 6 16 16 ' 16 ' 16 or 313, 104, 104 and 35 (approx) respectively. x2 = ~ (fo- f\"e) 2 = (315-313)2 + (101-104)2 ~ 313 104 +(108-104)2 +(32-35)2 =0.511 104 35 Also, the no. of degrees of freedom, u = (n- 1) = (4- 1) = 3 From tables, x~_05 =7.815, for u = 3. The calculated value is much less than this value and so this calculated value is not significant and so there is a very high degree of agreement between theory and experiment. Problem 13.6 A group of 11 students selected at random secured the grade points : 1.5, 2.2, 0.9, 1.3, 2.0, 1.6, 1.8, 1.5, 2.0, 1.2 and 1.7 (out of3.0). Use the sign testtotestthe hypothesis that intelligence is a random function (with a median of 1.8) at 5% level of significance. Solution {i) H0 : jl = 1.8 (ii) ~ : jl :;C 1.8

Business Statistics-H (iii) a= 0.05 NOTES p= t(iv) Test statistic :Binomial variable X with (v) Computations: Replacing each value by the symbol\"+\" ifit exceeds 1.8, by the symbol \"-\"ifit is less than 1.8, and discarding the one measurement that equals 1.8, we obtain the sequence - +-- +-- +-- for which n = 10, x= 3 and ~ = 5. Therefore, from table C.8(ii), the computed P-value is p=t)P= 2P( Xs;3 when ±{= 2 ~)x; 10, = 2(0.1719) = 0.3438 > 0.05 x=O . (vi) Decision: Do not reject the null hypothesis and conclude that the median score is not significantly different from 1.8. Remark One can also use the sign test to test the null hypothesis (ji1 - ji2) = d0 for paired observations. Here, we replace each difference di' with a plus or minus sign depending on whether the adjusted difference, (~- dJ, is positive or negative. Throughout this section, we have assumed that the populations are symmetric. However, even if populations are skewed we can carry out the same test procedure, but the hypotheses refer to the population medians rather than the means. Problem 13.7 Solve Example (1) by using the Signed-Rank Test. Make use of the following table segment. Critical vaues for the Signed-Rank Test n One-sided a - 0.01 One sided a - 0.025 One sided a - 0.05 THO-sided a = 0.02 THO-sided a = 0.05 THO-sided a = 0.10 5- - 1 6- 2 70 1 4 82 2 6 93 4 8 10 5 6 11 8 Solution (i) H0 : f.1 = 1.8 (it) ~ : fl'~ 1.8 (iii) a= 0.05 (iv) Critical region: Since n= 10, after discarding the one measurement that equals 1.8, the preceding table shows the critical region to be ws; 8. (v) Computations: Subtracting 1.8 from each measurement and then ranking the differences without regard to sign, we have 336 Self-InstructionalMlterial

d. -0.3 0.4 -0.9 -0.5 02 -0.2 -0.3 02 -0.6 -0.1 Nm-Paranr:tric 7ests I NOTES Ranks 5.5 7 10 8 3 3 5.5 3 9 1 Self-Instructional Mlterial 337 wNow, = 13 and w = 42 so that w= 13, the smaller of wand w. (vi) Decision: As before, do not reject H0 and conclude that the average score is not significantly different from 1.8. Remark The Signed-Rank test can also be used to test the null hypothesis that (!l1 - 112) = d0 • In this case, the populations need not be symmetric. As with the sign test we subtract d0 from each difference, rank the adjusted differences without regard to sign, and apply the same procedure as above. Problem 13.8 In a factory, three different incentive schemes were being offered to the workers. On the data give below based on a sample from each ofthe groups, test the hypothesis that there is no difference in the productivity of the workers. Productivity (Average per day output) Group A 2.48 3.25 3.94 3.45 3.00 4.00 3.60 3.87 GroupB 2.84 3.10 3.50 2.27 3.88 2.87 3.27 2.80 Groupe 3.40 3.17 2.85 2.46 3.15 2.69 2.88 3.44 Use Kruskal-Wallis Test Null Hypothesis: H0 : 111 = 112 = 113 (Mean productivities of the three groups) * *~ : Ill 112 113 0 =0.05 Critical region : h > x~.os = 5.991 for v= 2 degrees of freedom Computations: Ranks for Productivities ofDifferent Groups Group A 3 14 23 18 10 24 20 21 rA= 133 GroupB 6 11 19 1 22 8 15 5 r8 =87 Groupe 16 13 7 2 12 4 9 17 rr=80 From the preceding table, the sum of ranks, rA= 133; r 0 = 87; and rc= 80. tH=[ 12 rl_3(n+l)l n(n+l) i=l ni where, n =total no. of elements of ksamples. Thus, H= [ 12 {133 2 + 87 2 + 80 2 }_ 3 (24 +o] 24(24+1) 8 8 8 = 4.145 As the calculated value of His less than the critical value of x2, distribution for 2 degrees of freedom at 5% level ofsignificance (4.145 < 5.991), the null hypothesis is accepted. Thus, productivity of the workers is not affected by the incentive scheme.

Business S11ltistics-H Problem 13.9 The phosphorus content in two alternative drugs, measured in milligrams, was found to be as follows: NOTES .D-qg-A 21 4.0 63 5.4 4.8 3.7 6.1 33 .D-qg-B 4.1 0.6 3.1 25 4.0 62 1.6 22 1.9 5.4 Test the hypothesis, at the 0.05 level ofsignificance, that the average phosphorus contents ofthe two drugs are equal against the alternative that they are unequal. Make use ofthe following table segment. Critical Values for the Rank-sum Test One-tailed test at a = 0.025 or TMO tailed test at a = 0.05 ~ 456 7 8 9 10 1 2 0 0 0 3 2 3 4 01 1 2 4 5 5 7 8 6 0 12 3 4 10 11 7 12 14 8 23 5 6 15 17 9 17 20 56 8 23 10 8 10 13 Solution *(i) Ho : J.11 = J.12 (ii) ~ : J.11 1-12 (iii) a= 0.05 (iv) Critical region: u~ 17 (from the preceding table)- (v) Computations: The observations are arranged in ascending order and ranks from 1 to 18 assigned. 338 Self-InstructionalMlterial Origi111ll Dlta Ranks Drug Sample 0.6 1 B 1.6 2 B 1.9 3 B 21 4 A 22 5 B 2.5 6 B 3.1 7 B 33 8 A 3.7 9 A 4.0 10.5 A 4.0 10.5 B 4.1 12 .· B 4.8 13 5.4 14.5 A 5.4 14.5 A 6.1 16 B 62 17 B 18 B 6.3 A

Now, WI =(4+8+9+10.5+13+14.5+18)=93 Nm-Param:tric Jests (sum ofranks pertaining to drug-A) NOTES [Cand Wz = 8;19)-93] =78 u1 = [93-(8;9)] =57 e~ = [78 - 0; 11)J = 23 (vi) Decision: Do not reject/{, and conclude that there is no significant difference in the average phosphorus contents of the two alternative drugs. - Problem 13.10 It is claimed that an Engineering student in the final year can increase his overall annual score by at least 50 marks ifhe is provided with sample problems in advance. To test this claim, 20 final year Engineering students are divided into 10 pairs such that each matched pair has almost the same overall track-record ofperformance for their first three years ofEngineering. Sample problems and answers are provided at random to one member ofeach pair-one week prior to the examination. The following examination scores were recorded: Pair I u m IV v w VII vm IX X With Sample 531 621 663 579 451 660 591 719 543 575 Problems Without 509 540 688 502 424 683 568 748 530 524 Sample Problems Test the null hypothesis at the 0.05 level of significance that sample problems increase the scores by 50 marks against the alternative hypothesis that the increase is less than 50 marks. Make use of the table segment given in example-2. Solution Let 1-11 and 1-11 represent the mean score of all students taking the test in question with and without sample problems, respectively. (i) H0 : (1-11 - 1-12) =50 (ii) ~ : (1!1 - 1-12) < 50 (iii) a= 0.05 (iv) Critical region: Since n= 10, the pertinent table shows the critical region to be w-~11 (v) Computations: Pair I u m IV v w VII vm IX X d. 22 81 -25 77 27 -23 23 -29 13 51 1 (~-dJ -28 31 -75 27 -23 -73 -27 -79 -37 1 Ranks 5 6 9 3.5 2 8 3.5 10 7 1 wNow, we find that = (6 + 3.5 + 1) = 10.5 (vi) Decision: Reject ~and conclude that sample problems do not, on an average, increase one's overall annual score by as much as 50 marks. Self-Instructional Mucrial 339

Business Statistics-H Problem 13.11 The customers arriving at the booking counter of a Road Traffic Corporation, sex.wise are expected to follow a random sequence. The position in respect NOTES of 30 passengers on a day was as follows: 340 Self-InstructionalMlterial MMFFFMFFMMFFFFMMMFFMMFMMMFFFMM Comment whether the arrival pattern is random. Runs test is applied to test the randomness ofthe arrival pattern. H0 : The arrival pattern is random. Runs: MM FFF M FF MM FFFF MMM FF MM F MMM FFF MM. Runs (r}= 13 n1 (M) = 15 ~(F)= 15 (n1 + ~) = 30 ~~ = Ir-a~(r) I l(r)-{~~~ +1}1 = --;=:=====:=='========= (2Dt~)(2n1 ~ -Dt -~) (n1 + ~) 2 (11t + n2 -1) 113 _(2x15x15 + 1)1 = 15 + 15 = 1.11 r===::::l::::============::!:::::== (2 X 15 X 15) X (2 X 15 X 15 - 15 - 15) (15 + 15)2 X (15 + 15 - 1) As the value ofI~i.e., 1.11 < 1.96 (the critical value of2J at 5% level ofsignificance, the null hypothesis is accepted and the arrival pattern of the passengers at the RTC counter may be considered as random. Problem 13.12 A car manufacturing firm is exploring the possibility ofreplacing the regular belted tyres by radial tyres. Sixteen cars are equipped with radial tyres and driven over a prescnbed test course. Without changing the cha:ffeurs, the same cars are then equipped with the regular belted tyres and driven once again over the test course. The petrol consumption, in kilometres per litre was recorded as follows: Car Radial 1jTes Belted 1jTes 1 42 4.1 2 4.7 4.9 3 6.6 62 4 7.0 6.9 5 6.7 6.8 6 4.5 4.4 7 5.7 5.7 8 6.0 5.8 9 7.4 6.9 10 4.9 4.9 11 6.1 6.0 12 52 4.9 13 5.7 5.3 14 6.9 65 15 6.8 7.1 16 4.9 4.8