MCM 602 Quantitative Techniques for Managers

Hypothesis Testing Part - I 295 Where P = n1 p1  n2 p2 = 400  400 = 0.73 n  n2 500  600 Q = 1 – P = 0.27 0.8 – 0.67 Z = = 4.81 L O0.73  0.27  1  1 MN PQ500 600 For 5% level of significance Z > 1.645 and for 1% level of significance Z > 2.33. Calculate value of Z is even greater than 3 (one tail test). Hence it is highly significant change in tea drinking. The null hypothesis is rejected. Problem 6 The average number of defective articles in a certain factory is claimed to be less than the average for all the factories. The average for all the factories is 30.5. A random sample of 100 defective articles showed the following distribution. Class Limits Number 16 – 20 12 21 – 24 22 26 – 30 20 31 – 35 30 36 – 40 16 Calculate the mean and the standard deviation of the sample and use it to test the claim that the average is less than the figure for all the factories at 5% level of significance. Given Z(–1.645) = 0.95. Solution: Null hypothesis H0 :  = 30.5. Alternative H1 :  < 30.5 CU IDOL SELF LEARNING MATERIAL (SLM)

296 Quantitative Techniques for Managers Table for Computing Mean and SD of the Sample Class Limits Number Mid pt x  28 fd fd2 d= 5 16 – 20 (f) (x) 21 – 25 –2 –24 48 26 – 30 12 18 31 – 35 22 23 –1 –22 22 36 – 40 20 28 30 33 00 000 00 16 38 f = N = 100 01 30 30 02 32 64 fd = 16 fd2 = 164 5  16 Hence x = 28 + 100 = 28.8 GF JIs = i ×  fd 2   fd 2 H KN N FGH IKJ= 5 × 164  16 2 100 100 = 5 × 1.27 = 6.35 Since sample is very large 2 = s2   = 6.35 For large sample, we can use normal variate relationship x   28.8 – 30.5  Z=  / = = –2.67 n 6.35 / 100 Since Z < –1.96, the difference is significant at 5% level of significance for a single tail test. Hence we reject the null hypothesis which means that the claim of the factory about its low defectives level is valid. CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - I 297 Problem 7 The mean height of 50 male students who showed above average participation in college athletics was 68.2 inches with a standard deviation of 2.5 inches while 50 male students who showed no interest in such participaation had a mean height of 67.5 inches with a standard deviation of 2.8 inches. Test the hypothesis that male students who participated in college athletics are taller than other male students. Solution: Given here, n1 = 50, x = 68.2, Sx = 2.5 n2 = 50, y = 67.5, Sy = 2.8 Null hypothesis H0 : x = y Alternative hypothesis H0 : x > y (right tail test) Calculating the value of Z, Z= xy at N(0.1) Sx  Sy n1 n2 68.2 – 67.5 = = 1.3188 GF IJ6.25  7.84 H K50 50 Since Z < 1.96, the difference in the hights of student of two categories is not significant at 5% level of significance. Hence the Null hypothesis is accepted as the data is consistent with Null hypothesis. 10.9 Summary The unit is summarised by some of its important points as below:  Alpha (): The Probability of a type 1 error.  Alternative hypothesis: The conclusion drawn when the given data does not support the provision of null-hypothesis. CU IDOL SELF LEARNING MATERIAL (SLM)

298 Quantitative Techniques for Managers  Beta (): The probability of a type II error.  Hypothesis: An assumption adopted regarding the population parameter.  Null Hypothesis: The assumptions about the population parameter to be tested for it to be true.  Power of the test: A measure to minimise the type II error. 10.10 Key Words/Abbreviations  Rejection region: “The acceptance of a statistical hypothesis is due to insufficient evidence provided by the sample to reject it and does not necessarily imply that it is true”.  Type I error: Rejecting a nult hypothesis when it is true.  Type II error: Accepting a nult hypothesis when it is false. 10.11 Learning Activity 1. In a random sample of 400 persons from a large population, 120 were females, can it be said that males and females are in the ratio 5:3 in the population? Use 1% level of significance. 2.. In a certain district A, 450 persons were considered regular consumers of tea out of a sample of 1,000 persons. In another district B, 400 were regular consumers of tea out of a sample of 800 persons. Do these facts reveal a significant difference between the two districts as far as tea drinking habit is concerned? Use 5% level. 10.12 Unit End Questions (MCQ and Descriptive) A. Descriptive Type Questions 1. Explain the following with reference to testing of hypothesis (i) Type 1 error and Type II errors (ii) Critical region (iii) Power of a test (iv) Most powerful test CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - I 299 2. Explain the assumptions in large sample theory. Distinguish clearly between large sample and small sample tests of significance. Are small sample tests valid for large samples? 3. Outline the procedure for large sample tests and discuss their theoretical basis. Comment on the assumptions made. 4. Discuss the large sample test for testing the equality of two population means. 5. A random sample of 500 pineapples was taken from a large consignment and 65 of them were found to be bad. Show that the standard error of proportion of bad ones in a sample of this size is 0.015 and deduce that the percentage of bad pineapples in the consignment almost certainly lies between 8.5 and 17.5. 6. A random sample of 700 units from a large consignment showed that 200 were damaged. Find (i) 95% and (ii) 99% confidence limits for the proportion of damaged units in the consignment. 7. Out of 20,000 customers ledger accounts, a sample of 600 accounts was taken to test the accuracy of posting and balancing wherein 45 mistakes were found. Assign limits within which the number of defective cases can be expected at 95% level. 8. In a marketing survey for the introduction of a new product in a town, a sample of 400 persons was taken. When they were approached for sale, 80 of them purchased the product. Find 95% confidence limits for the percentage of persons who would buy the product in the town. 9. In a sample of 1,000 TV viewers, 340 watched a particular programme. Find 99% confidence limits for the percentage of all viewers who watch this programme. 10. In a sample of 400 oranges from a large consignment, 40 were considered bad. Estimate the percentage of defective oranges in the whole consignment and assign limits within which the percentage will probably lie. 11. A life insurance company has 1,500 policies averaging Rs. 2,000 on lives at age 40. From the experience table; it is found that of 1,00,000 alive at the age of 30, as many as 90,000 were alive at age of 31. Find the lower and upper values of the amount the company will have to pay out in insurance during the year. 12. A factory is producing 50,000 pairs of shoes daily. From a sample of 500 pairs, 2% were found to be of sub-standard quality. Estimate the number of pairs that can be reasonably expected to be spotted in the daily production and assign limits at 95% of level of confidence. CU IDOL SELF LEARNING MATERIAL (SLM)

300 Quantitative Techniques for Managers B. Multiple Choice/Objective Type Questions 1. A statement made about a population for testing purpose is called ___________. (a) Statistic (b) Hypothesis (c) Level of significance (d) Test-statistic 2. If the assumed hypothesis is tested for rejection considering it to be true is called? (a) Null hypothesis (b) Statistical hypothesis (c) Simple hypothesis (d) Composite hypothesis 3. Type I error occurs when we ___________. (a) reject a false null hypothesis (b) do not reject a false null hypothesis (c) reject a true null hypothesis (d) do not reject a true null hypothesis 4. If a one – tail test for the population mean, if the null hypothesis is not rejected when the alternative hypothesis is true, then ___________. (a) Type I error is commited (b) a correct decision is made (c) a two – tail test should be used of a one – tail test (d) Type II error is commited 5. By taking a level of significance of 5% it is the same as saying ___________. (a) We are 95% confident that the result have not occurred by change (b) we are 5% confident the result have not occurred by change (c) we are 95% confident that result have occurred by change (d) None of these Answers 1. (b), 2. (a), 3. (c), 4. (d), 5. (a) 10.13 References References of this unit have been given at the end of the book. CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 301 UNIT 11 HYPOTHESIS TESTING PART - II Structure: 11.0 Learning Objectives 11.1 Introduction 11.2 Tests for Equality of Population Mean 11.3 Tests of Significance based on t, F and Z Distribution 11.4 Solved Problems 11.5 Summary 11.6 Key Words/Abbreviations 11.7 LearningActivity 11.8 Unit End Questions (MCQ and Descriptive) 11.9 References 11.0 Learning Objectives After studying this unit, you will be able to:  Explain the test of significance based on t, F and Z distribution.  Discuss about fundamental Assumption of t distribution.  Elaborate t-test for difference of means. CU IDOL SELF LEARNING MATERIAL (SLM)

302 Quantitative Techniques for Managers 11.1 Introduction So far we have studied the inference of the population characteristics based on some samples drawn from it in a scientific manner, because census method is time consuming, costly, and at times unpractical. These concepts have already been discussed in chapter on ‘Sampling Theory’. The samples are expected to give close results regarding the population provided samples and drawn to make them representative of the population. These results can now be generalised if we know how much these generalisations condition are valid. We then can estimate the population parameter with the degree of confidence. The techniques used for the purpose are dealt with in statistics by “Statistical Inference”, which is classified into two main categories: 1. Theory of Estimation 2. Testing of Hypothesis Estimation theory is a branch of statistics that deals with estimating the values of parameters based on measured empirical data that has a random component. The parameters describe an underlying physical setting in such a way that their value affects the distribution of the measured data. An estimator is a statistic that estimates some fact about the population. You can also think of an estimator as the rule that creates an estimate. For example, the sample mean(x ) is an estimator for the population mean, ì. The quantity that is being estimated (i.e. the one you want to know) is called the estimand. While discussing the central limit theory in the proceeding paragraphs, we had the case of samples being of large size. It was discussed that the sampling distribution of the sample statistic is normal or can he approximated to normal. For this purpose either the population from which the sample has been chosen is normal or that the sample size is large enough to represent the population in true way. Only in such cases, we have been able to use central limit theory. As per this theory, we have discussed the distribution of the sample mean to be normally distributed regardless of the sample size. Generally as sample of size n > 30 is considered to be a large mean or sample proportion respectively and if we apply central limit theory, we can use the distribution sample variance S2 (when s2 is not known) to be normal when the population is normal and the sample size is large. In the last chapter, we have established the relationship. X – Z = in order to define the standard normal variate. But in case ot a situation, when / n either the population variance s2 is not known or the sample size is small, then we cannot use S2 as the unbiased estimator of s2, In case of sample size being small, it is seen that the sample distribution CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 303 will be t-distribution. In this case, sample size n being less than 30, the central limit theorem will not be applicable and sample variance S2 can no longer be used as an unbiased estimator of s2 i.e. the population variance. The student’s t - distribution was obtained and published by WS Gosset. but under a changed name as “student”. In addition, we also have F - distribution. Z - distributions and c2 -distribution in addition to Binomial and Poisson distributions for small sample applications. Binomial and Poisson Distributions have been discussed in earlier chapters. Now we will be discussing the others. 11.2 Tests for Equality of Population Mean In this case, the null hypothesis will be formulated H0 : 1 = 2 Where 1 is the population mean of population 1 and 2 that of population 2. Then various relationship for the test statistic will be x1  x2 (i) Z = 12  22 with usual notations n1 n2 If1 = 2 = , then the value can be written as x1  x2 (ii) Z=  1  1 n1 n2 However, if is not known, then it has to be estimated from the sample standard deviation (s) as follows n1  1 s12  n2  1 s22 s = n1  n2  2 and then test of statistic is calculated as CU IDOL SELF LEARNING MATERIAL (SLM)

304 Quantitative Techniques for Managers x1  x2 t= s  1  1   n1 n2  This has t-distribution with (n1 + n2 – 2) degrees of freedom. 11.3 Tests of Significance based on t, F and Z Distribution The tests discussed so far pertain to business decisions where large samples can be obtained and their characteristics can be utilised to draw inferences about the population characteristics, from which such samples are drawn. But when the sample size is small, say n  30, then the sample may not represent the population in its true characteristics and the normal tests may not be useful for such decisions. We, therefore, now discuss certain specific tests known as Exact Sample Tests, which have been developed to deal with such situations. The initial and important research and developmental work on Exact Sample Theory was done by William S. Gosset, an Irish brewery employee. He had been publishing his work under the name \"Student\" and hence the statistic developed was named as “student” t-distribution (1908). The concept was later refined by Prof. R. A. Fisher. Now we shall discuss these distributions by the name of t, F and Z-distributions and tests on the basis of these distributions. FundamentalAssumptions For discussions on the Exact Sample Tests, we make and follow the under given fundamental assumptions. 1. The parent population from which the sample is drawn is Normally Distributed. 2. The sample drawn is random and independent. Students’ t-Distribution For the large sample tests for mean, we had discussed the statistic as Z= x  N (0,1) with usual notations and the curve of distribution was asymptotic. / n When the population variance s2 was not known, we replaced s2 by S2 (the sample variance for small a fsamples) and hence obtained S2 = 1 n 1  xx 2 or ns2 = (n – 1) S2. We had then replaced2 by S2 CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 305 and applied Normal tests. But student t-distribution devised by Gosset established that the sampling distribution of the statistic. x S/ n was applicable to small samples, but it was not near Normal Distribution. He gave a x statistic as t = S/ n where = xx n  1 2 and S2 = n 1 x  x is an unbiased estimate of the population variance 2 Gosset indicated that t-statistic follows student’s t-distribution with  = (n – 1) degrees of freedom with probability density function. 1  t2  ( 1) / 2 1+   p(t) = Const. ; – < t < . The constant in this equation is calculated by the area under the probability curve p(t), (total  area = 1), given by  p(t)dt =1  1 Hence Constant =    1 ,   2 2 mn Where  (m, n) is the Beta function defined by (m, n) = Tmn ; m > 0, n > 0. Where Gamma m(m) is given by m = (m – 1) (m – 1) and m = (m – 1)! if m is a positive integer. CU IDOL SELF LEARNING MATERIAL (SLM)

306 Quantitative Techniques for Managers Properties of t-Distribution p(t) t-curve 1. Moments 1 (about origin) = 0 Normal (2r+1) t 11 (about origin) = 0, Mean = 0 Fig. 11.1 Comparison between Normal and t- 1 (about origin) = 1 (2r+1) (2r+1) (about mean) 2r 12r  3...5.3.1 2r =    2   4....   2r 2r 1  =   2r r 1 = 13 = 0, 2 =       . 12 2     2. The curve is uninodal with Mean = p(t) Median = Mode = 0. 3. t-distribution depends only on ‘Sample Rejection region Rejection size  = n – 1, nor on population parameters for v large, ( > 30), t-distribution approximates to –t t t Normal distribution as can be seen from Fig. 11.2.  4. t-distribution varies from –  to as in Fig. 11.2 Two tailed critical values of t case of normal distribution 5. t-distribution is also bell-shaped and symmetrical about mean zero as in the case of normal distribution. 6. It is more platikurtic them the normal distribution 7. It has more dispersion than the standard normal distribution Critical Values of t-Distribution The critical values of t-distribution at level of significance  and required degree of freedom . For two tail test are given by CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 307 P[|t|] > tv () =  or P[|t|  tv ()] = 1 –  The values of t-distribution have been tabulated by Fisher and Yates for degrees of freedom  and different levels of significance . These values figure at the end of this book. If the calculated value of |t| is found greater than the book estimated value for a given degree of freedom, the hypothesis is rejected Applications of t-Distribution This distribution can be used in business decisions for the following purposes 1. t-test for significance of single mean when population variance is not known. 2. t-test for significance of the difference between two sample means, when population variances are equal but not known. 3. t-test for significance of an observed sample correlation coefficient. t-Tests for Single Mean We formulate the null hypothesis H0 :  = 0 (Poulation mean = 0) i.e., when there is significant difference between sample mean and the population mean and the fluctuation is sample fluctuation. Under H0, we test the t-statistic t = x  0 S/ n x 1 2  n xx  x S2  where = and = n 1 . It follows t-distribution with (n – 1) degrees of freedom. The calculated value of t can then be compared with the tabulated values of t for (n – 1) degrees of freedom at certain level of significance. If the calculated value of |t| is greater than the tabulated value of t, the difference is significant and H0 is rejected and vice-versa. Prof. R.A. Fisher, later defined hiw own t-distribution as CU IDOL SELF LEARNING MATERIAL (SLM)

308 Quantitative Techniques for Managers X t = Y where X ~ N(0, 1) n and Y is an independent 2-variate with n degrees of freedom. It follows the student’s t-distribution with n degrees of freedom and 1 p(t) = n   1 , n  t2  n1 / 2 ; –< t <.  2 2  1  n  Confidence limits for  are given by S  = x ± ta () × n These can be used for testing of hypotheses. We can use the t-statistic in numerical problem as x  0 S 2 / n  1 ~ t(n–1) t=  x  0 = S 2 /n t-Test for Difference of Means When we use two independent random samples as x1, x2, ... xn and y1, y2, .... yn from the given Normal populations having same mean, we set the null hypothesis as H0 : x = y. Under the assumptions 2x  2y  2 . The test-statistic under these situation is x y t = S/ 1 + 1 n1 n2     1  2  and S2 = n1  n2  2  2 y y is an unbiased estimate of common xx  population variance 2 of the two samples. We can now compare the calculated value of |t| for (n1 + n2 – 2) degrees of freedom at a given CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 309 level of significance. Null hypothesis H0 can be accepted if calculated value of |t| is within the acceptance zone or else rejected. Paired t-Test for Difference of Means A very special situation can be considered when the sample sizes of two samples drawn from a Normal population with equal means are same i.e., n1 = n2 = n and the two samples are not completely independent. In such cases, we take di = xi – yi. Under the null hypothesis that the increments are just by chance and not attributable to a particular cuase. Then H0 : x = y The best statistic dd t = S / n = S2 / n Where di = xi – yi and = d =n1 n di i=1    and 1  2 1  d 2  d d   S2 = n 1 n 1  d 2 n    t-Test for Significance of Observed Sample Correlation Coefficient When we have a random sample (xi, yi) of size n drawn from a bivariate Normal Distribution and let r be the correlation coefficient, then our problem is to find if this correlation coefficient r is significant of any correlation between the variables in the population. Prof. R.A. Fisher proved that under the Null hypothesis H0 :  = 0, i.e., the variables are un- correlated in the population. We use statistic  r t = 1 r2 n  2 ~ tn–2 t-follows the student’s t-distribution with (n – 2) degrees of freedom when n is the sample size. 95% confidence limit for , CU IDOL SELF LEARNING MATERIAL (SLM)

310 Quantitative Techniques for Managers 1 r2  r ± t0.05(n – 2) × SE(r) = r ± t0.05 (n – 2) ×  n  Fisher's Z-Distribution We have already utilised the t-test for null hypothesis Ho :  = 0, to establish that the variables are uncorrelated in the population. In sampling from a bivariate Normal population, in which variables are correlated, the distribution of statistic t is not normal even for large samples, Hence R.A. Fisher suggested a transformation from r to a new variable Z (Hence we call it as Z-Transformation also), such that Z = 1 loge  1 + r  2  1  r  Prof. Fisher proved that even for small samples, the distribution of Z will approximately be normal with mean 1 1+   = 2 log 1   and variance 1/(n – 3) i.e., Z = 1 log  1 +   N  , 1  2  1    n  3 This transformation can be used for testing 1. If the correlation coefficient in the population has a specified value Z 2. If r differes significantly from = 0, then t-test may be preferred i.e., U = 1/ n 3 ~ N(0,1) 3. If two independent sample correlation coefficients r1 and r2 differ significantly Z1  Z2 Z= 1 1 ~ N(0,1)  n1  3   n2  3 CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 311 By comparing the value of Z with 1.96 or with 2.58, we may accept or reject the null hypothesis H0 at 5% or 1% level of significance respectively. Students t-distribution : While discussing the sampling distribution of mean, two assumptions were made : (a) the mean of the population (m) is known, and (b) the standard deviation of the population (s) is also known. This means that the sample size n is large enough to justify the above assumption and that s can be replaced by the sample mean (s). If we are not in a position to know m and s. i.e. if the sample size is not large, we can define a new variable (Students t-variable), given by the relationship. x t= s/ n where  2 (SD of the sample) s = xi  x n 1 x When we compare t-distribution with standard normal variate Z = / n we can observe that t differs from Z, such that t contains sample deviation whereas Z contains the population deviation. By this explanation, we can draw a benefit that t-distribution can be used for analysis even with the knowledge of sample mean (s) only. In practice t-distribution is very close to Z-distribution, except when sample size n is very small. Similar to Z-tables, we have t-distribution tables also available, indicating various values of t for different degrees of freedom. The number of degrees of freedom for a sample size n is (n - 1). The probability density curve of the t-distribution is uninodal and symmetric about t = 0. The variability of t is higher than that of Z and it decreases as the degrees of freedom increase. In that case, with increasing degrees of freedom, t-distribution tends to coincide with the normal distribution curve, as given in Fig. 11.3 CU IDOL SELF LEARNING MATERIAL (SLM)

312 Quantitative Techniques for Managers Fig. 11.3. Normal and t-distribution curve This distribution has been used for testing of hypothesis. 11.4 Solved Problems Problem 1 While checking the quality of a product, one particular dimension was varying slightly due to changes in the machine setting, though  and were not much at variation. The target value of mean  was 0 = 50 and  = 2.5 Dimensions checked by the inspector based on sample procedure were 43, 51, 50, 41, 53, 52, 47, 54, 51, 45, 48 and 47. Formulate the null hypothesis and test the same. Solution: The target value of mean is 50. Hence null hypothesis will be H0 : = 50 If the production is to be guarded against decreasing value of , the alternative hypothesis will be H1 :  < 50 For testing the hypothesis, we work out the Standard Normal variate Z. Z= x  0 / n Here x = x = 43  51 50  41......47 n 12 = 48.5 Now x = 48.5, n = 12 and  = 2.5 CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 313 Hence 48.5  50 Z = 2.5/ 12 = – 2.078 For a 5% level of significance, the critical region will be R : Z  – 1.645 But Z = – 2.078 < – 1.645 This means that  is significantly less than expected. Hence at 5% level of significance H0 is rejected. Problem 2 In a factory, the following null hypothesis is formulated for its defect levels. H0 :   60 H1 : > 60 From an inspection report, the samples showed the following values, n = 16, = 12 and x = 62. Test the hypothesis. Solution For the given values, Normal Variate x Z = / n 62  60 2  4 = 12 / 16 = 12 8 = 12 = 0.667 At 5% level of significance Rejection region will be R : 1.64. The value of Z being less than the Rejection region, the value falls in the acceptance region. Hence H0 is accepted. CU IDOL SELF LEARNING MATERIAL (SLM)

314 Quantitative Techniques for Managers Problem 3 In a restaurant, the average sales of Pizzas is 200 per day. Due to a new office building in the vicinity, the sales increased during first 27 days, and these were found to be 205, 215, 216, 220, 225, 236, 240, 241, 245, 250, 216, 240, 238, 204, 217, 219, 225, 235, 196, 193, 215, 168, 190, 216, 218, 222, 219. Discuss that the sales of Pizzas have increased. Solution The average sales figures are 200 per day. The hypothesis are H0 :  = 200 H1 : > 200 For the sample sales for 27 days x = x = 5924 27 n = 219. The standard deviation of sample 2 xi  x  s = n 1 196  16  9  1  36  289  441  484  676  961  9  441  361  225  4 = 0  36  256  529  676  16  2601 841  9  1  9  0 27 1 = 18.73 The calculated value of the test statistic will be x   219  200 t = s / n = 18.73/ 27 CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 315 19  5.2 = 18.73 = 5.27 At 5% level of significance, the rejection region is |t| 1.711 (table for t-distribution) Since value of t falls beyond the rejection region, the hypothesis is rejected in favour of null hypothesis. Problem 4 From a certain process, it was concluded that on the average there are 15 per cent defectives. The new material purchased was used in the process and it was noticed that out of total output of 400 units, 48 were found to be defective. Would you accept the new material? Solution 48 From given data p = 400 = 0.16 The test statistic PP Z = pq n 0.15  0.16  20 = 0.16  0.84 0.2 = 0.367 = – 0.545 For 5% level of significance, the rejection region is R : Z – 1.65. The tested statistic value is not within this range of rejection. Hence, the hypothesis is accepted. CU IDOL SELF LEARNING MATERIAL (SLM)

316 Quantitative Techniques for Managers Problem 5 Before an increase in excise duty on tea, 400 people out of a sample of 500 persons were found to be tea drinkers. After an increase in duty, 400 people were tea drinkers in a sample of 600 people. Using standard error of proportion, state whether there is a significant decrease in the consumption of tea. [Delhi University, M.Com., 1978, C.A. (Inter), May 1977] Solution With Standard Notations, given are n1 = 500 and n2 = 600 400 400 p1 = 500 = 0.8 and p2 = 600 = 0.67 The null hypothesis is H0 : p1 = p2 Alternative hypothesis H1 : p1 > p2 or H1 : p1 < p2 Under the null hypothesis, P1  P2 P1  P2 pq  1 + 1   n1 n2  Z = SE  P1  P2  = Where p= n1 p1  n2 p2 400 + 400 q= n + n2 = 500 + 600 = 0.73 1 – p = 0.27 0.8  0.67 Z= 0.73  0.27   1  1   500 600  = 4.81 For 5% level of significance Z > 1.645 and for 1% level of significance Z > 2.33 Calculate value of Z is even greater than 3 (one tail test). Hence it is highly significant change in tea drinking. The null hypothesis is rejected. CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 317 Problem 6 The average number of defective articles in a certain factory is claimed to be less than the average for all the factories. The average for all the factories is 30.5. A random sample of 100 defective articles showed the following distribution. Class Limits Number 16 - 20 12 21 - 25 22 26 - 30 20 31 - 35 30 36 - 40 16 Calculate the mean and the standard deviation of the sample and use it to test the claim that the average is less than the figure for all the factories at 5% level of significance. Given Z(– 1.645) = 0.95. [C.A. (Inter), May 1979] Solution : Null hypothesis H0 : m = 30.5. Alternative H1 : m < 30.5 TABLE FOR COMPUTING MEAN AND SD OF THE SAMPLE. Class limits Number Mid pt d  x  28 fd fd2 (f) (x) 5 16 - 20 12 18 –2 –24 48 21 - 25 22 23 –1 –22 22 26 - 30 20 28 0 00 31 - 35 30 33 1 30 30 36 - 40 16 38 2 32 64 f = 100 = n fd =16 fd2 = 164 CU IDOL SELF LEARNING MATERIAL (SLM)

318 Quantitative Techniques for Managers Hence 5  16 x = 28 – 100 = 28.8 s = i×  fd    fd 2  n n   164  16  2 100  100 = 5×  = 5 × 1.27 = 6.35 Since sample is very large 2 = s2   = 6.35 For large sample, we can use normal variate relationship  Z= x  28.8  30.5 = – 2.67 / n 6.35/ 100 Since Z < –1.96, the difference is significant at 5% level of significance for a single tail test. Hence we reject the null hypothesis which means that the claim of the factory about its low defectives level is valid. Problem 7 The mean height of 50 male students who showed above average participation in college athletics was 68.2 inches with a standard deviation of 2.5 inches while 50 male students who showed no interest in such participation had a mean height of 67.5 inches with a standard deviation of 2.8 inches. Test the hypothesis that male students who participated in college athletics are taller than other male students. [Delhi University, M.A. (Eco.), 1981] Solution Given here, n1 = 50, x = 68.2, Sx = 2.5 CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 319 n2 = 50, y = 67.5, Sy = 2.8 Null hypothesis Ho : x = y Alternative hypothesis H1 : x > y (right tail test) Calculating the value of Z, Z= x y Sx2  S y2 at N (0,1) n1 n2 68.2  67.5 =  6.25 + 7.84  = 1.3188  50 50  Since, Z < 1.96, the difference in the heights of students of two categories is not significant at 5% level of significance. Hence, the Null hypothesis is accepted as the data is consistent with Null hypothesis. Problem 8 In a class, there are 30 boys and 20 girls. These students are selected for getting into the bus for the picnic according to their pattern of arrival as given below. G, B, G, G, G, B, B, B, G, B, G, B, B, G, G, G, B, G, G, B, B, G, B, B, B, G, B, B, G, G, B, B, G, G, B, B, B, G, G, B, B, B, B, G, B, B, B, B, B, B. From this sequence of arrival, can we conclude, if the arrival pattern is random? Solution In this case, we have to establish the null hypothesis that the students arrive in a random manner. Here n1 = 30 Boys n2 = 20 Girls. r = number of runs = 24 CU IDOL SELF LEARNING MATERIAL (SLM)

320 Quantitative Techniques for Managers Since, n1 and n2 are greater than 10, the normal approximation can be used for sample distribution. E(r) = 2n1n2 n1 + n2 +1 2  30  20 = 30  20 +1 = 25 SD(r) = Var(r) 2n1n2 (2n1n2  n1  n2 ) =  n1  n2 2 (n1  n2  1) 2  30  20 2  30  20  30  20 = 30 + 202 (30  20  1) 1200  1150 = 2500  49 = 3.35 Z statistic = r  E(r) = 24  25 Var(r) 3.35 = – 0.299 For 5% level of significance, acceptance region is – 1.645 Z. Hence, the calculated value of Z falls in acceptance region. Hence, H0 is accepted. Problem 9 Ten Cartons are taken at random from an automatic filling machine. The mean net weight of the 10 cartons is 11.8 oz and the standard deviation is 0.15 oz. Does the sample mean differ significantly from the intended weight of 12 oz? you are given for v = 9, t0.05 = 2.26. [Delhi University, MBA, '77] Solution CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 321 Given in the problem, we have n = 10, x = 11.8 oz, s = 0.15 oz and m = 12 oz. We formulate the null hypothesis as H0 :  = 12 oz. Alternative hypothesis H1 :   12 oz. For testing, we work out t-statistic (small sample size) x  t = S2 /n x = S2 / n 1 11.8  12 = 0.15/ 10 1 = – 4.0 The tabulated value of t for (n – 1) i.e., v = 9 degrees of freedom is given to us as 2.26 at 5% level of significance. Since calculated |t| is greater than the tabulated value, this is outside the acceptance zone. Hence H0, the null hypothesis is rejected at 5% level of significance, which means that the sample mean differs significantly from the population mean,  = 12 oz. Problem 10 The means of two random smaples of size 9 and > are 196.42 and 198.82 respectively. The sum of the squares of the deviations from the mean are 26.94 and 18.73 respectively. Can the samples be considered to have been drawn from the same normal population? [AIMA Dip. in Mgmt., August 1979] Solution The given information can be written as a fn1 = 9; x = 196.42 and  x  x 2 = 26.94 CU IDOL SELF LEARNING MATERIAL (SLM)

322 Quantitative Techniques for Managers a fn2 = 7; y = 198.82 and  y  y 2 = 18.73 Now Null hypothesis is formualted as Null hypothesis H0 : x = y Alternative hypothesis H1 : x  y Under the null hypothesis, the t-statistic is given by x y t = S/ 1  1 n1 n2 1 2 2  S2 = n1 + n2  2  xx      and y y = 9 + 1 2 26.94  18.73 7 = 3.26 196.42 198.82 Hence t= 3.26 /  1  1   9 7  = – 2.64 Tabulated value of t for (n1  n2  2) = 14 degrees of freedom at 5% level of significance is 2.15 Since calculated |t| is greater than tabulated value of t, the null hypothesis is rejected. Problem 11 An IQ test was adminstered to 5 persons before and after they were trained. The results are given below Candidates :I II III IV V IQ before training : 110 120 123 132 125 CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 323 IQ after training : 120 118 125 136 121 Test whether there is any change in IQ after the training programmes given t0.01(n) = 4.6 [C.A. (Inter), May 1980] Solution We have two dependent small samples of sizes n1 = n2 = 5 and these are paired together due to cause of training. We wish to test the null hypothesis H0 :1 = 2. d The test statistic t =  S/ n Given here d1 = 110 – 120 = – 10 Similarly d2 = 2, d3 = –2, d4 = –4, d5 = 4  d =  d  10  2  2  4  4 n5 = –2  d =2  and 1  d 2  d 2  S2 =  n 1  n     LNM QPO= 1 140  100 = 30 45 |d| 2  |t| = S / n = 30 / 5 = 0.816 The calcualted value of t is 0.816, which is much smaller than the tabulated value given for 1% level of significance at 4 degrees of freedom. Hence the Null hypothesis is accepted i.e., there is no significant change in the IQ before and after the training. Problem 12 A random sample of 27 pairs of obsrevations from a normal population gives a correlation coefficient of 0.42. Is it likely that the variables in the population are uncorrelated? [Bombay University, B.Com., 1976] CU IDOL SELF LEARNING MATERIAL (SLM)

324 Quantitative Techniques for Managers Solution Given in the problem n = 27 and r = 0.42 We get null hypothesis as H0 : = 0 and H1 :  0. We calculate test statistic t= r n2  0.42 25 = 2.31 1 r2 1  (0.42)2 From the tables, the t0.05 for 5% level of significance at 25 degrees of freedom is 2.06, which is less than the calculated t-value. Hence the null hypothesis is rejected at 5% level of significance. Problem 13 Using the data in problem 15.25, test the hypothesis using Wilcoxon Signed rank test. Solution The given data can be organised in the following manner. xi : 5 7 15 10 8 20 16 18 12 14 Di : –10 3 –3 –1 Ranks : –8 0 –5 –7 5 1 3.5 3.5 1.5 9 8 5.5 7 5.5 1.5 Now s+ = 5.5 + 1.5 + 3.5 = 10.5 s– = 9 + 8 + 5.5 + 7 + 3.5 + 1.5 = 34.5 Here s = smaller value of s+ i.e. s = 10.5 Since n(n +1) s+ < s– , Us++ or Us = 4 z 9(9 +1) Since = 4 = 22.5 s  U s– = Ts– = 22.5 Ts2 = n(n +1)(2n +1) 9(9  1)(18  1)  24 24 CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 325  Ts = 7.8 3.35 – 22.5  z = 7.8 = 1.53 Since value of z at 0.05 level of significance is 1.64 the statistic completed is written the acceptance zone and hence H0 is accepted Problem 14 If we have the following data from a sample-number of products manufactured during the day each hour = 55, 57, 65, 62, 58, 64, 63, 70. If the mean production is 60, calculate the value of t. Solution Given x1 = 55, x2 = 57, x3 = 65 and so on. x =  x1 + x2 + ...... x8   8  55  57  65  62  58  64  63  70 = 8 = 61.75  2 s = xi  x n 1 6.752  4.752  3.252  (0.25)2  3.752  2.252  1.252  (8.25)2 = 81 45.56  22.56  10.56  0.625  14.06  5.06 1.56  68.05 = 7 = 4.9 x t= S/ n CU IDOL SELF LEARNING MATERIAL (SLM)

326 Quantitative Techniques for Managers 61.75  60 = 4.9/ 8 1.75  2.82 = 4.9 = 1.00 The degrees of freedom = n – 1 = 8 – 1 = 7 Problem 15 If we have the following data from a sample-number of products manufactured during the day each hour as 55, 57, 65, 62, 58, 64, 63, 70. If the mean production is 60, calculate the value of t. Solution Given x1 = 55, x2 = 57, x3 = 65 and so on. x =  x1 + x2 + ...... x8   8  55  57  65  62  58  64  63  70 = 8 = 61.75  2 s = xi  x n 1 6.752  4.752  3.252  (0.25)2  2.752  1.252  (8.25)2 = 81 45.56  22.56  10.56  0.625  7.56  5.06 1.56  68.05 = 7 = 4.9 x t = S/ n CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 327 61.75  60 = 4.8/ 8 1.75  2.82 = 4.8 = 1.02 The degrees of freedom = n – 1 = 8 – 1 = 7 11.5 Summary The unit is summarised by some of its important points as below:  Alpha (a): The Probability of a type 1 error.  Alternative hypothesis :The conclusion drawn when the given data does not support the provision of null-hypothesis.  Beta (b): The probability of a type II error.  Confidence Interval: A range of values that have some specified probability of including the true population parameter value.  Confidence level: The probability that is associated with the interval estimate of a population statistic, including how reliable these are for inclusion in the population parameter.  Confidence limits: The upper and lower boundaries of a confidence interval.  Critical value: The value of the standard statistic, beyond which the null hypothesis is rejected.  Degree of Freedom: The number of values in a sample we can specify freely once something is known about the sample.  Estimate: A specific value of the estimator.  Estimator: A sample statistic used to estimate a, population parameter.  Hypothesis: An assumption adopted regarding the population parameter.  Significance level: A value denoting the percentage of sample values outside certain limits, assuming that the null-hypothesis is correct. CU IDOL SELF LEARNING MATERIAL (SLM)

328 Quantitative Techniques for Managers  Students t - distribution: It is a probability distribution used when sample size is small (x < 30) and the population standard deviation is not known.  Type I error: Rejecting a nult hypothesis when it is true.  Type II error: Accepting a nult hypothesis when it is false.  Two - sample test: Hypothesis test based on samples taken from two populations in order to compare their means or proportions.  Unbiased estimator: An estimator of a population parameter that, on average, assumes values above the population parameter and to some extent even below the population parameter. 11.6 Key Words/Abbreviations  Alpha (a): The Probability of a type 1 error.  Beta (b): The probability of a type II error.  Confidence limits: The upper and lower boundaries of a confidence interval.  Degree of Freedom: The number of values in a sample we can specify freely once something is known about the sample.  Estimate: A specific value of the estimator.  Estimator: A sample statistic used to estimate a, population parameter. 11.7 Learning Activity 1. Two laboratories carry out independent estimates of particular chemical in a medicine produced by a certain firm. A sample is taken from each batch, halved and the seperate halves sent to the two laboratories. The following data is obtained. No. of samples 10 Mean value of the difference of estimates 0.6 Sum of the squares of the differences (from their mean) 20 Is the difference significant? (Value of t at 5% level for 9 d.f is 4.262) 2. A correlation coefficient of 0.2 is discovered in a sample of 28 pairs. Use Z test to find out, if this is significantly different from zero. CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 329 11.8 Unit End Questions (MCQ and Descriptive) A. Descriptive Type Questions 1. Define Student’s t-Statistic and write its probability density functions. 2. (a) Explain the t-test for testing the significance of the difference between two sample means. State the assumptions while formulating it. (b) Explain the t-test for testing the significance of an observed sample correlation. 3. A population consists of five numbers (2, 3, 6, 8, 11). Consider all possible samples of size two which can be drawn with replacement from this population. Calculate the standard error of the sample mean. 4. A random sample of 100 items taken from a large batch of articles contains 5 defective items. (a) set up 96 percent confidence limits for the proportion of defective items in the batch. (b) If batch contains 2,696 items, set up 95% confidence interval for the proportion of defective items. 5. In a sample of 400 oranges from a large consignment 40 were considered bad. Estimate the percentage of defective oranges in the whole consignment and assign limits within which the percentage will probably lie. 6. A life insurance company has 1,500 policies averaging Rs. 2,000 on lives at age 40. From the experience table, it is found that of 1,00,000 alive at 30, 90,000 were alive at age 31. Find the lower and upper values of the amount the company will have to pay out in insurance during the year. 7. In a sample of 900 stockholders of companies, 400 stated that their major aim in holding stocks is capital appreciation. What is the 90% confidence range within which lies the population proportion of stockholders who hold stocks for capital appreciation. (Area between t = 0 and t = 1.64 is 45% where t is Standard Normal Variate). 8. In a random sample of 400 persons from a large population, 120 were females, can it be said that males and females are in the ratio 5:3 in the population? Use 1% level of significance. CU IDOL SELF LEARNING MATERIAL (SLM)

330 Quantitative Techniques for Managers 9. In a sample of 400 parts manufactured by a factory, the number of defective parts was found to be 30. The company, however, claimed that only 5% of their product is defective. Is the claim reasonable. 10. A die was thrown 9,000 times and of these, 3,220 yielded a 3 or 4. Is this consistent with the hypothesis that the die was unbiased. 11. If a coin is tossed at random 400 times and tail turns up 160 times, can the coin be regarded as unbiased? 12. A dice is thrown 49,152 times and of these 25,145 yielded either 4 or 5 or 6. Is this consistent with the hypothesis that the die must be unbiased? 13. A candidate at an election claims 90% of support of all voters in a locality. Verify his claim if in a random sample of 400 voters from the locality, 320 supported his candidature. Use 5% level of significance. 14. In a sample of 600 students of a certain college, 400 are found to use dot pens. In another college, from a sample of 900 students, 450 were found to use dot pens. Test whether the two colleges are significantly different with respect to the habit of using dot pens. (Null and alternative hypothesis should be stated clearly). 15. A machine produced 20 defective articles in a batch of 400. After overhauling, it produced 10 defectives in a batch of 300. Has the machine improved? 16. While throwing 5 dice 30 times, a person obtained success 23 times, securing a 6 which was considered a success. Can we consider the difference between the observed and the expected results as being significantly different? 3 17. A firm found with the help of a sample survey of a city (size of the sample 900), that th of 4 the population consumes things produced by them. A firm then advertised the goods in paper and on radio. After one year, a sample of size 1,000 reveals that proportion of 4 consumers of the goods produced by the firm is 5 th. Is this rise significant to indicate that the advertisement was 18. A candidate for an election from a large constituency thinks that he will win the election if at least 45% of the electorate vote for him. He therefore, conducts the sample survey to enable him to decide whether he should stand for the election or not. The survey covers 10,000 voters and it is found that 4,420 voted for him. Advise him as to whether he should CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 331 stand for the election, stating clearly the level of significance and other assumptions on which you base your conclusion. 19. The sizes of components produced by a machine are normally distributed. It is required that the size should be between 15.63 cm and 15.84 cm and it is known that 2.872% of the production is rejected for being oversize and 1.072 of the production is rejected for being undersized. (a) Calculate the mean of the component sizes. (b) Obtain the SD of the component sizes. (c) Calculate the range that would contain the middle 50% of the production. 20. A random sample of 400 items is found to have a mean of 82 and standard deviation of 18. Find 95% confidence limits for the mean of the population from which the sample is drawn. 21. The average number of defective articles in a certain factory is claimed to be less than the average for all the factories. The average of all the factories is 30.5. A random sample of 100 defective articles showed the following distribution. Class limits Number 16 - 20 12 21 - 25 22 26 - 30 20 31 - 35 30 36 - 40 16 Calculate the mean and standard deviation of the sample and use it to test the claim that the average is less than the figure for all the factories at 5% level of significance. Given Z (– 1.645) = 0.05. 22. A sample of 400 male students is found to have a mean height of 171.38 cm. Can it be reasonably regarded as a sample from a large population with mean height of 171.17 cm. and standard deviation of 3.30 cm. 23. A sample of size 400 was drawn and the sample mean was found to be 99. Test whether this sample could have come from a normal population with mean 100 and variance 64 at 5% level of significance. CU IDOL SELF LEARNING MATERIAL (SLM)

332 Quantitative Techniques for Managers 24. The mean breaking strength of the cables supplied by a manufacturer is 1,800 with a standard deviation 100. By the new technique in the manufacturing process, it is claimed that the breaking strength of the cables have increased. In order to test the claim, a sample of 50 cables is tested. It is found that the mean breaking strength is 1850. Can we support the claim at 1% level of significance? 25. An educator claims that the average IQ of American College Students is at most 110, and that in the study made to test this claim, 150 American college students selected at random had an average IQ of 111.2 with standard deviation of 7.2. Use a level of significance of 0.01 to test the claim of the educator. 26. A sample of 100 workers in a large plant gave a mean assembly time of 294 seconds, with a standard deviation of 12 seconds in a time and motion study. Provide a 95% confidence interval for the mean assembly time for all the workers in the plant. 27. A machine is designed to produce insulating washers for electrical devices of average thickness 0.025 cm. A random sample of 10 washers was found to have an average thickness of 0.024 cm. with a standard deviation of 0.002 cm. Test the significance of the deviation. Values of t for 9 degrees of freedom at 5% level is 2.262. 28. Two types of batteries are tested for their length of life and the following data are obtained. No. of samples Mean life in Hours Variance TypeA : 9 600 121 Type B : 8 640 144 Is there a significant difference in the two means? Value of t for 15 degrees of freedom at 5% level is 2.131. 29. A group of 5 patients treated with medicine A, weigh 42, 39, 48, 60 and 41 kgs. Second group of 7 patients from the same hospital treated with medicine B weigh 38, 42, 56, 64, 68, 69 and 62 kgs. Do you agree with the claim that medicine B increases the weight significantly? (The value of t at 5% level of significance for 10 d.f is 2.2281). 30. A certain stimulus administered to each of 12 patients resulted in the following changes in blood pressure : 5, 3, 8, –1, 3, 0, –2, 1, 5, 0, 4, 6. Can it be concluded that the stimulus will in general be accompanied by an increase in blood pressure? (Given for 11 degrees of freedom, t0.05 = 2.201) CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 333 31. Two working designs were under consideration for adoption in a plant. A time and motion study shows that 12 workers using design A have a mean assembly time of 300 seconds with a standard deviation of 12 seconds and that 15 workers using design B have mean assembly time of 335 seconds with a standard deviation of 15 seconds. Is the difference in the mean assembly time between the two working designs significant at 1 percent level of significance? The following table gives some t-values, which may be used. Level of Significance Degrees of Freedom 25 26 27 a = 0.05 2.06 2.06 2.06 a = 0.01 2.79 2.78 2.77 32. A company is interested in knowing if there is a difference in the average salary received by foremen in two divisions. Accordingly samples of 12 foremen in the first division and 10 foremen in the second division were selected at random. Based upon experience, foremen's salaries are known to be approximately normally distributed and the standard deviations are about the same. First Division Second Division Sample size 12 10 Average monthly salary of foremen (in Rs.) 1,050 980 Standard Deviation of salaries (Rs.) 68 74 The table value of t for 20 d.f at 5% level of significance is 2.086. 33. Measurements performed on random samples of two kinds of cigarettes yielded the following results on their nicotine content (in milligrams) Brand A : 21.4 23.6 24.8 22.4 26.3 Brand B : 22.4 27.7 23.5 29.1 25.8 Use the 1 percent level of significance to check on the claim that Brand B has a higher average nicotine content than Brand A. Memory capacity of 10 students was tested before and after training. State whether training was effective or not from the following scroes: Roll No. : 1 2 3 4 56 7 8 9 10 05 6 Before training : 12 14 11 8 7 10 3 CU IDOL SELF LEARNING MATERIAL (SLM)

334 Quantitative Techniques for Managers After training : 15 16 10 7 5 12 10 2 3 8 (For n = 9, t0.05 = 2.26) 34. The diameter of the shaft is on the average 20 cm. We select random samples of 15 shafts choosing a shaft at every hour of working. If the measurements obtained are 20.56, 20.89, 20.5, 19.69, 19.90, 20.35, 19.58, 20.05, 19.96 and 20.85, calculate the t-values. B. Multiple Choice/Objective Type Questions 1. Approximately what percentage of people would have scores lower than an individual with a z-score of 1.65 in a normally distributed sample? (a) 95% (b) 98% (c) It is not possible to calculate this unless the mean and standard deviation are given. (d) 1% 2. In a situation where the population standard deviation is known and we wish to estimate the population mean with 90 percent confidence, what is the appropriate critical value to use? (a) z = 1.96 (b) z = 2.33 (c) z = 1.645 (d) Can't be determined without knowing the degrees of freedom 3. In developing a confidence interval estimate for the population mean, the t-distribution is used to obtain the critical value when: (a) the sample contains some extreme values that skew the results. (b) the population standard deviation is unknown. (c) the sampling that is being used is not a statistical sample. (d) the confidence level is low 4. A study was recently conducted to estimate the mean cholesterol for adult males over the age of 55 years. From a random sample of n = 10 men, the sample mean was found to be 242.6 and the sample standard deviation was 73.33. To find the 95% confidence interval estimate for the mean, the correct critical value to use is: CU IDOL SELF LEARNING MATERIAL (SLM)

Hypothesis Testing Part - II 335 (a) 1.96 (b) 2.2281 (c) 2.33 (d) 2.2622 5. A concern of Major League Baseball is that games last too long. Some executives in the league's headquarters believe that the mean length of games this past year exceeded 3 hours (180 minutes). To test this, the league selected a random sample of 80 games and found the following results: minutes and s = 16 minutes. Based on these results, if the null hypothesis is tested using an alpha level equal to 0.10, which of the following is true? (a) The null hypothesis should be rejected if . (b) The test statistic is t = 1.2924. (c) Based on the sample data, the null hypothesis cannot be rejected. (d) It is possible that when the hypothesis test is completed that a Type II statistical error has been made Answers 1. (a), 2. (c), 3. (b), 4. (d), 5. (a) 11.9 References References of this unit have been given at the end of the book. CU IDOL SELF LEARNING MATERIAL (SLM)

REFERENCES 1. Stigler, Cambidge M.A., “The History of Statistics”, 1986, Belknap Press. 2. Grant, L.E. and R.C. Leavenworth, “Statistical Quality Control”, 1996, McGraw-Hill Book Co. 3. Tufte, E.R., “The Visual Display of Quantitative Information”, 1983, Graphics Press. 4. Rowntree, D., “Probability” 1984, Charles Scribner’s Sons. 5. Levin, R.I. and D.S. Rubin, “Statistics for Management”, 1997, Prentice Hall (India). 6. Gupta, S.C., “Fundamentals of Statistics”, Himalaya Publishing House, Mumbai.