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Hypothesis Testing Keller’s powerpnt modified by Tony WONG 1 Inference About the Description of a Single Population 2 11.1 Introduction • In this chapter we utilize the approach developed before for making statistical inference about populations. – Identify the parameter to be estimated or tested . – Specify the parameter’s estimator and its sampling distribution. – Construct an interval estimator or perform a test. 3 • We will develop techniques to estimate and test three population parameters. – The expected value m – The variance s2 – The population proportion p (for qualitative data) • Examples – A bank conducts a survey to estimate the number of times customer will actually use ATM machines. – A random sample of processing times is taken to test the mean production time and the variance of production time on a production line. 4 11.2 Inference About a Population Mean When the Population Standard Deviation Is Unknown • Recall that when s is known x is normally distributed – If the sample is drawn from a normal population, or if – the population is not normal but the sample is sufficiently large. • When s is unknown, we use its point estimator s, and the Z statistic is replaced then by the t-statistic 5 ZZZt tm t t t ZZ x Z ttt t t tt t x m s n ss ss s ss ss s sssss n ss s s When the sampled population is normally distributed, the statistic t is Student t distributed. The “degrees of freedom”, The t distribution is mound-shaped, a function of the sample size and symmetrical around zero. determines how spread the distribution is (compared to the d.f. = n2 normal distribution) d.f. = n1 n1 < n2 0 6 Probability calculations for the t distribution – The t table provides critical value for various probabilities of interest. – The form of the probabilities that appear in table 4 Appendix B are: P(t > tA, d.f.) = A – For a given degree of freedom, and for a predetermined right hand tail probability A, the entry in the table is the corresponding tA. – These values are used in computing interval estimates and performing hypotheses tests. 7 A = .05 tA Degrees of Freedom t.100 t.05 t.025 t.01 t.005 1 3.078 6.314 12.706 31.821 63.657 2 1.886 2.92 4.303 6.965 9.925 . . . . . . . . . . . . 20 1.325 1.725 2.086 2.528 2.845 . . . . . . . . . . . . 200 1.286 1.653 1.972 2.345 2.601 1.282 1.645 1.96 2.326 2.576 8 Testing the population mean when the population standard deviation is unknown • If the population is normally distributed, the test statistic for m when s is unknown is t. x m t s n • This statistic is Student t distributed with n-1 degrees of freedom. 9 • Example 11.1Trainees productivity 10 • Example 11.1 Trainees productivity – In order to determine the number of workers required to meet demand, the productivity of newly hired trainees is studied. – It is believed that trainees can process and distribute more than 450 packages per hour within one week of hiring. – Can we conclude that this belief is correct, based on productivity observation of 50 trainees, See file XM11-01. 11 • Solution – The problem objective is to describe the population of the number of packages processed in one hour. – The data are quantitative. H0:m = 450 H1:m > 450 – The t statistic x m t s n d.f. = n - 1 = 49 12 – Solving by hand • The rejection region is t > ta,n - 1 • ta,n - 1 = t.05,49 = approximately to 1.676. • From the data we have x i 23,019 x i2 10,671,357, thus 23,019 x 460.38, and 50 x 2 s2 x i2 n i 1507.55. n 1 s 1507.55 38.83 13 Rejection region • The test statistic is 1.676 1.89 x m 460.38 450 t 1.89 s n 38.83 50 • Since 1.89 > 1.676 we reject the null hypothesis in favor of the alternative. • There is sufficient evidence to infer that the mean productivity of trainees one week after being hired is greater than 450 packages at .05 significance level. 14 Test of Hypothesis About MU (SIGMA Unknown) Data 505 Test of MU = 450 Vs MU greater than 450 400 Sample standard deviation = 38.8271 499 Sample mean = 460.38 415 Test Statistic: t = 1.8904 .05 418 P-Value = 0.0323 . .0323 . . • Since .0323 < .05, we reject the null hypothesis in favor of the alternative. • There is sufficient evidence to infer that the mean productivity of trainees one week after being hired is greater than 450 packages at .05 significance level. 15 Estimating the population mean when the population standard deviation is unknown • Confidence interval estimator of m when s is unknown s x ta 2 d.f . n 1 n 16 • Example 11.2 – An investor is trying to estimate the return on investment in companies that won quality awards last year. – A random sample of 50 such companies is selected, and the return on investment is calculated had he invested in them. – Construct a 95% confidence interval for the mean return. 17 • Solution – The problem objective is to describe the population of annual returns from buying shares of quality award-winners. – The data are quantitative. – Solving by hand 2 • From the data we determine x 14.75 s 66.90 s 8.18 s 8.18 x ta 2 14.75 2.009 12.43,17.07 n 50 18 0.95 Confidence Interval Estimate of MU (SIGMA Unknown) Data 18.58 Sample mean = 14.7522 10.35 Sample standard deviation = 8.1793 22.41 Lower confidence limit = 12.4277 4.51 Upper confidence limit = 17.0767 . . . [12,42, 17.07] 19 Checking the required conditions • We need to check that the population is normally distributed, or at least not extremely non-normal. • There are statistical methods to test for normality (to be introduced later in the book). • Currently, we can plot the histogram of the data set. 20 14 A Histogram for XM11- 01 12 10 8 6 4 2 0 400 425 450 475 500 525 550 575 More Packages 12 A Histogram for XM11- 02 10 8 6 4 2 0 5 10 15 20 25 30 35 More Returns 21 11.3 Inference About a Population Variance • Some times we are interested in making inference about the variability of processes. • Examples: – The consistency of a production process for quality control purposes. – Investors use variance as a measure of risk. • To draw inference about variability, the parameter of interest is s2. 22 • The sample variance s2 is an unbiased, consistent and efficient point estimator for s2. (n 1) s 2 • The statistic s2 has a distribution called Chi-squared, if the population is normally distributed. (n 1)s 2 2 2 d.f . n 1 d.f. = 1 s d.f. = 5 d.f. = 10 23 The 2 table A =.01 A =.01 1 - A =.99 21-A 2 A .990 .010 2.01,10 23.2093 Degrees of freedom .995 2.990 2.975 2.010 2.005 2 1 0.0000393 0.0001571 0.0009821 . . 6.6349 7.87944 . . 10 2.15585 2.55821 3.24697 . . 23.2093 25.1882 . . . . . . . . . . . . . . 24 Estimating the population variance • From the following probability statement P(21-a/2 < 2 < 2a/2) = 1-a we have (by substituting 2 = [(n - 1)s2]/s2.) (n 1)s 2 2 (n 1)s 2 s 2 / 2 a 2 1a / 2 25 • Example 11.3 (operation management application) – A container-filling machine is believed to fill 1 liter containers so consistently, that the variance of the filling will be less than 1 cc (.001 liter). – To test this belief a random sample of 25 1-liter fills was taken, and the results recorded. – The data are provided in file XM11-03. – Do these data support the belief that the variance is less than 1cc at 5% significance level? 26 • Solution – The problem objective is to describe the population of 1-liter fills from a filling machine. – The data are quantitative, and we are interested in the variability of the fills. – The complete test is: H0: s2 = 1 H1: s2 <1 The test statistic is 2 (n 1)s 2 2 . s 2 The rejection region is 2 1a,n1 27 – Solving by hand • Note that (n - 1)s2 = S(xi - x)2 = Sxi2 - Sxi/n • From the sample (data is presented in units of cc-1000 to avoid rounding) we can calculate Sxi = -3.6, and Sxi2 = 21.3. • Then (n - 1)s2 = 21.3 - (-3.6)2/25 = 20.8. • The complete test is shown next There is insufficient evidence (n 1)s 2 20.8 to reject the hypothesis that 2 2 20.8, the variance is equal to 1cc, s 2 1 in favor of the hypothesis that 1a ,n1 .95,251 13.8484. 2 2 it is smaller. Since 13.8484 20.8, do not reject the null hypothesis. 28 a = .05 1-a = .95 Rejection region 2 13.8484 13.8484 20.8 2 .295,251 Do not reject the null hypothesis 29 11.4 Inference About a Population Proportion • When the population consists of qualitative or categorical data, the only inference we can make is about the proportion of occurrence of a certain value. • The parameter “p” was used before to calculate probabilities using the binomial distribution. 30 • Statistic and sampling distribution – the statistic employed is x p ˆ where n x the number of successes. n sample size. – Under certain conditions, [np > 5 and n(1-p) > 5], ˆ p is approximately normally distributed, with m = p and s2 = p(1 - p)/n. 31 • Test statistic for p p p ˆ Z p(1 p) / n where np 5 and n(1 p) 5 • Interval estimator for p (1-a confidence level) p za / 2 p(1 p) / n ˆ ˆ ˆ provided np 5 and n(1 p) 5 ˆ ˆ 32 • Example 11.5 (marketing application) – For a new newspaper to be financially viable, it has to capture at least 12% of the Toronto market. – In a survey conducted among 400 randomly selected prospective readers, 58 participants indicated they would subscribe to the newspaper if its cost did not exceed $20 a month. – Can the publisher conclude that the proposed newspaper will be financially viable at 10% significance level? 33 • Solution – The problem objective is to describe the population of newspaper readers in Toronto. – The responses to the survey are qualitative. – The parameter to be tested is “p”. – The hypotheses are: H0: p = .12 H1: p > .12 We want to prove that the newspaper is financially viable 34 – Solving by hand • The rejection region is z > za = z.10 = 1.28. ˆ • The sample proportion is p 58 400 .145 • The value of the test statistic is ˆ pp .145 .12 Z 1.54 p(1 p) / n .12(1 .12) / 400 • The p-value is = P(Z>1.54) = .0618 There is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. At 10% significance level we can argue that at least 12% of Toronto’s readers will subscribe to the new newspaper. 35 Test of p = 0.12 Vs p greater than 0.12 Sample Proportion = 0.145 Test Statistic = 1.5386 P-Value = 0.0619 36 • Example 11.6 (marketing application) – In a survey of 2000 TV viewers at 11.40 p.m. on a certain night, 226 indicated they watched “The Tonight Show”. – Estimate the number of TVs tuned to the Tonight Show in a typical night, if there are 100 million potential television sets. Use a 95% confidence level. – Solution ˆ ˆ ˆ p z a / 2 p(1 p) / n .113 1.96 .113(.887) / 2000 .113 .014 37 Selecting the Sample Size to Estimate the Proportion • The interval estimator for the proportion is ˆ ˆ ˆ p z a / 2 p(1 p) / n • Thus, if we wish to estimate the proportion to within W, ˆ ˆ we can write W z a / 2 p(1 p) / n • The required sample size is 2 z a / 2 p(1 p) / n ˆ ˆ n W 38 • Example – Suppose we want to estimate the proportion of customers who prefer our company’s brand to within .03 with 95% confidence. – Find the sample size needed to guarantee that this requirement is met. 2 – Solution 1.96 p(1 p) ˆ ˆ n W = .03; 1 - a = .95, .03 therefore a/2 = .025, Since the sample has not yet so z.025 = 1.96 been taken, the sample proportion is still unknown. We proceed using either one of the following two methods: 39 • Method 1: ˆ – There is no knowledge about the value of p ˆ • Let p .5 , which results in the largest possible n needed ˆ for a p W 1-a confidence interval. • If the sample proportion does not equal .5, the actual W will be narrower than .03. • Method 2: ˆ – There is some idea about the value of p ˆ • Use the value of p to calculate the sample size 2 1.96 .5(1 .5) 2 n 1,068 1.96 .2(1 .2) .03 n 683 .03 40 Chapter 12 Inference about the Comparison of Two Populations 41 12.1 Introduction • Variety of techniques are presented whose objective is to compare two populations. • We are interested in: – The difference between two means. – The ratio of two variances. – The difference between two proportions. 42 12.2 Inference about the Difference between Two Means: Independent Samples • Two random samples are drawn from the two populations of interest. • Because we are interested in the difference between the two means, we shall build the statistic x for each sample (and support the analysis by the statistic S2 as well). 43 The Sampling Distribution x x of 1 2 x1 x 2 is normally distributed if the (original) population distributions are normal . x1 x 2 is approximately normally distributed if the (original) population is not normal, but the sample size is large. Expected value of x 1 x 2 is m1 - m2 The variance of x 1 x 2 is s12/n1 + s22/n2 44 • If the sampling distribution of x1 x 2 is normal or approximately normal we can write: ( x 1 x 2 ) ( m1 m 2 ) Z s12 s 2 2 n1 n2 • Z can be used to build a test statistic or a confidence interval for m1 - m2 45 • Practically, the “Z” statistic is hardly used, because the population variances are not known. ( x 1 x 2 ) ( m1 m 2 ) t Z s1122 s222 ? S2 S ? n1 n2 • Instead, we construct a “t” statistic using the sample “variances” (S12 and S22). 46 • Two cases are considered when producing the t-statistic. – The two unknown population variances are equal. – The two unknown population variances are not equal. 47 Case I: The two variances are equal • Calculate the pooled variance estimate by: (n1 1)s12 (n2 1)s 22 Sp 2 n1 n2 2 n2 = 15 n1 = 10 S 2 Sp 2 S2 2 1 Example: S12 = 25; S22 = 30; n1 = 10; n2 = 15. Then, (10 1)(25) (15 1)(30) Sp 2 28.04347 10 15 2 48 • Construct the t-statistic as follows: ( x1 x 2 ) (m1 m 2 ) t 2 1 1 sp ( ) n1 n2 d.f . n1 n2 2 • Perform a hypothesis test Build an interval estimate H0: m1 m2 = 0 H1: m1 m2 > 0; 1 1 ( x1 x 2 ) t a 2 sp ( ) 2 n1 n2 or < 0; or 0 where 1 a is the confidence level. 49 Case II: The two variances are unequal ( x1 x2 ) ( m1 m 2 ) t 2 s12 s2 ( ) n1 n2 ( s12 n1 s2 / n2 ) 2 2 d.f. 2 2 2 2 ( s n1 ) ( s n2 ) 1 2 n1 1 n2 1 50 Run a hypothesis test as needed, or, build an interval estimate Estimator 2 s1 s2 (x 1 x 2 ) t a 2 2 n1 n 2 where 1 a is the confidence level. 51 52 • Example 12.1 – Do people who eat high-fiber cereal for breakfast consume, on average, fewer calories for lunch than people who do not eat high-fiber cereal for breakfast? – A sample of 150 people was randomly drawn. Each person was identified as a consumer or a non-consumer of high-fiber cereal. – For each person the number of calories consumed at lunch was recorded. 53 Calories consumed at lunch Consmers Non-cmrs 568 705 498 819 589 706 Solution: 681 509 540 613 646 582 • The data are quantitative. 636 601 739 608 539 787 • The parameter to be tested is 596 573 the difference between two means. 607 428 529 754 637 741 • The claim to be tested is that 617 628 633 537 mean caloric intake of consumers (m1) 555 748 is less than that of non-consumers (m2). . . . . . . . . 54 • Identifying the technique –The hypotheses are: H0: (m1 - m2) = 0 H1: (m1 - m2) < 0 m1 < m2) – To check the relationships between the variances, we use a computer output to find the samples’ standard deviations. We have S1 = 64.05, and S2 = 103.29. It appears that the variances are unequal. – We run the t - test for unequal variances. 55 Calories consumed at lunch Consmers Non-cmrs t-Test: Two-Sample Assuming 568 705 498 819 Unequal Variances 589 706 Nonconsumers Consumers 681 509 Mean 604.023 633.234 540 613 Variance 4102.98 10669.8 646 582 Observations 43 107 636 601 0 Hypothesized Mean Difference 739 608 df 123 539 787 t Stat -2.09107 596 573 P(T<=t) one-tail 0.01929 607 428 t Critical one-tail 1.65734 529 754 P(T<=t) two-tail 0.03858 637 741 t Critical two-tail 1.97944 617 628 633 537 555 748 . . • At 5% significance level there is . . . . sufficient evidence to reject the null . . hypothesis. 56 • Solving by hand – The interval estimator for the difference between two means is s2 s2 (x x ) t ( 1 2) 1 2 a 2 n n 1 2 64.05 2 103.29 2 (604.02 633.239) 1.9796 43 107 29.21 27.65 57 • Example 12.2 – Do job design (referring to worker movements) affect worker’s productivity? – Two job designs are being considered for the production of a new computer desk. – Two samples are randomly and independently selected • A sample of 25 workers assembled a desk using design A. • A sample of 25 workers assembled the desk using design B. • The assembly times were recorded – Do the assembly times of the two designs differs? 58 Assembly times in Minutes Design-A Design-B 6.8 5.2 5.0 6.7 7.9 5.7 5.2 6.6 Solution 7.6 8.5 5.0 6.5 5.9 5.9 • The data are quantitative. 5.2 6.7 6.5 6.6 . . • The parameter of interest is the difference . . between two population means. . . . . • The claim to be tested is whether a difference between the two designs exists. 59 • Solving by hand (6.288 6.016) 0 t 0.93 1 1 –The hypotheses test is: 1.075( ) 25 25 H0: (m1 - m2) = 0 d.f . 25 25 2 48 H1: (m1 - m2) 0 – To check the relationship between the two variances calculate the value of S1 and S2. We have S1= 0.92, and S2 =1.14. We can infer that the two variances are equal to one another. – To calculate the t-statistic we have: Let us determine the x1 6.288 x 2 6.016 s 0.8481 s 1.2996 2 2 1 2 rejection region (25 1)(0.8481) (25 1)(1.2996) Sp 2 1.075 25 25 2 60 • The rejection region is t t a 2,d.f. t 0.025,48 2.009 Notice the absolute value For a = 0.05 • The test: Since t= 0.93 < 2.009, there is |t| insufficient evidence to reject the null hypothesis. .025 Rejection region .093 2.009 61 • Conclusion: From this experiment, it is unclear at 5% significance level if the two job designs are different in terms of worker’s productivity. .025 Rejection region .093 2.009 62 Design-A Design-B 6.8 5.2 The Excel printout 5.0 6.7 7.9 5.7 5.2 6.6 7.6 8.5 5.0 6.5 5.9 5.9 t-Test: Two-Sample Assuming Equal Variances 5.2 6.7 6.5 6.6 Design-A Design-B . . Mean 6.288 6.016 . . Variance S 1 0.847766667 2 1.3030667 S2 2 . . Observations 25 25 . . Pooled Variance 1.075416667 Sp 2 0 Hypothesized Mean Difference Degrees of freedom df 48 m1 m 2 t - statistic t Stat 0.927332603 P-value of the one tail test P(T<=t) one-tail 0.179196744 t Critical one-tail 1.677224191 P-value of the two tail test P(T<=t) two-tail 0.358393488 t Critical two-tail 2.01063358 63 A 95% confidence interval for m1 - m2 is calculated as follows: 1 1 ( x1 x 2 ) t a 2 sp ( ) 2 n1 n2 1 1 6.288 6.016 2.0106 1.075( ) 25 25 0.272 0.5896 [ 0.3176, 0.8616 ] Thus, at 95% confidence level -0.3176 < m1 - m2 < 0.8616 Notice: “Zero” is included in the interval 64 Checking the required Conditions for the equal variances case (example 12.2) 12 Design A The distributions are not 10 bell shaped, but they 8 seem to be approximately 6 normal. Since the technique 4 is robust, we can be confident 2 about the results. 0 5 5.8 6.6 7 7.4 8.2 Design B More 6 5 4 3 2 1 0 4.2 5 5.8 6.6 7.4 More 65 12.4 Matched Pairs Experiment • What is a matched pair experiment? • Why matched pairs experiments are needed? • How do we deal with data produced in this way? The following example demonstrates a situation where a matched pair experiment is the correct approach to testing the difference between two population means. 66 67 Example 12.3 • To determine whether a new steel-belted radial tire lasts longer than a current model, the manufacturer designs the following experiment. – A pair of newly designed tires are installed on the rear wheels of 20 randomly selected cars. – A pair of currently used tires are installed on the rear wheels of another 20 cars. – Drivers drive in their usual way until the tires worn out. – The number of miles driven by each driver were recorded. See data next. 68 Solution New-Design Exstng-Dsn • Compare two populations of 70 47 83 65 quantitative data. 78 59 46 61 74 75 56 65 74 52 73 85 • The parameter is m1 - m2 99 97 57 84 77 72 84 39 72 98 72 91 The hypotheses are: 81 63 64 63 H0: (m1 - m2) = 0 88 79 69 74 H1: (m1 - m2) > 0 54 76 97 43 m1 Mean distance driven before worn out occurs for the new design tires m2 Mean distance driven before worn out occurs for the existing design tires 69 • The hypotheses are t-Test: Two-Sample Assuming Equal Variances H0: m1 - m2 = 0 New Dsgn Exstng dsgn Mean 73.6 69.2 H1: m1 - m2 > 0 Variance 243.4105263 226.8 Observations 20 20 Pooled Variance 235.1052632 The test statistic is Hypothesized Mean Difference0 df 38 x1 x 2 ( m1 m 2 ) t Stat 0.907447484 t 1 1 P(T<=t) one-tail 0.184944575 s( ) 2 p t Critical one-tail 1.685953066 n1 n1 P(T<=t) two-tail 0.36988915 t Critical two-tail 2.024394234 We run the t test, and We conclude that there is insufficient obtain the following evidence to reject H0 in favor of H1. Excel results. 70 7 New design 6 5 4 3 2 1 0 45 60 75 90 105 More 12 Existing design 10 8 6 4 2 0 45 60 75 90 105 More While the sample mean of the new design is larger than the sample mean of the existing design, the variability within each sample is large enough for the sample distributions to overlap and cover about the same range. It is therefore difficult to argue that one expected value is different than the other. 71 • Example 12.4 Car 1 New-Dsn Exst-Dsn 57 48 – to eliminate variability 2 3 64 102 50 89 among Sample t-Test: Paired Twoobservations within 4 62 56 each for Meanssample the experiment 5 81 78 6 87 75 was redone. New-Dsn Exst-Dsn 7 61 50 Mean 73.6 69.05 8 62 49 – Variance One tire of each type was 316.366 242.779 9 74 70 10 62 66 Observationsinstalled on the rear wheel of 20 20 20 11 100 98 Pearson Correlation 0.91468 randomly selected Hypothesized Mean Differencecars (each 12 90 86 0 13 83 78 df car was sampled twice, 19 thus 14 84 90 t Stat creating a pair of observations). P(T<=t) one-tail 2.81759 0.0055 15 86 98 16 62 58 – The number of miles until t Critical one-tail 1.72913 17 67 58 18 40 41 wear-out was recorded P(T<=t) two-tail t Critical two-tail 0.01099 2.09302 19 71 61 20 77 82 72 The range of observations sample A So what really The happened here? values each sample consists of might markedly vary... The range of observations sample B 73 Differences ...but the differences between pairs of observations might be quite close to one another, resulting in a small variability. The range of the differences 0 74 Observe the statistic t shown below and notice how a small variability of the differences (small sD) helps in rejecting the null hypothesis. 75 • Solving by hand – Calculate the difference for each xi – Calculate the average differences and the standard deviation of the differences – Build the statistics as follows: xD m D t sD nD – Run the hypothesis test using t distribution with nD - 1 degrees of freedom. 76 – The hypotheses test for this problem is H0: mD = 0 New-Dsn Exst-Dsn Difference 57 48 9 H1: mD > 0 region is: The rejection 64 50 14 102 89 13 = t > ta with d.f. = 20-1 56 19. 62 6 If a = .05, t.05,19 = 1.729. 81 87 78 75 3 12 The statistic is 61 62 50 49 11 13 xD m D Since 2.817 > 1.729, there 74 70 4 t 62 100 66 is sufficient evidence in the data 98 -4 2 sD nD to reject the null hypothesis in 90 83 86 78 4 5 favor of the alternative hypothesis. 4.55 0 84 90 -6 86 98 -12 62 67 58 58 4 9 7.22186 20 Conclusion: At 5% significance 40 41 -1 level the new type tires last longer 71 61 10 2.817 77 than the current type. 82 Average = -5 4.55 Standard Deviation = 7.2218677 Estimating the mean difference Interval Estimator of m D sD x D t a / 2, n D 1 nD The 95% confidence int erval of the mean difference 7.22 in Example 12.4 is 4.55 2.093 4.55 3.38 20 78 Checking the required conditions for the paired observations case • The validity of the results depends on the normality of the differences. 8 6 4 2 0 -12 -6 0 6 12 More 79 12.5 Inferences about the ratio of two variances • In this section we discuss how to compare the variability of two populations. • In particular, we draw inference about the ratio of two population variances. • This question is interesting because: – Variances can be used to evaluate the consistency of processes. – The relationships between variances determine the technique used to test relationships between mean values 80 • Point estimator of s12/s22 – Recall that S2 is an unbiased estimator of s2. – Therefore, it is not surprising that we estimate s12/s22 by S12/S22. • Sampling distribution for s12/s22 – The statistic [S12/s12] / [S22/s22] follows the F distribution. – The test statistic for s12/s22 is derived from this statistic. 81 • Testing s12 / s22 – Our null hypothesis is always H0: s12 / s22 = 1 S12/s12 – Under this null hypothesis the F statistic F = becomes S22/s22 S12 F= S22 82 83 Calories consumed at lunch Example 12.5 Consmers Non-cmrs 568 705 498 819 (see example 12.1) 589 706 The hypotheses are: 681 509 In order to2perform a 540 613 s test regarding average H0: 1 1 646 582 636 601 2 s2 consumption of 739 608 2 539 787 at calories s1people’s 596 573 lunch 1in s 2 1 to the Two-Sample for Variances 607 H : relation F-Test 529 428 754 2 inclusion of high-fiber 637 617 Consumers Nonconsumers 741 628 cereal in their Mean 604.0232558 633 633.2336449 537 Variance 4102.975637 555 10669.76565 748 breakfast, the variance Observations 43 . 107 . . 106 . ratio of two samplesF df 42 0.384542245 . . has to be tested first. 0.000368433 P(F<=f) one-tail 0.637072617 F Critical one-tail . . 84 • Solving by hand – The rejection region is F>Fa/2,n1,n2 or F<1/Fa2,n2,n1 which becomes (for a=0.05)... F Fa / 2,n1,n 2 F.025,42,106 F.025,40,120 1.61 F 1/ Fa / 2,n 2,n1 1/ F.025,106,42 1/ F.025,120,40 .63 – The F statistic value is F=S12/S22 = .3845 – Conclusion: Because .3845<.63 we can reject the null hypothesis in favor of the alternative hypothesis. – There is sufficient evidence in the data to argue at 5% significance level that the variance of the two groups differ. 85 Estimating the Ratio of Two Population Variances • From the statistic F = [S12/s12] / [S22/s22] we can isolate s12/s22 and build the following interval estimator: s1 2 1 s1 s 1 2 2 2 2 Fa / 2,n 2,n1 s2 F s2 s2 2 a / 2,n1,n 2 w here n1 n 1 and n 2 n2 1 86 • Example 12.6 – Determine the 95% confidence interval estimate of the ratio of the two population variances in example 12.1 – Solution • we find Fa/2,v1,v2 = F.025,40,120 = 1.61 (approximately) Fa/2,v2,v1 = F.025,120,40 = 1.72 (approximately) • LCL = (s12/s22)[1/ Fa/2,v1,v2 ] = (4102.98/10,669.770)[1/1.61]= .2388 • UCL = (s12/s22)[ Fa/2,v2,v1 ] = (4102.98/10,669.770)[1.72]= .6614 87 12.6 Inference about the difference between two population proportions • In this section we deal with two populations whose data are qualitative. • When data are qualitative we can (only) ask questions regarding the proportions of occurrence of certain outcomes. • Thus, we hypothesize on the difference p1-p2, and draw an inference from the hypothesis test. 88 • Sampling Distribution of the Difference ˆ ˆ p1 p 2 Between Two sample proportions – Two random samples are drawn from two populations. – The number of successes in each sample is recorded. – The sample proportions are computed. Sample 1 Sample size n1 Sample 2 Number of successes x1 Sample size n2 Sample proportion Number of successes x2 x1 p1 ˆ Sample proportion n1 x2 ˆ p2 n2 89 ˆ ˆ – The statistic p1 p 2 is approximately normally distributed if n1p1, n1(1 - p1), n2p2, n2(1 - p2) are all equal to or greater than 5. Because p1, p2, are unknown, we use their estimates instead. ˆ ˆ – The mean of p1 p 2 is p1 - ˆ ˆ ˆ ˆ p2. Thus, n1p1,n1q1,n2p2 ,n2q2 are all equal to or greater than 5. ˆ ˆ – The variance of p1 p 2 is p1(1-p1) /n1)+ (p2(1-p2)/n2) The statistic ˆ ˆ (p1 p 2 ) (p1 p 2 ) Z p1 (1 p1 ) p 2 (1 p 2 ) n1 n2 is approximately normally distribute d 90 • Testing the Difference between Two Population p1 p 2 Proportions – We hypothesize on the difference between the two proportions, p1 - p2. – There are two cases to consider: Case 1: Case 2: H0: p1-p2 =0 H0: p1-p2 =D (D is not equal to 0) Calculate the pooled proportion Do not pool the data x1 x 2 x1 x2 ˆ p ˆ p1 ˆ p2 n1 n 2 n1 n2 Then Then ˆ ˆ (p1 p 2 ) (p1 p 2 ) ˆ ˆ (p1 p 2 ) D Z Z 1 1 ˆ ˆ ˆ ˆ p1 (1 p1 ) p 2 (1 p 2 ) ˆ ˆ p(1 p)( ) n1 n2 n1 n2 91 • Example 12.7 – A research project employing 22,000 American physicians was conduct to discover whether aspirin can prevent heart attacks. – Half of the participants in the research took aspirin, and half took placebo. – In a three years period,104 of those who took aspirin and 189 of those who took the placebo had had heart attacks. – Is aspirin effective in preventing heart attacks? 92 • Solution – Identifying the technique • The problem objective is to compare the population of those who take aspirin with those who do not. • The data is qualitative (Take/do not take aspirin) • The hypotheses test are Population 1 - aspirin takers H0: p1 - p2 = 0 Population 2 - placebo takers H1: p1 - p2 < 0 • We identify here case 1 so ˆ ˆ (p1 p 2 ) (p1 p 2 ) Z 1 1 p ˆ ˆ (1 p)( ) n1 n2 93 – Solving by hand • For a 5% significance level the rejection region is z < -za = -z.05 = -1.645 - 5.02 < - 1.645, so reject The sample proportion s are the null hypothesis. ˆ ˆ p1 104 11,000 .00945, and p 2 189 11,000 .01718 The pooled proportion is ˆ p ( x1 x 2 ) (n1 n2 ) (104 189) (11,000 11,000) .01332 The z statistic becomes ˆ ˆ ( p1 p 2 ) ( p1 p 2 ) .009455 .01718 Z 5.02 1 1 1 1 ˆ ˆ p(1 p)( ) .01332(.98668)( ) n1 n2 11,000 11,000 94 • Example 12.8 (Marketing application) – Management needs to decide which of two new packaging designs to adopt, to help improve sales of a soap. – A study is performed in two communities: • Design A is distributed in Community 1. • Design B is distributed in Community 2. • The old design packages is still offered in both communities. – For design A to be financially viable it has to outsell design B by at least 3%. 95 – Summary of the experiment results • Community 1 - 580 packages with new design A sold 324 packages with old design sold • Community 2 - 604 packages with new design B sold 442 packages with old design sold – Use 1% significance level and perform a test to find which type of packaging to use. 96 • Solution – Identifying the technique • The problem objective is to compare two populations, consisting of the values “purchase of the new design”, and “purchase of the old design”. • Data are qualitative. We need to test p1 - p2.. • The hypotheses to test are H0: p1 - p2 = .03 H1: p1 - p2 > .03 • We have to perform case 2 of the test for difference in proportions (the difference is not equal to zero). 97 • Solving by hand ˆ ˆ (p1 p 2 ) D Z ˆ ˆ ˆ ˆ p1 (1 p1 ) p 2 (1 p 2 ) n1 n2 580 604 .03 580 324 604 442 1.58 .642(1 .642) .577(1 .577) 904 1046 .642 The rejection region is z > za = z.01 = 2.33. Conclusion: Do not reject the null hypothesis. There is insufficient evidence to infer that packaging with design A will outsell design B by 3% or more. 98 • Estimating the Difference Between Two Population Proportions ˆ ˆ ˆ ˆ p1 (1 p1 ) p 2 (1 p 2 ) ˆ ˆ (p1 p 2 ) n1 n2 • Example 12.9 Estimate with 95% the proportion of men who would avoid a heart attack if they take aspirin regularly. .009455(.999545 ) .01718(.98282 ) (.009455 .01718 ) 1.96 11,000 11,000 [ .010753, .004697 ] 99 12.7 Market Segmentation (Optional) • Marketing Segmentation is a statistical analysis aimed at determining the differences that exist between buyers and non-buyers of a company’s product. • Statistics plays a major role in market segmentation. – Surveys are used to gather the relevant data. – Statistical tests are used to differentiate among segments. – Sales and profit estimates are derived. 100 • Example 12.10 – A new company in the market offers no-wait services for car oil and filter change. – The company wants to make decisions about where to advertise, and the nature of the advertisement. – A sample of 1000 car owners was selected. The drivers were asked to report whether or not they used a no-wait station, as well as several characteristics of their lives (including age). 101 – The research should reveal whether differences in age exist between customers of no-wait service and customers of other types of facilities (see file XM12-10) • Solution – Identifying the technique • The problem objective is to compare the population of ages of no-wait customers, to the population of ages of other facility users. • Data are quantitative. • Samples are independent. • The parameter to be tested is m1 - m2., (m represents mean age) 102 – The hypotheses are H0: m1 - m2 = 0 H1: m1 - m2 = 0 – When testing for the relationship between the two variances we get the following results F-Test Two-Sample for Variances No-Wait Other Mean 47.78331 44.03448 Variance 77.17323 60.09721 We run the test for m1 - m2 Observations 623 377 with two equal variances df 622 376 F 1.28414 0.003822 P(F<=f) one-tail 1.166224 F Critical one-tail 103 Tutorial • 11.1, 11.9, 11.27, 11.33 • 11.41, 11.45, 11.49 • 11.59, 11.63, 11.69 • 12.1, 12.11, 12.25 • 12.39, 12.43, 12.51, 12.71, 12.77 104

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