Quantitative Methods for Business chapter 17 potx

CHAPTER High performance – statistical inference for comparing population means and bivariate data 17 Chapter objectives This chapter will help you to: ■ test hypotheses on the difference between two population means using independent samples and draw appropriate conclusions ■ carry out tests of hypotheses about the difference between two population means using paired data and draw appropriate conclusions ■ test differences between population means using analysis of variance analysis (ANOVA) and draw appropriate conclusions 540 Quantitative methods for business Chapter 17 ■ conduct hypothesis tests about population correlation coeffi- cients and draw appropriate conclusions ■ produce interval estimates using simple linear regression models ■ perform contingency analysis and interpret the results ■ use the technology; test differences between sample means, apply correlation and regression inference, and contingency analysis in EXCEL, MINITAB and SPSS ■ become acquainted with the business use of contingency analysis In the previous chapter we looked at statistical inference in relation to univariate data, estimating and testing single population parameters like the mean using single sample results. In this chapter we will con- sider statistical inference methods that enable us to compare means of two or more populations, to test population correlation coefficients, to make predictions from simple linear regression models and to test for association in qualitative data. 17.1 Testing hypotheses about two population means In section 16.3 of the previous chapter we looked at tests of the popu- lation mean based on a single sample mean. In this section we will con- sider tests designed to assess the difference between two population means. In businesses these tests are used to investigate whether, for instance, the introduction of a new logo improves sales. To use these tests you need to have a sample from each of the two populations. For the tests to be valid the samples must be random, but they can be independent or dependent. Independent samples are selected from each population separately. Suppose a domestic gas supplier wanted to assess the impact of a new charging system on customers’ bills. The company could take a ran- dom sample of customers and record the size of their bills under the existing charging system then, after the new system is introduced, take another random sample of customers and record the size of their bills. These samples would be independent. Dependent samples consist of matched or paired values. If the gas supplier took a random sample of customers and recorded the size of their bills both before and after the introduction of the new charging system they would be using a paired or dependent sample. The choice of independent or dependent samples depends on the context of the test. Unless there is a good reason for using paired data it is better to use independent samples. We will begin by looking at tests for use with independent samples and deal with paired samples later in this section. As with single sample tests, the size of the samples is important because it determines the nature of the sampling distribution. In this section we will assume that the population standard deviations are not known. 17.1.1 Large independent samples The null hypothesis we use in comparing population means is based on the difference between the means of the two populations, ␮ 1 Ϫ ␮ 2 . The possible combinations of null and alternative hypotheses are shown in Table 17.1. The hypotheses listed in Table 17.1 all assume that the focus of the test is that there is no difference between the population means. This is very common but the same formats can be used to test whether the difference between two population means is a non-zero constant, e.g. H 0 : ␮ 1 Ϫ ␮ 2 ϭ 6. If both samples contain 30 or more items the difference between their means, x – 1 Ϫ x – 2 , belongs to the sampling distribution of X — 1 Ϫ X — 2 . This sampling distribution is normally distributed with a mean of ␮ 1 Ϫ ␮ 2 , and a standard error of: where ␴ 1 and ␴ 2 are the standard deviations of the first and second populations, and n 1 and n 2 are the sizes of the samples from the first and second populations. ␴ ␴ 1 2 1 2 2 2 nn ϩ Chapter 17 Statistical inference: population means and bivariate data 541 Table 17.1 Types of hypotheses for comparing population means Null hypothesis Alternative hypothesis Type of test H 0 : ␮ 1 Ϫ ␮ 2 ϭ 0H 1 : ␮ 1 Ϫ ␮ 2 ϶ 0 Two-sided H 0 : ␮ 1 Ϫ ␮ 2 р 0H 1 : ␮ 1 Ϫ ␮ 2 Ͼ 0 One-sided H 0 : ␮ 1 Ϫ ␮ 2 у 0H 1 : ␮ 1 Ϫ ␮ 2 Ͻ 0 One-sided We will assume that the population standard deviations are not known, in which case the estimated standard error of the sampling distribution is: The test statistic is: If the null hypothesis suggests that the difference between the popu- lation means is zero, we can simplify this to: Once we have calculated the test statistic we need to compare it to the appropriate critical value from the Standard Normal Distribution. z xx nn ( ) s s 12 ϭ Ϫ ϩ 1 2 1 2 2 2 z xx nn ( ) ( ) s s 12 12 ϭ ϪϪϪ ϩ ␮␮ 1 2 1 2 2 2 s s 1 2 1 2 2 2 nn ϩ 542 Quantitative methods for business Chapter 17 Example 17.1 A national breakdown recovery service has depots at Oxford and Portsmouth. The mean and standard deviation of the times that it took for the staff at the Oxford depot to assist each of a random sample of 47 motorists were 51 minutes and 7 minutes respectively. The mean and standard deviation of the response times recorded by the staff at the Portsmouth depot in assisting a random sample of 39 customers were 49 minutes and 5 minutes respectively. Test the hypothesis that there is no difference between the mean response times of the two depots. Use a 5% level of significance. H 0 : ␮ 1 Ϫ ␮ 2 ϭ 0H 1 : ␮ 1 Ϫ ␮ 2 ϶ 0 This is a two-tail test using a 5% level of confidence so the critical values are Ϯz 0.025 . Unless the test statistic is below Ϫ1.96 or above ϩ1.96 the null hypothesis cannot be rejected. The test statistic, 1.541, is within Ϯ1.96 so we cannot reject H 0 . The population mean response times of the two breakdown services could be equal. Test statistic, 51 4 2/1.298 1.541 z ϭ Ϫ ϩ ϭϭ 9 7 47 5 39 22 Notice that in Example 17.1 we have not said anything about the dis- tributions of response times. The Central Limit Theorem allows us to use the same two-sample z test whatever the shape of the populations from which the samples were drawn as long as the size of both samples is 30 or more. At this point you may find it useful to try Review Questions 17.1 to 17.3 at the end of the chapter. 17.1.2 Small independent samples If the size of the samples you want to use to compare population means is small, less than 30, you can only follow the procedure outlined in the previous section if both populations are normal and both population standard deviations known. In the absence of the latter it is possible to test the difference between two population means using small inde- pendent samples but only under certain circumstances. If both populations are normal and their standard deviations can be assumed to be the same, that is ␴ 1 ϭ ␴ 2 , we can conduct a two-sample t test. We use the sample standard deviations to produce a pooled esti- mate of the standard error of the sampling distribution of X — 1 Ϫ X — 2 , s p . The test statistic is We then compare the test statistic to the appropriate critical value from the t distribution. The number of degrees of freedom for this test is n 1 ϩ n 2 Ϫ 2, one degree of freedom is lost for each of the sample means. t xx s nn ( ) * 12 p ϭ Ϫ ϩ 11 12 s ns ns nn p 11 2 22 2 2 () () 2 ϭ ϪϩϪ ϩϪ 11 1 Chapter 17 Statistical inference: population means and bivariate data 543 Example 17.2 A cereal manufacturer claims to use no more oats, the cheapest ingredient, in produc- ing packets of ‘own-brand’ muesli for a supermarket chain than they use to produce their own premium brand. The mean and standard deviation of the oat content by weight of a random sample of 14 ‘own-brand’ packets are 34.9% and 1.4% respectively. The mean and standard deviation of the oat content of a random sample of 17 premium At this point you may find it useful to try Review Questions 17.4 to 17.6 at the end of the chapter. 17.1.3 Paired samples If you want to test the difference between population means using dependent or paired samples the nature of the data enables you to test the mean of the differences between all the paired values in the popu- lation, ␮ d . This approach contrasts with the methods described in the earlier parts of this section where we have tested the difference between population means, ␮ 1 Ϫ ␮ 2 . The procedure involved in testing hypotheses using paired samples is very similar to the one-sample hypothesis testing we discussed in section 16.3 of Chapter 16. We have to assume that the differences between the paired values are normally distributed with a mean of ␮ d , and a standard deviation of ␴ d . The sampling distribution of sample mean 544 Quantitative methods for business Chapter 17 brand packets are 33.4% and 1.1% respectively. Test the hypothesis that the mean oat content of the premium brand is no greater than the mean oat content of the ‘own-brand’ muesli using a 1% level of significance. We will define ␮ 1 as the population mean of the ‘own-brand’ and ␮ 2 as the population mean of the premium product. H 0 : ␮ 1 Ϫ ␮ 2 р 0H 1 : ␮ 1 Ϫ ␮ 2 Ͼ 0 First we need the pooled estimate of the standard error: Now we can calculate the test statistic: This is a one-tail test so the null hypothesis will only be rejected if the test statistic exceeds the critical value. From Table 6 on page 623 in Appendix 1, t 0.01,29 is 2.462. Since the test statistic is greater than the critical value we can reject the null hypothesis at the 1% level. The difference between the sample means is very significant. t 34.9 33.4 1.243 * 1 14 1 17 3.344 ϭ Ϫ ϩ ϭ s p 22 (14 )1.4 (17 )1.1 17 2 1.243 ϭ ϪϩϪ ϩϪ ϭ 11 14 differences will also be normally distributed with a mean of ␮ d and a standard error of ␴ d /√n, where n is the number of differences in the sample. Since we assume that ␴ d is unknown we have to use the estimated standard error s d /√n, where s d is the standard deviation of the sample differences. Typically samples of paired data tend to be small so the benchmark distribution for the test is the t distribution. The test is therefore called the paired t test. Table 17.2 lists the three possible combinations of hypotheses. The test statistic is: where x – d is the mean of the sample differences. We then compare the test statistic to the appropriate critical value from the t distribution with n Ϫ 1 degrees of freedom. t x sn dd d 0 ϭ Ϫ ␮ √ Chapter 17 Statistical inference: population means and bivariate data 545 Table 17.2 Types of hypotheses for the mean of the population of differences Null hypothesis Alternative hypothesis Type of test H 0 : ␮ d ϭ ␮ d0 H 1 : ␮ d ϶ ␮ d0 (not equal) Two-sided H 0 : ␮ d р ␮ d0 H 1 : ␮ d Ͼ ␮ d0 (greater than) One-sided H 0 : ␮ d у ␮ d0 H 1 : ␮ d Ͻ ␮ d0 (less than) One-sided In this table ␮ d0 represents the value of the population mean that is to be tested. Example 17.3 A Business School claims that, on average, people who take their MBA programme will enhance their annual salary by at least £8000. Each of a random sample of 12 graduates of the programme were asked for their annual salary prior to beginning the programme and their current annual salary. Use the sample data to test whether the mean difference in annual earnings is £8000 or more using a 10% level of significance. H 0 : ␮ d у 8.00 H 1 : ␮ d Ͻ 8.00 To conduct the test we first need to find the mean and standard deviation of the salary differences in the sample. Graduate 1 2 3 4 5 6 7 8 9 10 11 12 Prior salary (£000) 22 29 29 23 33 20 26 21 25 27 27 29 Current salary (£000) 31 38 40 29 41 25 29 26 31 37 41 36 Salary difference (£000) 9 9 11 6 8 5 3 5 6 10 14 7 546 Quantitative methods for business Chapter 17 The mean and standard deviation of the sample differences are 7.75 and 3.05, to 2 decimal places. The test statistic is: From Table 6 on page 623, t 0.10,11 is 1.363. The alternative hypothesis is that the population mean salary difference is less than £8000 so the critical value is Ϫ1.363. A sample mean that produces a test statistic this low or lower would lead us to reject the null hypothesis. In this case, although the sample mean is less than £8000, the test statistic, Ϫ0.28, is not less than the critical value and the null hypothesis cannot be rejected. The population mean of the salary differences could well be £8000. t 7.75 8.00 12 0.25 0.88 0.284ϭ Ϫ ϭ Ϫ ϭϪ 305. √ At this point you may find it useful to try Review Questions 17.7 to 17.9 at the end of the chapter. 17.2 Testing hypotheses about more than two population means – one-way ANOVA In some investigations it is important to establish whether two random samples come from a single population or from two populations with different means. The techniques we looked at in the previous section enable us to do just that. But what if we have three or more random samples and we need to establish whether they come from populations with different means? You might think that the obvious answer is to run t tests using each pair of random samples to establish whether the first sample came from the same population as the second, the first sample came from the same population as the third and the second sample came from the same population as the third and so on. In doing this you would be testing the hypotheses: H 0 : ␮ 1 ϭ ␮ 2 H 0 : ␮ 1 ϭ ␮ 3 H 0 : ␮ 2 ϭ ␮ 3 etc. Although feasible, this is not the best way to approach the investiga- tion. For one thing the more random samples that are involved the greater the chance that you miss out one or more possible pairings. For another, each test you conduct carries a risk of making a type 1 error, wrongly rejecting a null hypothesis because you happen to have a sample result from an extreme end of its sampling distribution. The chance of this occurring is the level of significance you use in conducting the test. The problem when you conduct a series of related tests is that the probability of making a type 1 error increases; if you use a 5% level of significance then the probability of not making a type 1 error in a sequence of three tests is, using the multiplication rule of probability, 0.95 * 0.95 * 0.95 or 0.857. This means the effective level of signifi- cance is 14.3%, considerably greater than you might have assumed. To establish whether more than two samples come from populations with different means we use an alternative approach, analysis of variance, usually abbreviated to ANOVA (analysis of variance). At first sight it seems rather odd to be using a technique based on variance, a measure of spread, to assess hypotheses about means, which are measures of loca- tion. The reason for doing this is that it enables us to focus on the spread of the sample means, after all the greater the differences between the sample means the greater the chance that they come from populations with different means. However, we have to be careful to put these dif- ferences into context because, after all, we can get different samples from the same population. Using ANOVA involves looking at the bal- ance between the variance of the sample means and the variance in the sample data overall. Example 17.4 illustrates why this is important. Chapter 17 Statistical inference: population means and bivariate data 547 Example 17.4 The Kranilisha Bank operates cash dispensing machines in Gloucester, Huddersfield and Ipswich. The amounts of cash dispensed (in £000s) at a random sample of machines during a specific period were: These are independent samples and so the fact that one sample (Huddersfield) con- tains fewer values does not matter. The sample data are shown in the form of boxplots in Figure 17.1. The distributions in Figure 17.1 suggest that there are differences between the amounts of cash dispensed at the machines, with those in Ipswich having the largest turnover and those in Huddersfield having the smallest. The sample means, which are represented by the dots, bear out this impression: 34 for Gloucester, 25 for Huddersfield and 41 for Ipswich. Gloucester 25 30 32 39 44 Huddersfield 17 25 27 31 Ipswich 29 34 44 47 51 In Example 17.4 the sample means are diverse enough and the dis- tributions shown in Figure 17.1 distinct enough to indicate differences between the locations, but is it enough to merely compare the sample means? 548 Quantitative methods for business Chapter 17 Figure 17.1 Cash dispensed at machines in Gloucester, Huddersfield and Ipswich IpswichHuddersfieldGloucester 50 40 30 20 Place Cash dispensed (£000) Example 17.5 IpswichHuddersfieldGloucester 65 55 45 35 25 15 Place Cash dispensed (£000) Figure 17.2 Revised amounts of cash dispensed at machines in Gloucester, Huddersfield and Ipswich [...]... differences, one for each section or cell of the contingency table The total we get is the test statistic, ␹2, for the sample results The procedure can be represented using the following formula ␹2 ϭ ∑(O Ϫ E)2/E 572 Quantitative methods for business Chapter 17 Example 17. 19 Find the value of ␹2 for the sample data in Example 17. 16 We can work from the contingency table produced in Example 17. 18 Men Too... regression model for the data in Example 17. 7 (temperature and cans of soft drinks sold) Residuals versus the fitted values (response is sales) 10 Residual 5 0 Ϫ5 Ϫ10 Ϫ15 10 20 30 40 Fitted value of sales Figure 17. 8 A residual plot for the model in Example 17. 8 50 60 568 Quantitative methods for business Chapter 17 There is no apparent systematic pattern to the scatter in Figure 17. 8 so we would conclude... are useless So, if you know that you want to construct intervals based on 566 Quantitative methods for business Chapter 17 specific values of x, try to ensure that these values are within the range of x values in your sample At this point you may find it useful to try Review Questions 17. 16 to 17. 20 at the end of the chapter 17. 3.4 When simple linear models won’t do the job So far we have concentrated... Questions 17. 13 to 17. 15 at the end of the chapter 17. 3.2 Testing regression models The second bivariate quantitative technique we looked at in Chapter 7 was simple linear regression analysis This allows you to find the equation of the line of best fit between two variables, X and Y Such a line has two distinguishing features, its intercept and its slope In the standard 558 Quantitative methods for business. .. outside the range of x values in the sample it will be both wide and unreliable 564 Quantitative methods for business Chapter 17 Example 17. 11 Construct a 95% confidence interval for the mean sales of cans of soft drinks which the shopkeeper in Example 17. 7 can expect on days when the temperature is 35° The point estimate for the mean, y ϭ 0.74 ϩ 2.38 (35) ϭ 84.04 Confidence interval ϭ y Ϯ t ␣ 2, nϪ2 *... Example 17. 4 so the mean sum of squares between the samples is unchanged, 284.465 We need to calculate the sum of squares within the samples for the amended data: n SS G ϭ ∑(xi Ϫ x G )2 ϭ (14 Ϫ 34)2 ϩ (21 Ϫ 34)2 ϩ (32 Ϫ 34)2 ϩ (49 Ϫ 34)2 ϩ (54 Ϫ 34)2 iϭ1 ϭ 1198 n SS H ϭ ∑ (x i Ϫ x H )2 ϭ (15 Ϫ 25.0)2 ϩ (18 Ϫ 25.0)2 ϩ (27 Ϫ 25.0)2 ϩ (40 Ϫ 25.0)2 ϭ 378 i ϭ1 554 Quantitative methods for business Chapter 17. .. dispensers in Example 17. 4, i.e town centre, supermarket, or garage ANOVA is a flexible technique that can be used to take more than one factor into account For more on its capabilities and applications see Roberts and Russo (1999) At this point you may find it useful to try Review Questions 17. 10 to 17. 12 at the end of the chapter 17. 3 Testing hypotheses and producing interval estimates for quantitative bivariate... instance we will use the original data from Example 17. 4: xϭ (5 * 34) ϩ (4 * 25) ϩ (5 * 41) 475 ϭ ϭ 33.929 14 14 The test statistic we will use is based on comparing the variation between the sample means with the variation within the samples One of the measures of variation or spread that we looked at in Chapter 6 550 Quantitative methods for business Chapter 17 was the sample variance, the square of the... is 5% so our decision rule is therefore to reject the null hypothesis if the test statistic is outside Ϯ2.306 Since the test statistic in this case is 5.02 we should reject H0 and conclude that the evidence suggests that the population slope is not zero 562 Quantitative methods for business Chapter 17 The implication of the sort of result we arrived at in Example 17. 8 is that the model, represented... test statistic in Example 17. 19 was 4.166, which is large enough (larger than 3.841) to lead us to reject the null hypothesis at the 574 Quantitative methods for business Chapter 17 5% level of significance The sample results suggest that there is association between gender and attitude to the amount of sport on TV The set of data we have used to illustrate chi-square tests for association is a simple . Y. t rn r 2 1 2 ϭ Ϫ Ϫ 556 Quantitative methods for business Chapter 17 At this point you may find it useful to try Review Questions 17. 13 to 17. 15 at the end of the chapter. 17. 3.2 Testing regression. Figure 17. 1 distinct enough to indicate differences between the locations, but is it enough to merely compare the sample means? 548 Quantitative methods for business Chapter 17 Figure 17. 1 Cash. 12 ϭ ϪϪϪ ϩ ␮␮ 1 2 1 2 2 2 s s 1 2 1 2 2 2 nn ϩ 542 Quantitative methods for business Chapter 17 Example 17. 1 A national breakdown recovery service has depots at Oxford and Portsmouth. The mean and standard