Ebook Elementary statistics (11E): Part 2

(BQ) Part 2 book Elementary statistics has contents: Hypothesis testing, inferences from two samples, correlation and regression, goodness of fit and contingency tables, goodness of fit and contingency tables, nonparametric statistics, projects, procedures, perspectives, statistical process control.

Find more at www.downloadslide.com 8-1 Review and Preview 8-2 Basics of Hypothesis Testing 8-3 Testing a Claim About a Proportion 8-4 Testing a Claim About a Mean: s Known 8-5 Testing a Claim About a Mean: s Not Known 8-6 Testing a Claim About a Standard Deviation or Variance Hypothesis Testing 390 Find more at www.downloadslide.com CHAPTER PROBLEM Does the MicroSort method of gender selection increase the likelihood that a baby will be a girl? Gender-selection methods are somewhat controversial Some people believe that use of such methods should be prohibited, regardless of the reason Others believe that limited use should be allowed for medical reasons, such as to prevent gender-specific hereditary disorders For example, some couples carry X-linked recessive genes, so that a male child has a 50% chance of inheriting a serious disorder and a female child has no chance of inheriting the disorder These couples may want to use a gender-selection method to increase the likelihood of having a baby girl so that none of their children inherit the disorder Methods of gender selection have been around for many years In the 1980s, ProCare Industries sold a product called Gender Choice The product cost only $49.95, but the Food and Drug Administration told the company to stop distributing Gender Choice because there was no evidence to support the claim that it was 80% reliable The Genetics & IVF Institute developed a newer gender-selection method called MicroSort The Microsort XSORT method is designed to increase the likelihood of a baby girl, and the YSORT method is designed to increase the likelihood of a boy Here is a statement from the MicroSort Web site: “The Genetics & IVF Institute is offering couples the ability to increase the chance of having a child of the desired gender to reduce the probability of X-linked diseases or for family balancing.” Stated simply, for a cost exceeding $3000, the Genetics & IVF Institute claims that it can increase the probability of having a baby of the gender that a couple prefers As of this writing, the MicroSort method is undergoing clinical trials, but these results are available: Among 726 couples who used the XSORT method in trying to have a baby girl, 668 couples did have baby girls, for a success rate of 92.0% Under normal circumstances with no special treatment, girls occur in 50% of births (Actually, the current birth rate of girls is 48.79%, but we will use 50% to keep things simple.) These results provide us with an interesting question: Given that 668 out of 726 couples had girls, can we actually support the claim that the XSORT technique is effective in increasing the probability of a girl? Do we now have an effective method of gender selection? Find more at www.downloadslide.com 392 Chapter Hypothesis Testing 8-1 Review and Preview In Chapters and we used “descriptive statistics” when we summarized data using tools such as graphs, and statistics such as the mean and standard deviation Methods of inferential statistics use sample data to make an inference or conclusion about a population The two main activities of inferential statistics are using sample data to (1) estimate a population parameter (such as estimating a population parameter with a confidence interval), and (2) test a hypothesis or claim about a population parameter In Chapter we presented methods for estimating a population parameter with a confidence interval, and in this chapter we present the method of hypothesis testing In statistics, a hypothesis is a claim or statement about a property of a population A hypothesis test (or test of significance) is a procedure for testing a claim about a property of a population The main objective of this chapter is to develop the ability to conduct hypothesis tests for claims made about a population proportion p, a population mean m, or a population standard deviation s Here are examples of hypotheses that can be tested by the procedures we develop in this chapter: • Genetics The Genetics & IVF Institute claims that its XSORT method allows couples to increase the probability of having a baby girl A newspaper headline makes the claim that most workers get their jobs through networking • Business Medical researchers claim that when people with colds are treated with echinacea, the treatment has no effect • Medicine • Aircraft Safety The Federal Aviation Administration claims that the mean weight of an airline passenger (including carry-on baggage) is greater than 185 lb, which it was 20 years ago • Quality Control When new equipment is used to manufacture aircraft altimeters, the new altimeters are better because the variation in the errors is reduced so that the readings are more consistent (In many industries, the quality of goods and services can often be improved by reducing variation.) The formal method of hypothesis testing uses several standard terms and conditions in a systematic procedure Study Hint: Start by clearly understanding Example in Section 8-2, then read Sections 8-2 and 8-3 casually to obtain a general idea of their concepts, then study Section 8-2 more carefully to become familiar with the terminology CAUTION When conducting hypothesis tests as described in this chapter and the following chapters, instead of jumping directly to procedures and calculations, be sure to consider the context of the data, the source of the data, and the sampling method used to obtain the sample data (See Section 1-2.) Find more at www.downloadslide.com 8-2 Basics of Hypothesis Testing 8-2 Basics of Hypothesis Testing Key Concept In this section we present individual components of a hypothesis test In Part we discuss the basic concepts of hypothesis testing Because these concepts are used in the following sections and chapters, we should know and understand the following: • How to identify the null hypothesis and alternative hypothesis from a given claim, and how to express both in symbolic form • How to calculate the value of the test statistic, given a claim and sample data • How to identify the critical value(s), given a significance level • How to identify the P-value, given a value of the test statistic • How to state the conclusion about a claim in simple and nontechnical terms In Part we discuss the power of a hypothesis test Part 1: Basic Concepts of Hypothesis Testing The methods presented in this chapter are based on the rare event rule (Section 4-1) for inferential statistics, so let’s review that rule before proceeding Rare Event Rule for Inferential Statistics If, under a given assumption, the probability of a particular observed event is extremely small, we conclude that the assumption is probably not correct Following this rule, we test a claim by analyzing sample data in an attempt to distinguish between results that can easily occur by chance and results that are highly unlikely to occur by chance We can explain the occurrence of highly unlikely results by saying that either a rare event has indeed occurred or that the underlying assumption is not correct Let’s apply this reasoning in the following example Gender Selection ProCare Industries, Ltd provided a product called “Gender Choice,” which, according to advertising claims, allowed couples to “increase your chances of having a girl up to 80%.” Suppose we conduct an experiment with 100 couples who want to have baby girls, and they all follow the Gender Choice “easy-to-use in-home system” described in the pink package designed for girls Assuming that Gender Choice has no effect and using only common sense and no formal statistical methods, what should we conclude about the assumption of “no effect” from Gender Choice if 100 couples using Gender Choice have 100 babies consisting of the following? a 52 girls b 97 girls a We normally expect around 50 girls in 100 births The result of 52 girls is close to 50, so we should not conclude that the Gender Choice product is effective The result of 52 girls could easily occur by chance, so there isn’t sufficient evidence to say that Gender Choice is effective, even though the sample proportion of girls is greater than 50% continued 393 Aspirin Not Helpful for Geminis and Libras Physician Richard Peto submitted an article to Lancet, a British medical journal The article showed that patients had a better chance of surviving a heart attack if they were treated with aspirin within a few hours of their heart attacks Lancet editors asked Peto to break down his results into subgroups to see if recovery worked better or worse for different groups, such as males or females Peto believed that he was being asked to use too many subgroups, but the editors insisted Peto then agreed, but he supported his objections by showing that when his patients were categorized by signs of the zodiac, aspirin was useless for Gemini and Libra heartattack patients, but aspirin is a lifesaver for those born under any other sign This shows that when conducting multiple hypothesis tests with many different subgroups, there is a very large chance of getting some wrong results Find more at www.downloadslide.com 394 Chapter Hypothesis Testing b The result of 97 girls in 100 births is extremely unlikely to occur by chance We could explain the occurrence of 97 girls in one of two ways: Either an extremely rare event has occurred by chance, or Gender Choice is effective The extremely low probability of getting 97 girls suggests that Gender Choice is effective In Example we should conclude that the treatment is effective only if we get significantly more girls than we would normally expect Although the outcomes of 52 girls and 97 girls are both greater than 50%, the result of 52 girls is not significant, whereas the result of 97 girls is significant Gender Selection The Chapter Problem includes the latest results from clinical trials of the XSORT method of gender selection Instead of using the latest available results, we will use these results from preliminary trials of the XSORT method: Among 14 couples using the XSORT method, 13 couples had girls and one couple had a boy We will proceed to formalize some of the analysis in testing the claim that the XSORT method increases the likelihood of having a girl, but there are two points that can be confusing: Assume p ‫ ؍‬0.5: Under normal circumstances, with no treatment, girls occur in 50% of births So p = 0.5 and a claim that the XSORT method is effective can be expressed as p 0.5 Instead of P(exactly 13 girls), use P (13 or more girls): When determining whether 13 girls in 14 births is likely to occur by chance, use P (13 or more girls) (Stop for a minute and review the subsection of “Using Probabilities to Determine When Results Are Unusual” in Section 5-2.) Under normal circumstances the proportion of girls is p = 0.5, so a claim that the XSORT method is effective can be expressed as p 0.5 We support the claim of p 0.5 only if a result such as 13 girls is unlikely (with a small probability, such as less than or equal to 0.05) Using a normal distribution as an approximation to the binomial distribution (see Section 6-6), we find P (13 or more girls in 14 births) = 0.0016 Figure 8-1 shows that with a probability of 0.5, the outcome of 13 girls in 14 births is unusual, so we reject random chance as a reasonable explanation We conclude that the proportion of girls born to couples using the XSORT method is significantly greater than the proportion that we expect with random chance Here are the key components of this example: • Claim: The XSORT method increases the likelihood of a girl That is, p 0.5 assumption: The proportion of girls is p = 0.5 (with no effect from the XSORT method) • Working • The preliminary sample resulted in 13 girls among 14 births, so the sample proportion is pN = 13>14 = 0.929 • Assuming that p = 0.5, we use a normal distribution as an approximation to the binomial distribution to find that P (at least 13 girls in 14 births) = 0.0016 (Using Table A-1 or calculations with the binomial probability distribution results in a probability of 0.001.) • There are two possible explanations for the result of 13 girls in 14 births: Either a random chance event (with the very low probability of 0.0016) has occurred, Find more at www.downloadslide.com 8-2 Basics of Hypothesis Testing 395 or the proportion of girls born to couples using the XSORT method is greater than 0.5 Because the probability of getting at least 13 girls by chance is so small (0.0016), we reject random chance as a reasonable explanation The more reasonable explanation for 13 girls is that the XSORT method is effective in increasing the likelihood of girls There is sufficient evidence to support a claim that the XSORT method is effective in producing more girls than expected by chance 0.25 Figure 8-1 Probability Histogram for Numbers of Girls in 14 Births Probability 0.20 The probability of 13 or more girls is very small The probability of 13 or more girls is very small 0.15 0.10 0.05 0.00 10 11 12 13 14 Number of Girls in 14 births We now proceed to describe the components of a formal hypothesis test, or test of significance Many professional journals will include results from hypothesis tests, and they will use the same components described here Working with the Stated Claim: Null and Alternative Hypotheses • The null hypothesis (denoted by H 0) is a statement that the value of a population parameter (such as proportion, mean, or standard deviation) is equal to some claimed value (The term null is used to indicate no change or no effect or no difference.) Here is a typical null hypothesis included in this chapter: H 0: p = 0.5 We test the null hypothesis directly in the sense that we assume (or pretend) it is true and reach a conclusion to either reject it or fail to reject it alternative hypothesis (denoted by H or H a or HA) is the statement that the parameter has a value that somehow differs from the null hypothesis For the methods of this chapter, the symbolic form of the alternative hypothesis must use one of these symbols: 6, 7, Z Here are different examples of alternative hypotheses involving proportions: • The H1: p 0.5 H1: p 0.5 H1: p Z 0.5 Note About Always Using the Equal Symbol in H 0: It is now rare, but the symbols … and Ú are occasionally used in the null hypothesis H Professional statisticians Find more at www.downloadslide.com 396 Chapter Hypothesis Testing and professional journals use only the equal symbol for equality We conduct the hypothesis test by assuming that the proportion, mean, or standard deviation is equal to some specified value so that we can work with a single distribution having a specific value Note About Forming Your Own Claims (Hypotheses): If you are conducting a study and want to use a hypothesis test to support your claim, the claim must be worded so that it becomes the alternative hypothesis (and can be expressed using only the symbols 6, 7, or Z) You can never support a claim that some parameter is equal to some specified value For example, after completing the clinical trials of the XSORT method of gender selection, the Genetics & IVF Institute will want to demonstrate that the method is effective in increasing the likelihood of a girl, so the claim will be stated as p 0.5 In this context of trying to support the goal of the research, the alternative hypothesis is sometimes referred to as the research hypothesis It will be assumed for the purpose of the hypothesis test that p = 0.5, but the Genetics & IVF Institute will hope that p = 0.5 gets rejected so that p 0.5 is supported Supporting the alternative hypothesis of p 0.5 will support the claim that the XSORT method is effective Note About Identifying H and H 1: Figure 8-2 summarizes the procedures for identifying the null and alternative hypotheses Next to Figure 8-2 is an example using the claim that “with the XSORT method, the likelihood of having a girl is greater than 0.5.” Note that the original statement could become the null hypothesis, it could become the alternative hypothesis, or it might not be either the null hypothesis or the alternative hypothesis Figure 8-2 Identifying H0 and H1 Example: The claim is that with the XSORT method, the likelihood of having a girl is greater than 0.5 This claim in symbolic form is p 0.5 If p 0.5 is false, the symbolic form that must be true is p … 0.5 H1: p 0.5 H0: p = 0.5 Start Identify the specific claim or hypothesis to be tested, and express it in symbolic form Give the symbolic form that must be true when the original claim is false Using the two symbolic expressions obtained so far, identify the null hypothesis H0 and the alternative hypothesis H1: ? H1 is the symbolic expression that does not contain equality ? H0 is the symbolic expression that the parameter equals the fixed value being considered (The original claim may or may not be one of the above two symbolic expressions.) Find more at www.downloadslide.com 8-2 Basics of Hypothesis Testing Identifying the Null and Alternative Hypotheses Consider the claim that the mean weight of airline passengers (including carryon baggage) is at most 195 lb (the current value used by the Federal Aviation Administration) Follow the three-step procedure outlined in Figure 8-2 to identify the null hypothesis and the alternative hypothesis Refer to Figure 8-2, which shows the three-step procedure Step 1: Express the given claim in symbolic form The claim that the mean is at most 195 lb is expressed in symbolic form as m … 195 lb Step 2: If m … 195 lb is false, then m 195 lb must be true Step 3: Of the two symbolic expressions m … 195 lb and m 195 lb, we see that m 195 lb does not contain equality, so we let the alternative hypothesis H be m 195 lb Also, the null hypothesis must be a statement that the mean equals 195 lb, so we let H be m = 195 lb Note that in this example, the original claim that the mean is at most 195 lb is neither the alternative hypothesis nor the null hypothesis (However, we would be able to address the original claim upon completion of a hypothesis test.) Converting Sample Data to a Test Statistic The calculations required for a hypothesis test typically involve converting a sample statistic to a test statistic The test statistic is a value used in making a decision about the null hypothesis It is found by converting the sample statistic (such as the sample proportion pN , the sample mean x, or the sample standard deviation s) to a score (such as z, t, or x2) with the assumption that the null hypothesis is true In this chapter we use the following test statistics: pN - p Test statistic for proportion z = pq An x - m x - m Test statistic for mean z = s or t = s 2n 2n (n 1)s Test statistic for standard deviation x2 = s2 The test statistic for a mean uses the normal or Student t distribution, depending on the conditions that are satisfied For hypothesis tests of a claim about a population mean, this chapter will use the same criteria for using the normal or Student t distributions as described in Section 7-4 (See Figure 7-6 and Table 7-1.) Finding the Value of the Test Statistic Let’s again consider the claim that the XSORT method of gender selection increases the likelihood of having a baby girl Preliminary results from a test of the XSORT method of gender selection involved 14 couples who gave birth to 13 girls and boy Use the continued 397 Find more at www.downloadslide.com 398 Chapter Human Lie Detectors Researchers tested 13,000 people for their ability to determine when someone is lying They found 31 people with exceptional skills at identifying lies These human lie detectors had accuracy rates around 90% They also found that federal officers and sheriffs were quite good at detecting lies, with accuracy rates around 80% Psychology Professor Maureen O’Sullivan questioned those who were adept at identifying lies, and she said that “all of them pay attention to nonverbal cues and the nuances of word usages and apply them differently to different people They could tell you eight things about someone after watching a two-second tape It’s scary, the things these people notice.” Methods of statistics can be used to distinguish between people unable to detect lying and those with that ability Hypothesis Testing given claim and the preliminary results to calculate the value of the test statistic Use the format of the test statistic given above, so that a normal distribution is used to approximate a binomial distribution (There are other exact methods that not use the normal approximation.) From Figure 8-2 and the example displayed next to it, the claim that the XSORT method of gender selection increases the likelihood of having a baby girl results in the following null and alternative hypotheses: H 0: p = 0.5 and H 1: p 0.5 We work under the assumption that the null hypothesis is true with p = 0.5 The sample proportion of 13 girls in 14 births results in pN = 13>14 = 0.929 Using p = 0.5, pN = 0.929, and n = 14, we find the value of the test statistic as follows: z = pN - p 0.929 - 0.5 = = 3.21 pq (0.5) (0.5) An A 14 We know from previous chapters that a z score of 3.21 is “unusual” (because it is greater than 2) It appears that in addition to being greater than 0.5, the sample proportion of 13> 14 or 0.929 is significantly greater than 0.5 Figure 8-3 shows that the sample proportion of 0.929 does fall within the range of values considered to be significant because they are so far above 0.5 that they are not likely to occur by chance (assuming that the population proportion is p = 0.5) Figure 8-3 shows the test statistic of z = 3.21, and other components in Figure 8-3 are described as follows Unusually high sample proportions Critical region: Area of ␣ ϭ 0.05 used as criterion for identifying unusually high sample proportions p ϭ 0.5 or zϭ0 z ϭ 1.645 Critical value Sample proportion of: p ϭ 0.929 or z ϭ 3.21 Test Statistic Proportion of girls in 14 births Figure 8-3 Critical Region, Critical Value, Test Statistic Tools for Assessing the Test Statistic: Critical Region, Significance Level, Critical Value, and P-Value The test statistic alone usually does not give us enough information to make a decision about the claim being tested The following tools can be used to understand and interpret the test statistic Find more at www.downloadslide.com 8-2 Basics of Hypothesis Testing 399 • The critical region (or rejection region) is the set of all values of the test statistic that cause us to reject the null hypothesis For example, see the red-shaded critical region shown in Figure 8-3 significance level (denoted by a) is the probability that the test statistic will fall in the critical region when the null hypothesis is actually true If the test statistic falls in the critical region, we reject the null hypothesis, so a is the probability of making the mistake of rejecting the null hypothesis when it is true This is the same a introduced in Section 7-2, where we defined the confidence level for a confidence interval to be the probability - a Common choices for a are 0.05, 0.01, and 0.10, with 0.05 being most common • The •A critical value is any value that separates the critical region (where we reject the null hypothesis) from the values of the test statistic that not lead to rejection of the null hypothesis The critical values depend on the nature of the null hypothesis, the sampling distribution that applies, and the significance level of a See Figure 8-3 where the critical value of z = 1.645 corresponds to a significance level of a = 0.05 (Critical values were formally defined in Section 7-2.) Finding a Critical Value for Critical Region in the Right Tail Using a significance level of a = 0.05, find the critical z value for the alternative hypothesis H 1: p 0.5 (assuming that the normal distribution can be used to approximate the binomial distribution) This alternative hypothesis is used to test the claim that the XSORT method of gender selection is effective, so that baby girls are more likely, with a proportion greater than 0.5 Refer to Figure 8-3 With H 1: p 0.5, the critical region is in the right tail as shown With a right-tailed area of 0.05, the critical value is found to be z = 1.645 (by using the methods of Section 6-2) If the right-tailed critical region is 0.05, the cumulative area to the left of the critical value is 0.95, and Table A-2 or technology show that the z score corresponding to a cumulative left area of 0.95 is z = 1.645 The critical value is z = 1.645 as shown in Figure 8-3 Two-Tailed Test: z ϭ Ϫ1.96 Finding Critical Values for a Critical Region in Two Tails Using a significance level of a = 0.05, find the two critical z values for the alternative hypothesis H 1: p Z 0.5 (assuming that the normal distribution can be used to approximate the binomial distribution) Refer to Figure 8-4(a) With H 1: p Z 0.5, the critical region is in the two tails as shown If the significance level is 0.05, each of the two tails has an area of 0.025 as shown in Figure 8-4(a) The left critical value of z = -1.96 corresponds to a cumulative left area of 0.025 (Table A-2 or technology result in z = -1.96 by using the methods of Section 6-2) The rightmost critical value of z = 1.96 is found from the cumulative left area of 0.975 (The rightmost critical value is z 0.975 = 1.96.) The two critical values are z = -1.96 and z = 1.96 as shown in Figure 8-4(a) 0.025 0.025 zϭ0 z ϭ 1.96 (a) Left-Tailed Test: 0.05 z ϭ Ϫ1.645 z ϭ (b) Right-Tailed Test: 05 zϭ0 z ϭ 1.645 (c) Figure 8-4 Finding Critical Values Find more at www.downloadslide.com Symbol Table A complement of event A H Kruskal-Wallis test statistic H0 null hypothesis R H1 alternative hypothesis sum of the ranks for a sample; used in the Wilcoxon rank-sum test a alpha; probability of a type I error or the area of the critical region mR expected mean rank; used in the Wilcoxon rank-sum test b beta; probability of a type II error sR expected standard deviation of ranks; used in the Wilcoxon rank-sum test r sample linear correlation coefficient G number of runs in runs test for randomness rho; population linear correlation coefficient mG expected mean number of runs; used in runs test for randomness sG expected standard deviation for the number of runs; used in runs test for randomness mx mean of the population of all possible sample means x sx standard deviation of the population of all possible sample means x E margin of error of the estimate of a population parameter, or expected value r r coefficient of determination R2 multiple coefficient of determination rs Spearman’s rank correlation coefficient b1 point estimate of the slope of the regression line b0 point estimate of the y-intercept of the regression line yˆ predicted value of y d difference between two matched values Q1, Q2, Q3 quartiles d mean of the differences d found from matched sample data D1, D2, c, D9 deciles sd standard deviation of the differences d found from matched sample data P1, P2, c, P99 percentiles se standard error of estimate T rank sum; used in the Wilcoxon signedranks test x data value Find more at www.downloadslide.com Symbol Table f frequency with which a value occurs za>2 critical value of z S capital sigma; summation t t distribution Sx sum of the values ta>2 critical value of t Sx sum of the squares of the values df number of degrees of freedom square of the sum of all values F F distribution sum of the products of each x value multiplied by the corresponding y value x x2R right-tailed critical value of chi-square n number of values in a sample x2L left-tailed critical value of chi-square n! n factorial p N number of values in a finite population; also used as the size of all samples combined probability of an event or the population proportion q probability or proportion equal to p k number of samples or populations or categories pˆ sample proportion x mean of the values in a sample qˆ sample proportion equal to pˆ R mean of the sample ranges p proportion obtained by pooling two samples m mu; mean of all values in a population q proportion or probability equal to p s standard deviation of a set of sample values P(A) probability of event A s lowercase sigma; standard deviation of all values in a population P(A B) probability of event A, assuming event B has occurred s2 variance of a set of sample values nPr number of permutations of n items selected r at a time s2 variance of all values in a population nCr z standard score number of combinations of n items selected r at a time (Sx) Sxy chi-square distribution Find more at www.downloadslide.com TABLE A-3 Degrees of Freedom 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 45 50 60 70 80 90 100 200 300 400 500 1000 2000 Large t Distribution: Critical t Values 0.005 0.01 Area in One Tail 0.025 0.05 0.10 0.01 0.02 Area in Two Tails 0.05 0.10 0.20 63.657 9.925 5.841 4.604 4.032 3.707 3.499 3.355 3.250 3.169 3.106 3.055 3.012 2.977 2.947 2.921 2.898 2.878 2.861 2.845 2.831 2.819 2.807 2.797 2.787 2.779 2.771 2.763 2.756 2.750 2.744 2.738 2.733 2.728 2.724 2.719 2.715 2.712 2.708 2.704 2.690 2.678 2.660 2.648 2.639 2.632 2.626 2.601 2.592 2.588 2.586 2.581 2.578 2.576 31.821 6.965 4.541 3.747 3.365 3.143 2.998 2.896 2.821 2.764 2.718 2.681 2.650 2.624 2.602 2.583 2.567 2.552 2.539 2.528 2.518 2.508 2.500 2.492 2.485 2.479 2.473 2.467 2.462 2.457 2.453 2.449 2.445 2.441 2.438 2.434 2.431 2.429 2.426 2.423 2.412 2.403 2.390 2.381 2.374 2.368 2.364 2.345 2.339 2.336 2.334 2.330 2.328 2.326 12.706 4.303 3.182 2.776 2.571 2.447 2.365 2.306 2.262 2.228 2.201 2.179 2.160 2.145 2.131 2.120 2.110 2.101 2.093 2.086 2.080 2.074 2.069 2.064 2.060 2.056 2.052 2.048 2.045 2.042 2.040 2.037 2.035 2.032 2.030 2.028 2.026 2.024 2.023 2.021 2.014 2.009 2.000 1.994 1.990 1.987 1.984 1.972 1.968 1.966 1.965 1.962 1.961 1.960 6.314 2.920 2.353 2.132 2.015 1.943 1.895 1.860 1.833 1.812 1.796 1.782 1.771 1.761 1.753 1.746 1.740 1.734 1.729 1.725 1.721 1.717 1.714 1.711 1.708 1.706 1.703 1.701 1.699 1.697 1.696 1.694 1.692 1.691 1.690 1.688 1.687 1.686 1.685 1.684 1.679 1.676 1.671 1.667 1.664 1.662 1.660 1.653 1.650 1.649 1.648 1.646 1.646 1.645 3.078 1.886 1.638 1.533 1.476 1.440 1.415 1.397 1.383 1.372 1.363 1.356 1.350 1.345 1.341 1.337 1.333 1.330 1.328 1.325 1.323 1.321 1.319 1.318 1.316 1.315 1.314 1.313 1.311 1.310 1.309 1.309 1.308 1.307 1.306 1.306 1.305 1.304 1.304 1.303 1.301 1.299 1.296 1.294 1.292 1.291 1.290 1.286 1.284 1.284 1.283 1.282 1.282 1.282 Find more at www.downloadslide.com NEGATIVE z Scores z TABLE A-2 Standard Normal (z) Distribution: Cumulative Area from the LEFT z 00 01 02 03 Ϫ3.50 and lower Ϫ3.4 Ϫ3.3 Ϫ3.2 Ϫ3.1 Ϫ3.0 Ϫ2.9 Ϫ2.8 Ϫ2.7 Ϫ2.6 Ϫ2.5 Ϫ2.4 Ϫ2.3 Ϫ2.2 Ϫ2.1 Ϫ2.0 Ϫ1.9 Ϫ1.8 Ϫ1.7 Ϫ1.6 Ϫ1.5 Ϫ1.4 Ϫ1.3 Ϫ1.2 Ϫ1.1 Ϫ1.0 Ϫ0.9 Ϫ0.8 Ϫ0.7 Ϫ0.6 Ϫ0.5 Ϫ0.4 Ϫ0.3 Ϫ0.2 Ϫ0.1 Ϫ0.0 0001 0003 0005 0007 0010 0013 0019 0026 0035 0047 0062 0082 0107 0139 0179 0228 0287 0359 0446 0548 0668 0808 0968 1151 1357 1587 1841 2119 2420 2743 3085 3446 3821 4207 4602 5000 0003 0005 0007 0009 0013 0018 0025 0034 0045 0060 0080 0104 0136 0174 0222 0281 0351 0436 0537 0655 0793 0951 1131 1335 1562 1814 2090 2389 2709 3050 3409 3783 4168 4562 4960 0003 0005 0006 0009 0013 0018 0024 0033 0044 0059 0078 0102 0132 0170 0217 0274 0344 0427 0526 0643 0778 0934 1112 1314 1539 1788 2061 2358 2676 3015 3372 3745 4129 4522 4920 0003 0004 0006 0009 0012 0017 0023 0032 0043 0057 0075 0099 0129 0166 0212 0268 0336 0418 0516 0630 0764 0918 1093 1292 1515 1762 2033 2327 2643 2981 3336 3707 4090 4483 4880 NOTE: For values of z below Ϫ3.49, use 0.0001 for the area *Use these common values that result from interpolation: z score Area Ϫ1.645 Ϫ2.575 0.0500 0.0050 04 05 0003 0003 0004 0004 0006 0006 0008 0008 0012 0011 0016 0016 0023 0022 0031 0030 0041 0040 0055 0054 0073 0071 0096 0094 0125 0122 0162 0158 0207 0202 0262 0256 0329 0322 0409 0401 0505 * 0495 0618 0606 0749 0735 0901 0885 1075 1056 1271 1251 1492 1469 1736 1711 2005 1977 2296 2266 2611 2578 2946 2912 3300 3264 3669 3632 4052 4013 4443 4404 4840 4801 06 0003 0004 0006 0008 0011 0015 0021 0029 0039 0052 0069 0091 0119 0154 0197 0250 0314 0392 0485 0594 0721 0869 1038 1230 1446 1685 1949 2236 2546 2877 3228 3594 3974 4364 4761 07 08 0003 0003 0004 0004 0005 0005 0008 0007 0011 0010 0015 0014 0021 0020 0028 0027 0038 0037 0051 * 0049 0068 0066 0089 0087 0116 0113 0150 0146 0192 0188 0244 0239 0307 0301 0384 0375 0475 0465 0582 0571 0708 0694 0853 0838 1020 1003 1210 1190 1423 1401 1660 1635 1922 1894 2206 2177 2514 2483 2843 2810 3192 3156 3557 3520 3936 3897 4325 4286 4721 4681 09 0002 0003 0005 0007 0010 0014 0019 0026 0036 0048 0064 0084 0110 0143 0183 0233 0294 0367 0455 0559 0681 0823 0985 1170 1379 1611 1867 2148 2451 2776 3121 3483 3859 4247 4641 Find more at www.downloadslide.com POSITIVE z Scores TABLE A-2 z (continued) Cumulative Area from the LEFT z 00 01 02 03 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4 3.50 and up 5000 5398 5793 6179 6554 6915 7257 7580 7881 8159 8413 8643 8849 9032 9192 9332 9452 9554 9641 9713 9772 9821 9861 9893 9918 9938 9953 9965 9974 9981 9987 9990 9993 9995 9997 9999 5040 5438 5832 6217 6591 6950 7291 7611 7910 8186 8438 8665 8869 9049 9207 9345 9463 9564 9649 9719 9778 9826 9864 9896 9920 9940 9955 9966 9975 9982 9987 9991 9993 9995 9997 5080 5478 5871 6255 6628 6985 7324 7642 7939 8212 8461 8686 8888 9066 9222 9357 9474 9573 9656 9726 9783 9830 9868 9898 9922 9941 9956 9967 9976 9982 9987 9991 9994 9995 9997 5120 5517 5910 6293 6664 7019 7357 7673 7967 8238 8485 8708 8907 9082 9236 9370 9484 9582 9664 9732 9788 9834 9871 9901 9925 9943 9957 9968 9977 9983 9988 9991 9994 9996 9997 NOTE: For values of z above 3.49, use 0.9999 for the area *Use these common values that result from interpolation: 04 05 5160 5199 5557 5596 5948 5987 6331 6368 6700 6736 7054 7088 7389 7422 7704 7734 7995 8023 8264 8289 8508 8531 8729 8749 8925 8944 9099 9115 9251 9265 9382 9394 9495 * 9505 9591 9599 9671 9678 9738 9744 9793 9798 9838 9842 9875 9878 9904 9906 9927 9929 9945 9946 9959 9960 9969 9970 9977 9978 9984 9984 9988 9989 9992 9992 9994 9994 9996 9996 9997 9997 06 5239 5636 6026 6406 6772 7123 7454 7764 8051 8315 8554 8770 8962 9131 9279 9406 9515 9608 9686 9750 9803 9846 9881 9909 9931 9948 9961 9971 9979 9985 9989 9992 9994 9996 9997 07 08 5279 5319 5675 5714 6064 6103 6443 6480 6808 6844 7157 7190 7486 7517 7794 7823 8078 8106 8340 8365 8577 8599 8790 8810 8980 8997 9147 9162 9292 9306 9418 9429 9525 9535 9616 9625 9693 9699 9756 9761 9808 9812 9850 9854 9884 9887 9911 9913 9932 9934 9949 * 9951 9962 9963 9972 9973 9979 9980 9985 9986 9989 9990 9992 9993 9995 9995 9996 9996 9997 9997 09 5359 5753 6141 6517 6879 7224 7549 7852 8133 8389 8621 8830 9015 9177 9319 9441 9545 9633 9706 9767 9817 9857 9890 9916 9936 9952 9964 9974 9981 9986 9990 9993 9995 9997 9998 Common Critical Values z score Area Confidence Level Critical Value 1.645 0.9500 0.90 1.645 2.575 0.9950 0.95 1.960 0.99 2.575 Find more at www.downloadslide.com Formulas and Tables by Mario F Triola Copyright 2010 Pearson Education, Inc Ch 3: Descriptive Statistics Ch 7: Confidence Intervals (one population) ©x Mean n ©f # x xϭ Mean (frequency table) ©f xϭ sϭ sϭ sϭ ©1x - x22 B n - n 1n - 12 Standard deviation (shortcut) n ©1 f # x 224 - ©1 f # x242 n 1n - 12 B where E = z a>2 Standard deviation n 1©x 22 - 1©x22 B ˆp Ϫ E Ͻ p Ͻ ˆp ϩ E Standard deviation (frequency table) variance ϭ s Bn x - E m x + E Mean s where E = z a>2 (s known) 1n s or E = t a>2 (s unknown) 1n 1n - 12s xR2 1n - 12s s2 Variance x L2 Ch 7: Sample Size Determination Ch 4: Probability P 1A or B2 = P 1A2 + P 1B2 if A, B are mutually exclusive P 1A or B2 = P 1A2 + P 1B2 - P 1A and B2 if A, B are not mutually exclusive P 1A and B2 = P 1A2 # P 1B2 if A, B are independent P 1A and B2 = P 1A2 # P1B ƒ A2 if A, B are dependent P 1A2 = - P 1A2 Rule of complements n! Permutations (no elements alike) nPr = 1n - r2! n! Permutations (n1 alike, Á ) n 1! n 2! n k ! n! Combinations nCr = 1n - r2! r ! Ch 5: Probability Distributions m = ©x # P 1x2 Mean (prob dist.) s = 2©3x # P 1x24 - m2 Standard deviation (prob dist.) n! # p x # q n - x Binomial probability P 1x2 = 1n - x2! x ! Mean (binomial) m = n #p Variance (binomial) s2 = n # p # q Standard deviation (binomial) s = 2n # p # q x # -m Poisson distribution m e P 1x2 = Proportion pN qN x! where e Ϸ 2.71828 Ch 6: Normal Distribution x - m x - x Standard score or s s mx = m Central limit theorem n = n = n = 3z a>242 0.25 E2 3z a>242pN qN B E2 z a>2s E Proportion Proportion ( ˆp and qˆ are known) R Mean Ch 9: Confidence Intervals (two populations) 1pN - pN 22 - E 1p - p 22 1pN - pN 22 + E where E = z a>2 pN 1qN B n1 + pN 2qN n2 1x - x 22 - E 1m1 - m22 1x - x 22 + E s 21 s 22 + n2 Bn where E = t a>2 (df ϭ smaller of n1 Ϫ 1, n2 Ϫ 1) (s1 and s2 unknown and not assumed equal) sp2 sp2 + n2 Bn E = t a>2 sp2 = 1n - 12s 21 1df = n + n - 22 + 1n - 12s 22 1n - 12 + 1n - 12 (s1 and s2 unknown but assumed equal) s 12 s22 + n2 B n1 E = z a>2 ‹ z = (s1, s2 known) sx = d - E md d + E (Matched pairs) sd where E = t a>2 (df ϭ n Ϫ 1) 1n s 2n Central limit theorem (Standard error) ‹ ‹ (Indep.) Find more at www.downloadslide.com Formulas and Tables by Mario F Triola Copyright 2010 Pearson Education, Inc Ch 8: Test Statistics (one population) z = z = t = pN - p Proportion—one population pq Bn x - m x - m 1n - 12s 2n1©x 22 - 1©x22 2n1©y 22 - 1©y22 or r = Mean—one population (␴ unknown) s> 1n n©xy - 1©x21©y2 Correlation r = Mean—one population (␴ known) s> 1n x2 = Ch 10: Linear Correlation/Regression a Azx zyB n - b1 = Slope: Standard deviation or variance— one population s2 where z x = z score for x z y = z score for y n©xy - 1©x21©y2 n 1©x 22 - 1©x22 sy or b1 = r s x Ch 9: Test Statistics (two populations) z = t = 1pN - pN 22 - 1p - p 22 pq pq + B n1 n2 ‹ 1x - x 22 - 1m1 - m22 ‹ s 22 s 21 + n2 Bn y-Intercept: Two proportions x1 + x2 p = n1 + n2 df ϭ smaller of n1 Ϫ 1, n2 Ϫ Two means—independent; s1 and s2 unknown, and not assumed equal t = 1x - x 22 - 1m1 - m22 ‹ sp2 Bn + sp2 ‹ n2 (df ϭ n1 ϩ n2 Ϫ 2) sp2 = 1n - 12s 21 + 1n - 12s 22 n1 + n2 - Two means—independent; s1 and s2 unknown, but assumed equal z = 1x - x 22 - 1m1 - m22 s 12 B n1 t = F = d - md s 22 Two means—independent; ␴1, ␴2 known + n Two means—matched pairs (df ϭ n Ϫ 1) sd > 1n s 21 s22 Standard deviation or variance— two populations (where s 21 Ն s 22) or yN = b + b1x Estimated eq of regression line r2 = se = b0 = 1©y21©x 22 - 1©x21©xy2 b0 = y - b1x n 1©x 22 - 1©x22 explained variation total variation ©1y - yN22 B n - or ©y - b ©y - b1 ©xy B yN - E y yN + E where E = t a>2se n - Prediction interval + B n1x - x22 + n n1©x 22 - 1©x22 Ch 12: One-Way Analysis of Variance Procedure for testing H0: m1 = m2 = m3 = Á Use software or calculator to obtain results Identify the P-value Form conclusion: If P-value Յ a, reject the null hypothesis of equal means If P-value Ͼ a, fail to reject the null hypothesis of equal means Ch 12: Two-Way Analysis of Variance Ch 11: Goodness-of-Fit and Contingency Tables x2 = g x2 = g 1O - E22 E Goodness-of-fit (df ϭ k Ϫ 1) E Contingency table [df ϭ (r Ϫ 1)(c Ϫ 1)] 1O - E22 where E = 1row total21column total2 1grand total2 McNemar’s test for matched ƒ b - c ƒ - 12 x2 = pairs (df ϭ 1) b + c Procedure: Use software or a calculator to obtain results Test H0: There is no interaction between the row factor and column factor Stop if H0 from Step is rejected If H0 from Step is not rejected (so there does not appear to be an interaction effect), proceed with these two tests: Test for effects from the row factor Test for effects from the column factor Find more at www.downloadslide.com Formulas and Tables by Mario F Triola Copyright 2010 Pearson Education, Inc Ch 13: Nonparametric Tests z = z = 1x + 0.52 - 1n>22 1n>2 Wilcoxon signed ranks n 1n + 1212n + 12 (matched pairs and n Ͼ 30) 24 R - mR sR n 11n + n + 12 R- n 1n 21n + n + 12 = B H = Sign test for n Ͼ 25 T - n 1n + 12>4 B z = TABLE A-6 Wilcoxon rank-sum (two independent samples) 12 R 21 R 22 R 2k 12 a + + + b - 31N + 12 n2 nk N1N + 12 n Kruskal-Wallis (chi-square df ϭ k Ϫ 1) rs = - 6©d n1n - 12 Rank correlation acritical value for n 30: z = G - mG sG ; z b 1n - G - a 2n 1n + 1b n1 + n2 Runs test 12n 1n 2212n 1n - n - n 22 for n Ͼ 20 = B 1n + n 2221n + n - 12 Ch 14: Control Charts R chart: Plot sample ranges UCL: D4R Centerline: R LCL: D3R x chart: Plot sample means UCL: xx + A2R Centerline: xx n 10 11 12 13 14 15 16 17 18 19 20 25 30 35 40 45 50 60 70 80 90 100 a = 05 a = 01 950 878 811 754 707 666 632 602 576 553 532 514 497 482 468 456 444 396 361 335 312 294 279 254 236 220 207 196 990 959 917 875 834 798 765 735 708 684 661 641 623 606 590 575 561 505 463 430 402 378 361 330 305 286 269 256 NOTE: To test H0: r = against H1: r Z 0, reject H0 if the absolute value of r is greater than the critical value in the table Control Chart Constants LCL: xx - A2R p chart: Plot sample proportions pq UCL: p + B n Centerline: p LCL: p - Critical Values of the Pearson Correlation Coefficient r pq Bn Subgroup Size n A2 D3 D4 1.880 1.023 0.729 0.577 0.483 0.419 0.000 0.000 0.000 0.000 0.000 0.076 3.267 2.574 2.282 2.114 2.004 1.924 Find more at www.downloadslide.com General considerations • Context of the data • Source of the data • Sampling method • Measures of center • Measures of variation • Nature of distribution • Outliers • Changes over time • Conclusions • Practical implications FINDING P-VALUES HYPOTHESIS TEST: WORDING OF FINAL CONCLUSION Inferences about M: choosing between t and normal distributions t distribution: or Normal distribution: or s not known and normally distributed population s not known and n Ͼ 30 s known and normally distributed population s known and n Ͼ 30 Nonparametric method or bootstrapping: Population not normally distributed and n Յ 30 Find more at www.downloadslide.com NEGATIVE z Scores z TABLE A-2 Standard Normal (z) Distribution: Cumulative Area from the LEFT z 00 01 02 03 04 - 3.50 and lower - 3.4 - 3.3 - 3.2 - 3.1 - 3.0 - 2.9 - 2.8 - 2.7 - 2.6 - 2.5 - 2.4 - 2.3 - 2.2 - 2.1 - 2.0 - 1.9 - 1.8 - 1.7 - 1.6 - 1.5 - 1.4 - 1.3 - 1.2 - 1.1 - 1.0 - 0.9 - 0.8 - 0.7 - 0.6 - 0.5 - 0.4 - 0.3 - 0.2 - 0.1 0001 0003 0005 0007 0010 0013 0019 0026 0035 0047 0062 0082 0107 0139 0179 0228 0287 0359 0446 0548 0668 0808 0968 1151 1357 1587 1841 2119 2420 2743 3085 3446 3821 4207 4602 0003 0005 0007 0009 0013 0018 0025 0034 0045 0060 0080 0104 0136 0174 0222 0281 0351 0436 0537 0655 0793 0951 1131 1335 1562 1814 2090 2389 2709 3050 3409 3783 4168 4562 0003 0005 0006 0009 0013 0018 0024 0033 0044 0059 0078 0102 0132 0170 0217 0274 0344 0427 0526 0643 0778 0934 1112 1314 1539 1788 2061 2358 2676 3015 3372 3745 4129 4522 0003 0004 0006 0009 0012 0017 0023 0032 0043 0057 0075 0099 0129 0166 0212 0268 0336 0418 0516 0630 0764 0918 1093 1292 1515 1762 2033 2327 2643 2981 3336 3707 4090 4483 0003 0004 0006 0008 0012 0016 0023 0031 0041 0055 0073 0096 0125 0162 0207 0262 0329 0409 0505 0618 0749 0901 1075 1271 1492 1736 2005 2296 2611 2946 3300 3669 4052 4443 - 0.0 5000 4960 4920 4880 4840 NOTE: For values of z below - 3.49, use 0.0001 for the area *Use these common values that result from interpolation: z score - 1.645 - 2.575 Area 0.0500 0.0050 05 0003 0004 0006 0008 0011 0016 0022 0030 0040 0054 0071 0094 0122 0158 0202 0256 0322 0401 * 0495 0606 0735 0885 1056 1251 1469 1711 1977 2266 2578 2912 3264 3632 4013 4404 4801 06 07 0003 0004 0006 0008 0011 0015 0021 0029 0039 0052 0069 0091 0119 0154 0197 0250 0314 0392 0485 0594 0721 0869 1038 1230 1446 1685 1949 2236 2546 2877 3228 3594 3974 4364 0003 0004 0005 0008 0011 0015 0021 0028 0038 0051 0068 0089 0116 0150 0192 0244 0307 0384 0475 0582 0708 0853 1020 1210 1423 1660 1922 2206 2514 2843 3192 3557 3936 4325 4761 4721 08 0003 0004 0005 0007 0010 0014 0020 0027 0037 * 0049 0066 0087 0113 0146 0188 0239 0301 0375 0465 0571 0694 0838 1003 1190 1401 1635 1894 2177 2483 2810 3156 3520 3897 4286 4681 09 0002 0003 0005 0007 0010 0014 0019 0026 0036 0048 0064 0084 0110 0143 0183 0233 0294 0367 0455 0559 0681 0823 0985 1170 1379 1611 1867 2148 2451 2776 3121 3483 3859 4247 4641 Find more at www.downloadslide.com POSITIVE z Scores TABLE A-2 z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4 3.50 and up z (continued ) Cumulative Area from the LEFT 00 01 02 03 04 5000 5398 5793 6179 6554 6915 7257 7580 7881 8159 8413 8643 8849 9032 9192 9332 9452 9554 9641 9713 9772 9821 9861 9893 9918 9938 9953 9965 9974 9981 9987 9990 9993 9995 9997 9999 5040 5438 5832 6217 6591 6950 7291 7611 7910 8186 8438 8665 8869 9049 9207 9345 9463 9564 9649 9719 9778 9826 9864 9896 9920 9940 9955 9966 9975 9982 9987 9991 9993 9995 9997 5080 5478 5871 6255 6628 6985 7324 7642 7939 8212 8461 8686 8888 9066 9222 9357 9474 9573 9656 9726 9783 9830 9868 9898 9922 9941 9956 9967 9976 9982 9987 9991 9994 9995 9997 5120 5517 5910 6293 6664 7019 7357 7673 7967 8238 8485 8708 8907 9082 9236 9370 9484 9582 9664 9732 9788 9834 9871 9901 9925 9943 9957 9968 9977 9983 9988 9991 9994 9996 9997 5160 5557 5948 6331 6700 7054 7389 7704 7995 8264 8508 8729 8925 9099 9251 9382 9495 9591 9671 9738 9793 9838 9875 9904 9927 9945 9959 9969 9977 9984 9988 9992 9994 9996 9997 NOTE: For values of z above 3.49, use 0.9999 for the area *Use these common values that result from interpolation: z score Area 1.645 0.9500 2.575 0.9950 05 5199 5596 5987 6368 6736 7088 7422 7734 8023 8289 8531 8749 8944 9115 9265 9394 * 9505 9599 9678 9744 9798 9842 9878 9906 9929 9946 9960 9970 9978 9984 9989 9992 9994 9996 9997 06 07 5239 5636 6026 6406 6772 7123 7454 7764 8051 8315 8554 8770 8962 9131 9279 9406 9515 9608 9686 9750 9803 9846 9881 9909 9931 9948 9961 9971 9979 9985 9989 9992 9994 9996 9997 5279 5675 6064 6443 6808 7157 7486 7794 8078 8340 8577 8790 8980 9147 9292 9418 9525 9616 9693 9756 9808 9850 9884 9911 9932 9949 9962 9972 9979 9985 9989 9992 9995 9996 9997 08 5319 5714 6103 6480 6844 7190 7517 7823 8106 8365 8599 8810 8997 9162 9306 9429 9535 9625 9699 9761 9812 9854 9887 9913 9934 * 9951 9963 9973 9980 9986 9990 9993 9995 9996 9997 09 5359 5753 6141 6517 6879 7224 7549 7852 8133 8389 8621 8830 9015 9177 9319 9441 9545 9633 9706 9767 9817 9857 9890 9916 9936 9952 9964 9974 9981 9986 9990 9993 9995 9997 9998 Common Critical Values Confidence Critical Level Value 0.90 1.645 0.95 1.96 0.99 2.575 Find more at www.downloadslide.com TABLE A-3 t Distribution: Critical t Values 0.005 0.01 Area in One Tail 0.025 0.05 0.10 Degrees of Freedom 0.01 0.02 Area in Two Tails 0.05 0.10 0.20 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 45 50 60 70 80 90 100 200 300 400 500 1000 2000 Large 63.657 9.925 5.841 4.604 4.032 3.707 3.499 3.355 3.250 3.169 3.106 3.055 3.012 2.977 2.947 2.921 2.898 2.878 2.861 2.845 2.831 2.819 2.807 2.797 2.787 2.779 2.771 2.763 2.756 2.750 2.744 2.738 2.733 2.728 2.724 2.719 2.715 2.712 2.708 2.704 2.690 2.678 2.660 2.648 2.639 2.632 2.626 2.601 2.592 2.588 2.586 2.581 2.578 2.576 31.821 6.965 4.541 3.747 3.365 3.143 2.998 2.896 2.821 2.764 2.718 2.681 2.650 2.624 2.602 2.583 2.567 2.552 2.539 2.528 2.518 2.508 2.500 2.492 2.485 2.479 2.473 2.467 2.462 2.457 2.453 2.449 2.445 2.441 2.438 2.434 2.431 2.429 2.426 2.423 2.412 2.403 2.390 2.381 2.374 2.368 2.364 2.345 2.339 2.336 2.334 2.330 2.328 2.326 6.314 2.920 2.353 2.132 2.015 1.943 1.895 1.860 1.833 1.812 1.796 1.782 1.771 1.761 1.753 1.746 1.740 1.734 1.729 1.725 1.721 1.717 1.714 1.711 1.708 1.706 1.703 1.701 1.699 1.697 1.696 1.694 1.692 1.691 1.690 1.688 1.687 1.686 1.685 1.684 1.679 1.676 1.671 1.667 1.664 1.662 1.660 1.653 1.650 1.649 1.648 1.646 1.646 1.645 3.078 1.886 1.638 1.533 1.476 1.440 1.415 1.397 1.383 1.372 1.363 1.356 1.350 1.345 1.341 1.337 1.333 1.330 1.328 1.325 1.323 1.321 1.319 1.318 1.316 1.315 1.314 1.313 1.311 1.310 1.309 1.309 1.308 1.307 1.306 1.306 1.305 1.304 1.304 1.303 1.301 1.299 1.296 1.294 1.292 1.291 1.290 1.286 1.284 1.284 1.283 1.282 1.282 1.282 12.706 4.303 3.182 2.776 2.571 2.447 2.365 2.306 2.262 2.228 2.201 2.179 2.160 2.145 2.131 2.120 2.110 2.101 2.093 2.086 2.080 2.074 2.069 2.064 2.060 2.056 2.052 2.048 2.045 2.042 2.040 2.037 2.035 2.032 2.030 2.028 2.026 2.024 2.023 2.021 2.014 2.009 2.000 1.994 1.990 1.987 1.984 1.972 1.968 1.966 1.965 1.962 1.961 1.960 Find more at www.downloadslide.com Formulas and Tables by Mario F Triola Copyright 2010 Pearson Education, Inc Chi-Square (x2) Distribution TABLE A-4 Area to the Right of the Critical Value Degrees of Freedom 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 40 50 60 70 80 90 100 0.995 0.99 0.975 0.95 0.90 — 0.010 0.072 0.207 0.412 0.676 0.989 1.344 1.735 2.156 2.603 3.074 3.565 4.075 4.601 5.142 5.697 6.265 6.844 7.434 8.034 8.643 9.260 9.886 10.520 11.160 11.808 12.461 13.121 13.787 20.707 27.991 35.534 43.275 51.172 59.196 67.328 — 0.020 0.115 0.297 0.554 0.872 1.239 1.646 2.088 2.558 3.053 3.571 4.107 4.660 5.229 5.812 6.408 7.015 7.633 8.260 8.897 9.542 10.196 10.856 11.524 12.198 12.879 13.565 14.257 14.954 22.164 29.707 37.485 45.442 53.540 61.754 70.065 0.001 0.051 0.216 0.484 0.831 1.237 1.690 2.180 2.700 3.247 3.816 4.404 5.009 5.629 6.262 6.908 7.564 8.231 8.907 9.591 10.283 10.982 11.689 12.401 13.120 13.844 14.573 15.308 16.047 16.791 24.433 32.357 40.482 48.758 57.153 65.647 74.222 0.004 0.103 0.352 0.711 1.145 1.635 2.167 2.733 3.325 3.940 4.575 5.226 5.892 6.571 7.261 7.962 8.672 9.390 10.117 10.851 11.591 12.338 13.091 13.848 14.611 15.379 16.151 16.928 17.708 18.493 26.509 34.764 43.188 51.739 60.391 69.126 77.929 0.016 0.211 0.584 1.064 1.610 2.204 2.833 3.490 4.168 4.865 5.578 6.304 7.042 7.790 8.547 9.312 10.085 10.865 11.651 12.443 13.240 14.042 14.848 15.659 16.473 17.292 18.114 18.939 19.768 20.599 29.051 37.689 46.459 55.329 64.278 73.291 82.358 0.10 2.706 4.605 6.251 7.779 9.236 10.645 12.017 13.362 14.684 15.987 17.275 18.549 19.812 21.064 22.307 23.542 24.769 25.989 27.204 28.412 29.615 30.813 32.007 33.196 34.382 35.563 36.741 37.916 39.087 40.256 51.805 63.167 74.397 85.527 96.578 107.565 118.498 0.05 3.841 5.991 7.815 9.488 11.071 12.592 14.067 15.507 16.919 18.307 19.675 21.026 22.362 23.685 24.996 26.296 27.587 28.869 30.144 31.410 32.671 33.924 35.172 36.415 37.652 38.885 40.113 41.337 42.557 43.773 55.758 67.505 79.082 90.531 101.879 113.145 124.342 0.025 5.024 7.378 9.348 11.143 12.833 14.449 16.013 17.535 19.023 20.483 21.920 23.337 24.736 26.119 27.488 28.845 30.191 31.526 32.852 34.170 35.479 36.781 38.076 39.364 40.646 41.923 43.194 44.461 45.722 46.979 59.342 71.420 83.298 95.023 106.629 118.136 129.561 0.01 6.635 9.210 11.345 13.277 15.086 16.812 18.475 20.090 21.666 23.209 24.725 26.217 27.688 29.141 30.578 32.000 33.409 34.805 36.191 37.566 38.932 40.289 41.638 42.980 44.314 45.642 46.963 48.278 49.588 50.892 63.691 76.154 88.379 100.425 112.329 124.116 135.807 0.005 7.879 10.597 12.838 14.860 16.750 18.548 20.278 21.955 23.589 25.188 26.757 28.299 29.819 31.319 32.801 34.267 35.718 37.156 38.582 39.997 41.401 42.796 44.181 45.559 46.928 48.290 49.645 50.993 52.336 53.672 66.766 79.490 91.952 104.215 116.321 128.299 140.169 From Donald B Owen, Handbook of Statistical Tables, © 1962 Addison-Wesley Publishing Co., Reading, MA Reprinted with permission of the publisher Degrees of Freedom n - k - (r - 1)(c - 1) k - for confidence intervals or hypothesis tests with a standard deviation or variance for goodness-of-fit with k categories for contingency tables with r rows and c columns for Kruskal-Wallis test with k samples Find more at www.downloadslide.com Discover the Power of Real Data This Technology Update includes new instruction that covers major advancements in statistics software since the first printing of the Eleventh Edition In addition, new supplements are available to help students succeed in this course CHAPTER PROJECT Goodness-of-Fit and Contingency Tables StatCrunch Procedure for Goodness-of-Fit Sign into StatCrunch, then click on Open StatCrunch Click on Calculate and the results will be displayed The results include the chi-square test statistic and the P-value You must enter the observed frequencies in one column, and you must also enter the expected frequencies in another column Projects Click on Stat An experiment consists of rolling a die that is suspected of being loaded The outcomes of 1, 2, 3, 4, 5, occur with frequencies of 20, 12, 11, 8, 9, and 0, respectively Is there sufficient evidence to support the claim that the die is loaded? Click on Goodness-of-Fit, then click on the only option of Chi-Square test In the next window, select the column used for the observed frequencies and also select the column used for the expected frequencies Click on Calculate and the results will be displayed The results include the chi-square test statistic and the P-value See the accompanying display from Example in Section 11-2; the observed frequencies are in the first column, the expected frequencies are in the second column, and the results are in the window to the right The test statistic is x2 = 7.885 (rounded) and the P-value is 0.0485 StatCrunch Procedure for Contingency Tables Sign into StatCrunch, then click on Open StatCrunch Enter the row labels in a column, then enter the frequency counts in separate columns Use StatCrunch for the following StatCrunch integration includes instruction, interactive online examples, and projects for this powerful web-based statistics software NEW! StatCrunch Projects offer hands-on practice and in-depth analysis opportunities using this robust software Repeat Project using these outcomes from another die: 13, 12, 8, 12, 7, Click on Data, select Simulate data, then select Uniform and proceed to enter 100 for the number of rows, for the number of columns, for “a” and 10 for “b” Select Use single dynamic seed so that results are not all the same Click on Simulate Now use Data and Sort columns to sort the results Delete the deciyp mal portions of the numbers so that you have simulated whole numbers from to (Click on Data and select Transform data In the Y box select the column containing the data, select the floor(Y) function, where “floor” rounds the number down to a Why Do Doorways Have a Height of ft in.? The whole number, click on Set Expression, then click on Comtypical home doorway has a height of ft in., or 80 in Because men tend to be pute.) Use the goodness-of-fit test to test the claim that Stattaller than women, we will consider only men as we investigate the limitations of Crunch randomly selects the numbers so that they are equally that standard doorway height Given that heights of men are normally distributed likely with a mean of 69.0 in and a standard deviation of 2.8 in., find the percentage of Click on Stat Click on Tables, click on the option of Contingency, then click on the option of with summary In the next window, select all columns used for the observed frequencies; then in the next box select the column containing the row labels Use the data from Table 11-1 included with the Chapter Probmen who can fit through the standard doorway without bending or bumping lem and test the claim that deaths on shifts are independent of their head Is that percentage high enough to continue using 80 in as the standard whether the nurse was working Compare the results to those height? Will a doorway height of 80 in be sufficient in future years? found from Minitab as displayed on page 603 Refer to the “Data to Decision” project on the top of page 623 and use the given table to test for independence between surviving and passenger category Step 1: See Figure 6-12, which incorporates this information: Men have heights that are normally distributed with a mean of 69.0 in and a standard deviation of 2.8 in The shaded region represents the men who can fit through a doorway that has a height of 80 in Figure 6-12 Heights (in inches) of Men Area ϭ 9999 m ϭ 69 in zϭ0 x (height) x ϭ 80 in z scale z ϭ 93 Step 2: To use Table A-2, we first must use Formula 6-2 to convert from the nonstandard normal distribution to the standard normal distribution The height of 80 in is converted to a z score as follows: x - m 80 - 69.0 = 3.93 = z = s 2.8 continued Select examples are now mirrored in StatCrunch, giving students the opportunity for immediate practice Videos on DVD are included with all new copies of the book These videos feature chapter review exercises worked out by a professor This is an excellent resource for all students, particularly those enrolled in a distancelearning, individual-study, or self-paced learning program Find more at www.downloadslide.com MyStatLab Enhancements for Elementary Statistics • Increased exercise coverage provides even more opportunities to practice and gain mastery of statistics • NEW! Reading assessment questions allow students to check their understanding of the concepts • NEW! Conceptual Question Library helps students develop strong analytical skills and conceptual understanding with 1,000 new critical thinking questions • StatCrunch eText: this interactive, online textbook now includes StatCrunch—a powerful, web-based statistical software Embedded StatCrunch buttons allow users to open all data sets and tables from the book and immediately analyze with StatCrunch Name: Instructor: Date: Section: CHAPTER 2: SUMMARIZING AND GRAPHING DATA EXAMPLE 1: Constructing Various Displays of Data The number of college credits completed by a sample of 53 Elementary Statistics students is shown below Use this information to respond to the questions that follow 9 12 12 12 12 18 18 18 19 20 27 30 30 33 35 35 37 39 39 42 43 43 45 47 50 50 52 53 56 57 57 57 60 64 65 66 70 72 73 76 76 80 84 90 92 103 106 109 109 120 120 We want to create a frequency distribution with five classes What class width should we use? Class width = high data value - low data value total number of classes 120  111 22.5 o 23 Student Workbook Anne Landry Since our data values are whole numbers, our class width should also be a whole number We round up to the next highest whole number Class width = 23 What are the class limits of the five classes in our frequency distribution? The lower class limit of the first class is the lowest data value Use the class width to compute the class limits of each new class classes 9+23=32 32+23=55 55+23=78 78+23=101 frequency to accompany the Triola Statistics Series We fill in the upper class limits by making sure that the classes not overlap and that the classes have no gaps between them classes 9-31 32-54 55-77 78-100 101-123 Frequency Notice that the upper class limits are separated by the class width value of 23, just as the lower class limits were Also, notice that our highest data value belongs to the last class while our lowest data value belongs to the first class Copyright ©2011 Pearson Education, Inc Publishing as Addison-Wesley 2-1 M a r i o F Tr i o l a NEW! The Student Workbook supplements every chapter of the textbook with a substantial number of practice exercises and extra examples, many of which use real data This valuable resource is available to package with the textbook For more information, please contact your Pearson representative ... Left-tailed test; a = 0.10 21 a = 0.05; H is p Z 98.6°F 22 a = 0.01; H1 is p 0.5 23 a = 0.005; H is p 528 0 ft 24 a = 0.005; H is p Z 45 mm Finding Test Statistics In Exercises 25 28 , find the value of... distribution in this case Step The test statistic z = 22 .63 is calculated as follows: z = pN - p 0. 920 - 0.5 = = 22 .63 pq (0.5) (0.5) An A 726 We now find the P-value by using the following procedure,... An 25 Genetics Experiment The claim is that the proportion of peas with yellow pods is equal to 0 .25 (or 25 %) The sample statistics from one of Mendel’s experiments include 580 peas with 1 52 of