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Overview Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 1 Definitions In statistics, a hypothesis is a claim or statement about a property of a population. A hypothesis test (or test of significance) is a standard procedure for testing a claim about a property of a population. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 2 Rare Event Rule for Inferential Statistics If, under a given assumption, the probability of a particular observed event is exceptionally small, we conclude that the assumption is probably not correct. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 3 Key Concept This section presents individual components of a hypothesis test, and the following sections use those components in comprehensive procedures. The role of the following should be understood: null hypothesis alternative hypothesis test statistic critical region significance level critical value P-value Type I and II error Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 4 Components of a Formal Hypothesis Test Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 5 Null Hypothesis: H0 The null hypothesis (denoted by H0) is a statement that the value of a population parameter (such as proportion, mean, or standard deviation) is equal to some claimed value. We test the null hypothesis directly. Either reject H0 or fail to reject H0. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 6 Alternative Hypothesis: H1 The alternative hypothesis (denoted by H1 or Ha or HA) is the statement that the parameter has a value that somehow differs from the null hypothesis. The symbolic form of the alternative hypothesis must use one of these symbols: , <, >. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 7 Example: Use the given claims to express the corresponding null and alternative hypotheses in symbolic form. a) The proportion of drivers who admit to running red lights is greater than 0.5. b) The mean height of professional basketball players is at most 7 ft. c) The standard deviation of IQ scores of actors is equal to 15. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 8 a) The proportion of drivers who admit to running red lights is greater than 0.5. • • • In Step 1, we express the given claim as p > 0.5. In Step 2, we see that if p > 0.5 is false, then p 0.5 must be true. In Step 3, we see that the expression p > 0.5 does not contain equality, so we let the alternative hypothesis H1 be p > 0.5, and we let H0 be p = 0.5. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 9 b) The mean height of professional basketball players is at most 7 ft. • In Step 1, we express “a mean of at most 7 ft” in symbols as 7. • In Step 2, we see that if 7 is false, then µ > 7 must be true. • In Step 3, we see that the expression µ > 7 does not contain equality, so we let the alternative hypothesis H1 be µ > 7, and we let H0 be µ = 7. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 10 c) The standard deviation of IQ scores of actors is equal to 15. • In Step 1, we express the given claim as = 15. • In Step 2, we see that if = 15 is false, then 15 must be true. • In Step 3, we let the alternative hypothesis H1 be 15, and we let H0 be = 15. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 11 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 12 Test Statistic The test statistic is a value used in making a decision about the null hypothesis, and is found by converting the sample statistic to a score with the assumption that the null hypothesis is true. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 13 Test Statistic - Formulas z=p-p z= pq n x - µx Test statistic for proportions Test statistic for mean n 2 = (n – 1)s2 Test statistic 2 for standard deviation Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 14 Example: A survey of n = 880 randomly selected adult drivers showed that 56% (or p = 0.56) of those respondents admitted to running red lights. Find the value of the test statistic for the claim that the majority of all adult drivers admit to running red lights. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 15 Solution: The preceding example showed that the given claim results in the following null and alternative hypotheses: H0: p = 0.5 and H1: p > 0.5. we work under the assumption that the null hypothesis is true with p = 0.5, we get the following test statistic: z = p – p = 0.56 - 0.5 = 3.56 pq n (0.5)(0.5) 880 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 16 Interpretation: We know from previous chapters that a z score of 3.56 is exceptionally large. It appears that in addition to being “more than half,” the sample result of 56% is significantly more than 50%. See figure following. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 17 Critical Region, Critical Value, Test Statistic Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 18 Critical Region 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 region in the previous figure. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 19 Significance Level The significance level (denoted by ) is the probability that the test statistic will fall in the critical region when the null hypothesis is actually true. Common choices for are 0.05, 0.01, and 0.10. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 20 Two-tailed, Right-tailed, Left-tailed Tests The tails in a distribution are the extreme regions bounded by critical values. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 21 Two-tailed Test H0: = H1: is divided equally between the two tails of the critical region Means less than or greater than Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 22 Right-tailed Test H0: = H1: > Points Right Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 23 Left-tailed Test H0: = H1: < Points Left Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 24 P-Value The P-value (or p-value or probability value) is the probability of getting a value of the test statistic that is at least as extreme as the one representing the sample data, assuming that the null hypothesis is true. The null hypothesis is rejected if the P-value is very small, such as 0.05 or less. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 25 Decision Criterion Traditional method: Reject H0 if the test statistic falls within the critical region. Fail to reject H0 if the test statistic does not fall within the critical region. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 26 Decision Criterion - cont P-value method: Reject H0 if the P-value (where is the significance level, such as 0.05). Fail to reject H0 if the P-value > . Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 27 Decision Criterion - cont Confidence Intervals: Because a confidence interval estimate of a population parameter contains the likely values of that parameter, reject a claim that the population parameter has a value that is not included in the confidence interval. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 28 Procedure for Finding P-Values Figure 8.6 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 29 Procedure for Finding P-Values Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 30 Example: Finding P-values. First determine whether the given conditions result in a righttailed test, a left-tailed test, or a two-tailed test, then find the P-values and state a conclusion about the null hypothesis. a) A significance level of = 0.05 is used in testing the claim that p > 0.25, and the sample data result in a test statistic of z = 1.18. b) A significance level of = 0.05 is used in testing the claim that p 0.25, and the sample data result in a test statistic of z = 2.34. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 31 Example: Finding P-values. First determine whether the given conditions result in a righttailed test, a left-tailed test, or a two-tailed test, then find the P-values and state a conclusion about the null hypothesis. a) With a claim of p > 0.25, the test is right-tailed. Because the test is right-tailed, Figure 8-6 shows that the P-value is the area to the right of the test statistic z = 1.18. We refer to Table A-2 and find that the area to the right of z = 1.18 is 0.1190. The P-value of 0.1190 is greater than the significance level = 0.05, so we fail to reject the null hypothesis. The P-value of 0.1190 is relatively large, indicating that the sample results could easily occur by chance. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 32 Example: Finding P-values. First determine whether the given conditions result in a righttailed test, a left-tailed test, or a two-tailed test, then find the P-values and state a conclusion about the null hypothesis. b) With a claim of p 0.25, the test is two-tailed. Because the test is two-tailed, and because the test statistic of z = 2.34 is to the right of the center, Figure 8-6 shows that the P-value is twice the area to the right of z = 2.34. We refer to Table A-2 and find that the area to the right of z = 2.34 is 0.0096, so P-value = 2 x 0.0096 = 0.0192. The P-value of 0.0192 is less than or equal to the significance level, so we reject the null hypothesis. The small P-value o 0.0192 shows that the sample results are not likely to occur by chance. Slide 33 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Wording of Final Conclusion Figure 8-7 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 34 Accept Versus Fail to Reject Some texts use “accept the null hypothesis” We are not proving the null hypothesis. The sample evidence is not strong enough to warrant rejection (such as not enough evidence to convict a suspect). Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 35 Type I Error A Type I error is the mistake of rejecting the null hypothesis when it is true. The symbol (alpha) is used to represent the probability of a type I error. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 36 Type II Error A Type II error is the mistake of failing to reject the null hypothesis when it is false. The symbol (beta) is used to represent the probability of a type II error. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 37 Suppose that we are interested in the burning rate of a solid propellant. Now burning rate is a random variable that can be described by a probability distribution. we are interested in deciding whether or not the mean burning rate is 50 centimeters per second. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 38 How to Compute Type I and II Errors Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 39 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 40 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 41 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 42 When n = 10: Increasing n to 16: Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 43 Considering a mean value closer to the hypothetical mean (50.5) Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 44 Summary Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 45 Definition The power of a hypothesis test is the probability (1 - ) of rejecting a false null hypothesis, which is computed by using a particular significance level and a particular value of the population parameter that is an alternative to the value assumed true in the null hypothesis. That is, the power of the hypothesis test is the probability of supporting an alternative hypothesis that is true. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 46 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 47 Comprehensive Hypothesis Test – P-Value Method Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 48 Comprehensive Hypothesis Test – Traditional Method Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 49 Comprehensive Hypothesis Test - cont A confidence interval estimate of a population parameter contains the likely values of that parameter. We should therefore reject a claim that the population parameter has a value that is not included in the confidence interval. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 50 Testing a Claim About a Proportion Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 51 Test Statistic for Testing a Claim About a Proportion z= p–p pq n Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 52 P-Value Method Use the same method as described in Section 8-2 and in Figure 8-8. Use the standard normal distribution (Table A-2). Traditional Method Use the same method as described in Section 8-2 and in Figure 8-9. Confidence Interval Method Use the same method as described in Section 8-2 and in Table 8-2. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 53 Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > 0.5. The sample data are n = 880, and p = 0.56. np = (880)(0.5) = 440 5 nq = (880)(0.5) = 440 5 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 54 Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > 0.5. The sample data are n = 880, and p = 0.56. We will use the P-value Method. H0: p = 0.5 H1: p > 0.5 = 0.05 z= p–p = 0.56 – 0.5 pq (0.5)(0.5) n 880 = 3.56 Referring to Table A-2, we see that for values of z = 3.50 and higher, we use 0.9999 for the cumulative area to the left of the test statistic. The P-value is 1 – 0.9999 = 0.0001. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 55 Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > 0.5. The sample data are n = 880, and p = 0.56. We will use the P-value Method. H0: p = 0.5 H1: p > 0.5 = 0.05 z= p–p = 0.56 – 0.5 pq (0.5)(0.5) n 880 = 3.56 Since the P-value of 0.0001 is less than the significance level of = 0.05, we reject the null hypothesis. There is sufficient evidence to support the claim. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 56 Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > 0.5. The sample data are n = 880, and p = 0.56. We will use the P-value Method. H0: p = 0.5 H1: p > 0.5 = 0.05 z = 3.56 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 57 Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > 0.5. The sample data are n = 880, and p = 0.56. We will use the Traditional Method. H0: p = 0.5 H1: p > 0.5 = 0.05 z= p–p = 0.56 – 0.5 pq (0.5)(0.5) n 880 = 3.56 This is a right-tailed test, so the critical region is an area of 0.05. We find that z = 1.645 is the critical value of the critical region. We reject the null hypothesis. There is sufficient evidence to support the claim. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 58 Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > 0.5. The sample data are n = 880, and p = 0.56. We will use the confidence interval method. For a one-tailed hypothesis test with significance level , we will construct a confidence interval with a confidence level of 1 – 2. We construct a 90% confidence interval. We obtain 0.533 < p < 0.588. We are 90% confident that the true value of p is contained within the limits of 0.533 and 0.588. Thus we support the claim that p > 0.5. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 59 CAUTION When testing claims about a population proportion, the traditional method and the P-value method are equivalent and will yield the same result since they use the same standard deviation based on the claimed proportion p. However, the confidence interval uses an estimated standard deviation based upon the sample proportion p. Consequently, it is possible that the traditional and P-value methods may yield a different conclusion than the confidence interval method. A good strategy is to use a confidence interval to estimate a population proportion, but use the P-value or traditional method for testing a hypothesis. Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Slide 60