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Section 10.1 The Language of Hypothesis Testing Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objectives 1. Determine the null and alternative hypotheses 2. Explain Type I and Type II errors 3. State conclusions to hypothesis tests 10-2 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 • Determine the Null and Alternative Hypotheses 10-3 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. A hypothesis is a statement regarding a characteristic of one or more populations. In this chapter, we look at hypotheses regarding a single population parameter. 10-4 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Examples of Claims Regarding a Characteristic of a Single Population • In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. Source: ReadersDigest.com poll created on 2008/05/02 10-5 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Examples of Claims Regarding a Characteristic of a Single Population • In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. • According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. 10-6 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Examples of Claims Regarding a Characteristic of a Single Population • In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. • According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. • Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased. 10-7 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. CAUTION! We test these types of statements using sample data because it is usually impossible or impractical to gain access to the entire population. If population data are available, there is no need for inferential statistics. 10-8 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Hypothesis testing is a procedure, based on sample evidence and probability, used to test statements regarding a characteristic of one or more populations. 10-9 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Steps in Hypothesis Testing 1. Make a statement regarding the nature of the population. 2. Collect evidence (sample data) to test the statement. 3. Analyze the data to assess the plausibility of the statement. 10-10 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. The null hypothesis, denoted H0, is a statement to be tested. The null hypothesis is a statement of no change, no effect or no difference and is assumed true until evidence indicates otherwise. 10-11 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. The alternative hypothesis, denoted H1, is a statement that we are trying to find evidence to support. 10-12 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. In this chapter, there are three ways to set up the null and alternative hypotheses: 1.Equal versus not equal hypothesis (two-tailed test) H0: parameter = some value H1: parameter ≠ some value 10-13 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. In this chapter, there are three ways to set up the null and alternative hypotheses: 1. Equal versus not equal hypothesis (two-tailed test) H0: parameter = some value H1: parameter ≠ some value 2. Equal versus less than (left-tailed test) H0: parameter = some value H1: parameter < some value 10-14 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. In this chapter, there are three ways to set up the null and alternative hypotheses: 1. Equal versus not equal hypothesis (two-tailed test) H0: parameter = some value H1: parameter ≠ some value 2. Equal versus less than (left-tailed test) H0: parameter = some value H1: parameter < some value 3. Equal versus greater than (right-tailed test) H0: parameter = some value H1: parameter > some value 10-15 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. “In Other Words” The null hypothesis is a statement of status quo or no difference and always contains a statement of equality. The null hypothesis is assumed to be true until we have evidence to the contrary. We seek evidence that supports the statement in the alternative hypothesis. 10-16 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Parallel Example 2: Forming Hypotheses For each of the following claims, determine the null and alternative hypotheses. State whether the test is two-tailed, left-tailed or righttailed. a) In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. c) Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased. 10-17 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution a) In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. The hypothesis deals with a population proportion, p. If the percentage participating in charity work is no different than in 2008, it will be 0.62 so the null hypothesis is H0: p=0.62. Since the researcher believes that the percentage is different today, the alternative hypothesis is a two-tailed hypothesis: H1: p≠0.62. 10-18 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. The hypothesis deals with a population mean, μ. If the mean call length on a cellular phone is no different than in 2006, it will be 3.25 minutes so the null hypothesis is H0: μ = 3.25. Since the researcher believes that the mean call length has increased, the alternative hypothesis is: H1: μ > 3.25, a right-tailed test. 10-19 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution c) Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased. The hypothesis deals with a population standard deviation, σ. If the standard deviation with the new equipment has not changed, it will be 0.23 ounces so the null hypothesis is H0: σ = 0.23. Since the quality control manager believes that the standard deviation has decreased, the alternative hypothesis is: H1: σ < 0.23, a left-tailed test. 10-20 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 2 • Explain Type I and Type II Errors 10-21 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Four Outcomes from Hypothesis Testing 1. Reject the null hypothesis when the alternative hypothesis is true. This decision would be correct. 10-22 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Four Outcomes from Hypothesis Testing 1. Reject the null hypothesis when the alternative hypothesis is true. This decision would be correct. 2. Do not reject the null hypothesis when the null hypothesis is true. This decision would be correct. 10-23 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Four Outcomes from Hypothesis Testing 1. Reject the null hypothesis when the alternative hypothesis is true. This decision would be correct. 2. Do not reject the null hypothesis when the null hypothesis is true. This decision would be correct. 3. Reject the null hypothesis when the null hypothesis is true. This decision would be incorrect. This type of error is called a Type I error. 10-24 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Four Outcomes from Hypothesis Testing 1. Reject the null hypothesis when the alternative hypothesis is true. This decision would be correct. 2. Do not reject the null hypothesis when the null hypothesis is true. This decision would be correct. 3. Reject the null hypothesis when the null hypothesis is true. This decision would be incorrect. This type of error is called a Type I error. 4. Do not reject the null hypothesis when the alternative hypothesis is true. This decision would be incorrect. This type of error is called a Type II error. 10-25 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Parallel Example 3: Type I and Type II Errors For each of the following claims, explain what it would mean to make a Type I error. What would it mean to make a Type II error? a) In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. 10-26 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution a) In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. A Type I error is made if the researcher concludes that p ≠ 0.62 when the true proportion of Americans 18 years or older who participated in some form of charity work is currently 62%. A Type II error is made if the sample evidence leads the researcher to believe that the current percentage of Americans 18 years or older who participated in some form of charity work is still 62% when, in fact, this percentage differs from 62%. 10-27 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. A Type I error occurs if the sample evidence leads the researcher to conclude that μ > 3.25 when, in fact, the actual mean call length on a cellular phone is still 3.25 minutes. A Type II error occurs if the researcher fails to reject the hypothesis that the mean length of a phone call on a cellular phone is 3.25 minutes when, in fact, it is longer than 3.25 minutes. 10-28 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. α = P(Type I Error) = P(rejecting H0 when H0 is true) β = P(Type II Error) = P(not rejecting H0 when H1 is true) 10-29 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. The probability of making a Type I error, α, is chosen by the researcher before the sample data is collected. The level of significance, α, is the probability of making a Type I error. 10-30 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. “In Other Words” As the probability of a Type I error increases, the probability of a Type II error decreases, and vice-versa. 10-31 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 3 • State Conclusions to Hypothesis Tests 10-32 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. CAUTION! We never “accept” the null hypothesis, because, without having access to the entire population, we don’t know the exact value of the parameter stated in the null. Rather, we say that we do not reject the null hypothesis. This is just like the court system. We never declare a defendant “innocent”, but rather say the defendant is “not guilty”. 10-33 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Parallel Example 4: Stating the Conclusion According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. a) Suppose the sample evidence indicates that the null hypothesis should be rejected. State the wording of the conclusion. b) Suppose the sample evidence indicates that the null hypothesis should not be rejected. State the wording of the conclusion. 10-34 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution a) Suppose the sample evidence indicates that the null hypothesis should be rejected. State the wording of the conclusion. The statement in the alternative hypothesis is that the mean call length is greater than 3.25 minutes. Since the null hypothesis (μ = 3.25) is rejected, there is sufficient evidence to conclude that the mean length of a phone call on a cell phone is greater than 3.25 minutes. 10-35 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution b) Suppose the sample evidence indicates that the null hypothesis should not be rejected. State the wording of the conclusion. Since the null hypothesis (μ = 3.25) is not rejected, there is insufficient evidence to conclude that the mean length of a phone call on a cell phone is greater than 3.25 minutes. In other words, the sample evidence is consistent with the mean call length equaling 3.25 minutes. 10-36 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Section 10.3 Hypothesis Tests for a Population Mean Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objectives 1. Test hypotheses about a mean 2. Understand the difference between statistical significance and practical significance. 10-38 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 • Test Hypotheses about a Mean 10-39 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. To test hypotheses regarding the population mean assuming the population standard deviation is unknown, we use the t-distribution rather than the Z-distribution. When we replace σ with s, x s n follows Student’s t-distribution with n –1 degrees of freedom. 10-40 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Testing Hypotheses Regarding a Population Mean To test hypotheses regarding the population mean, we use the following steps, provided that: •The sample is obtained using simple random sampling. •The sample has no outliers, and the population from which the sample is drawn is normally distributed or the sample size is large (n ≥ 30). •The sampled values are independent of each other. 10-41 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Step 1: Determine the null and alternative hypotheses. Again, the hypotheses can be structured in one of three ways: 10-42 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Step 2: Select a level of significance, α, based on the seriousness of making a Type I error. 10-43 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Classical Approach Step 3: Compute the test statistic x 0 t0 s n which follows the Student’s t-distribution with n – 1 degrees of freedom. 10-44 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Classical Approach Use Student’s t - table to determine the critical value. 10-45 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Classical Approach Two-Tailed 10-46 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Classical Approach Left-Tailed 10-47 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Classical Approach Right-Tailed 10-48 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Classical Approach Step 4: Compare the critical value with the test statistic: Rejection Region: For a probability a of a Type-I error, we can reject H0 if 1. 2. 3. 10-49 t > ta t < - ta t > ta/2 or t < - ta/2 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. P-Value Approach The p-value associated with the observed value of the test statistic is the probability of getting the observed value or a value more extreme (in the direction of Ha), assuming H0 is true. To draw conclusions using p-value, you compare the p-value with aand draw your conclusions using the following rule. If the p-value < a,reject H0. If the p-value α, fail to reject H0. 10-50 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. P-Value Approach By Hand Step 3: Compute the test statistic x 0 t0 s n which follows the Student’s t-distribution with n – 1 degrees of freedom. 10-51 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. P-Value Approach Use Student’s t- table to approximate the Pvalue. 10-52 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. P-Value Approach Two-Tailed 10-53 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. P-Value Approach Left-Tailed 10-54 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. P-Value Approach Right-Tailed 10-55 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. P-Value Approach Technology Step 3: Use a statistical spreadsheet or calculator with statistical capabilities to obtain the P-value. The directions for obtaining the P-value using the TI-83/84 Plus graphing calculator, MINITAB, Excel, and StatCrunch, are in the Technology Step-by-Step in the text. 10-56 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. P-Value Approach Step 4: If the P-value < α, reject the null hypothesis. 10-57 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Step 5: State the conclusion. 10-58 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. The procedure is robust, which means that minor departures from normality will not adversely affect the results of the test. However, for small samples, if the data have outliers, the procedure should not be used. 10-59 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Parallel Example 1: Testing a Hypothesis about a Population Mean, Large Sample Assume the resting metabolic rate (RMR) of healthy males in complete silence is 5710 kJ/day. Researchers measured the RMR of 45 healthy males who were listening to calm classical music and found their mean RMR to be 5708.07 with a standard deviation of 992.05. At the α = 0.05 level of significance, is there evidence to conclude that the mean RMR of males listening to calm classical music is different than 5710 kJ/day? 10-60 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution We assume that the RMR of healthy males is 5710 kJ/day. This is a two-tailed test since we are interested in determining whether the RMR differs from 5710 kJ/day. Since the sample size is large (n>30), we follow the steps for testing hypotheses about a population mean for large samples. 10-61 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution Step 1: H0: μ = 5710 versus H1: μ ≠ 5710 Step 2: The level of significance is α = 0.05. Step 3: The sample mean is x = 5708.07 and the sample standard deviation is s = 992.05. The test statistic is 5708.07 5710 t0 0.013 992.05 45 10-62 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution: Classical Approach Step 4: Since this is a two-tailed test, we determine the critical values at the α = 0.05 level of significance with n –1 = 45 – 1 = 44 degrees of freedom to be approximately –t0.025 = –2.021 and t0.025 = 2.021. Step 5: Since the test statistic, t0 = –0.013, is between the critical values, we fail to reject the null hypothesis. 10-63 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution: P-Value Approach Step 4: Since this is a two-tailed test, the P-value is the area under the t-distribution with n –1 = 45 – 1 = 44 degrees of freedom to the left of –t0.025= –0.013 and to the right of t0.025 = 0.013. That is, Pvalue = P(t < -0.013) + P(t > 0.013) = 2 P(t > 0.013). 0.50 < P-value. Step 5: Since the P-value is greater than the level of significance (0.05<0.5), we fail to reject the null hypothesis. 10-64 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution Step 6: There is insufficient evidence at the α = 0.05 level of significance to conclude that the mean RMR of males listening to calm classical music differs from 5710 kJ/day. 10-65 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Parallel Example 3: Testing a Hypothesis about a Population Mean, Small Sample According to the United States Mint, quarters weigh 5.67 grams. A researcher is interested in determining whether the “state” quarters have a weight that is different from 5.67 grams. He randomly selects 18 “state” quarters, weighs them and obtains the following data. 5.70 5.67 5.73 5.61 5.70 5.67 5.65 5.62 5.73 5.65 5.79 5.73 5.77 5.71 5.70 5.76 5.73 5.72 At the α = 0.05 level of significance, is there evidence to conclude that state quarters have a weight different than 5.67 grams? 10-66 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution We assume that the weight of the state quarters is 5.67 grams. This is a two-tailed test since we are interested in determining whether the weight differs from 5.67 grams. Since the sample size is small, we must verify that the data come from a population that is normally distributed with no outliers before proceeding to Steps 1-6. 10-67 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution Step 1: H0: μ = 5.67 versus H1: μ ≠ 5.67 Step 2: The level of significance is α = 0.05. Step 3: From the data, the sample mean is calculated to be 5.7022 and the sample standard deviation is s = 0.0497. The test statistic is 5.7022 5.67 t0 2.75 .0497 18 10-68 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution: Classical Approach Step 4: Since this is a two-tailed test, we determine the critical values at the α = 0.05 level of significance with n – 1 = 18 – 1 = 17 degrees of freedom to be –t0.025 = –2.11 and t0.025 = 2.11. Step 5: Since the test statistic, t0 = 2.75, is greater than the critical value 2.11, we reject the null hypothesis. 10-69 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution: P-Value Approach Step 4: Since this is a two-tailed test, the P-value is the area under the t-distribution with n – 1 = 18 – 1 = 17 degrees of freedom to the left of –t0.025 = –2.75 and to the right of t0.025 = 2.75. That is, P-value = P(t < -2.75) + P(t > 2.75) = 2 P(t > 2.75). 0.01 < P-value < 0.02. Step 5: Since the P-value is less than the level of significance (0.02<0.05), we reject the null hypothesis. 10-70 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution Step 6: There is sufficient evidence at the α = 0.05 level of significance to conclude that the mean weight of the state quarters differs from 5.67 grams. 10-71 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 5 • Distinguish between Statistical Significance and Practical Significance 10-72 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. When a large sample size is used in a hypothesis test, the results could be statistically significant even though the difference between the sample statistic and mean stated in the null hypothesis may have no practical significance. 10-73 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Practical significance refers to the idea that, while small differences between the statistic and parameter stated in the null hypothesis are statistically significant, the difference may not be large enough to cause concern or be considered important. 10-74 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Parallel Example 7: Statistical versus Practical Significance In 2003, the average age of a mother at the time of her first childbirth was 25.2. To determine if the average age has increased, a random sample of 1200 mothers is taken and is found to have a sample mean age of 25.5 with a standard deviation of 4.8, determine whether the mean age has increased using a significance level of α = 0.05. 10-75 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution Step 1: To determine whether the mean age has increased, this is a right-tailed test with H0: μ = 25.2 versus H1: μ > 25.2. Step 2: The level of significance is α = 0.05. Step 3: Recall that the sample mean is 25.5. The test statistic is then t0 10-76 25.5 25.2 4.8 2.17 1200 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution Step 4: Since this is a right-tailed test, the critical value with α = 0.05 and 1200 – 1 = 1199 degrees of freedom is P-value = P(t > 2.17). From Table VI: .01 < P-value < .02 Step 5: Because the P-value is less than the level of significance, 0.05, we reject the null hypothesis. 10-77 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Solution Step 6: There is sufficient evidence at the 0.05 significance level to conclude that the mean age of a mother at the time of her first childbirth is greater than 25.2. Although we found the difference in age to be significant, there is really no practical significance in the age difference (25.2 versus 25.5). 10-78 Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Large Sample Sizes Large sample sizes can lead to statistically significant results while the difference between the statistic and parameter is not enough to be considered practically significant. 10-79 Copyright © 2013, 2010 and 2007 Pearson Education, Inc.