© 2010 Pearson Prentice Hall. All rights reserved Chapter Hypothesis Tests Regarding a Parameter 10.

© 2010 Pearson Prentice Hall. All rights reserved 10-106 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.

© 2010 Pearson Prentice Hall. All rights reserved 10-107 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

© 2010 Pearson Prentice Hall. All rights reserved 10-108 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. Examples of Claims Regarding a Characteristic of a Single Population

© 2010 Pearson Prentice Hall. All rights reserved 10-109 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 wonders if 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. Examples of Claims Regarding a Characteristic of a Single Population

© 2010 Pearson Prentice Hall. All rights reserved 10-110 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.

© 2010 Pearson Prentice Hall. All rights reserved 10-111 Hypothesis testing is a procedure, based on sample evidence and probability, used to test statements regarding a characteristic of one or more populations.

© 2010 Pearson Prentice Hall. All rights reserved 10-112 1.A statement is made regarding the nature of the population. 2.Evidence (sample data) is collected in order to test the statement. 3.The data is analyzed to assess the plausibility of the statement. Steps in Hypothesis Testing

© 2010 Pearson Prentice Hall. All rights reserved 10-113 The null hypothesis, denoted H 0, is a statement to be tested. The null hypothesis is a statement of no change, no effect or no difference. The null hypothesis is assumed true until evidence indicates otherwise. In this chapter, it will be a statement regarding the value of a population parameter.

© 2010 Pearson Prentice Hall. All rights reserved 10-114 The alternative hypothesis, denoted H 1, is a statement that we are trying to find evidence to support. In this chapter, it will be a statement regarding the value of a population parameter.

© 2010 Pearson Prentice Hall. All rights reserved 10-115 In this chapter, there are three ways to set up the null and alternative hypotheses: 1.Equal versus not equal hypothesis (two-tailed test) H 0 : parameter = some value H 1 : parameter ≠ some value

© 2010 Pearson Prentice Hall. All rights reserved 10-116 In this chapter, there are three ways to set up the null and alternative hypotheses: 1.Equal versus not equal hypothesis (two-tailed test) H 0 : parameter = some value H 1 : parameter ≠ some value 2.Equal versus less than (left-tailed test) H 0 : parameter = some value H 1 : parameter < some value

© 2010 Pearson Prentice Hall. All rights reserved 10-117 In this chapter, there are three ways to set up the null and alternative hypotheses: 1.Equal versus not equal hypothesis (two-tailed test) H 0 : parameter = some value H 1 : parameter ≠ some value 2.Equal versus less than (left-tailed test) H 0 : parameter = some value H 1 : parameter < some value 3.Equal versus greater than (right-tailed test) H 0 : parameter = some value H 1 : parameter > some value

© 2010 Pearson Prentice Hall. All rights reserved 10-118 “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. The claim that we are trying to gather evidence for determines the alternative hypothesis.

© 2010 Pearson Prentice Hall. All rights reserved 10-119 For each of the following claims, determine the null and alternative hypotheses. State whether the test is two-tailed, left-tailed or right-tailed. 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 wonders if 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. Parallel Example 2: Forming Hypotheses

© 2010 Pearson Prentice Hall. All rights reserved 10-120 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 H 0 : p=0.62. Since the researcher believes that the percentage is different today, the alternative hypothesis is a two-tailed hypothesis: H 1 : p≠0.62. Solution

© 2010 Pearson Prentice Hall. All rights reserved 10-121 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, . 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 H 0 :  =3.25. Since the researcher believes that the mean call length has increased, the alternative hypothesis is: H 1 :  > 3.25, a right-tailed test. Solution

© 2010 Pearson Prentice Hall. All rights reserved 10-122 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, . If the standard deviation with the new equipment has not changed, it will be 0.23 ounces so the null hypothesis is H 0 :  = 0.23. Since the quality control manager believes that the standard deviation has decreased, the alternative hypothesis is: H 1 :  < 0.23, a left-tailed test. Solution

© 2010 Pearson Prentice Hall. All rights reserved 10-125 1.We reject the null hypothesis when the alternative hypothesis is true. This decision would be correct. 2.We do not reject the null hypothesis when the null hypothesis is true. This decision would be correct. Four Outcomes from Hypothesis Testing

© 2010 Pearson Prentice Hall. All rights reserved 10-126 1.We reject the null hypothesis when the alternative hypothesis is true. This decision would be correct. 2.We do not reject the null hypothesis when the null hypothesis is true. This decision would be correct. 3.We 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. Four Outcomes from Hypothesis Testing

© 2010 Pearson Prentice Hall. All rights reserved 10-127 1.We reject the null hypothesis when the alternative hypothesis is true. This decision would be correct. 2.We do not reject the null hypothesis when the null hypothesis is true. This decision would be correct. 3.We 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.We 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. Four Outcomes from Hypothesis Testing

© 2010 Pearson Prentice Hall. All rights reserved 10-128 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. Parallel Example 3: Type I and Type II Errors

© 2010 Pearson Prentice Hall. All rights reserved 10-129 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%. Solution

© 2010 Pearson Prentice Hall. All rights reserved 10-130 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  >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. Solution

© 2010 Pearson Prentice Hall. All rights reserved 10-132 The probability of making a Type I error, , is chosen by the researcher before the sample data is collected. The level of significance, , is the probability of making a Type I error.

© 2010 Pearson Prentice Hall. All rights reserved 10-135 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”.

© 2010 Pearson Prentice Hall. All rights reserved 10-136 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. Suppose the sample evidence indicates that the null hypothesis should be rejected. State the wording of the conclusion. a)Suppose the sample evidence indicates that the null hypothesis should not be rejected. State the wording of the conclusion. Parallel Example 4: Stating the Conclusion

© 2010 Pearson Prentice Hall. All rights reserved 10-137 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 (  =3.25) is rejected, we conclude that there is sufficient evidence to conclude that the mean length of a phone call on a cell phone is greater than 3.25 minutes. Solution

© 2010 Pearson Prentice Hall. All rights reserved 10-138 b)Suppose the sample evidence indicates that the null hypothesis should not be rejected. State the wording of the conclusion. Since the null hypothesis (  =3.25) is not rejected, we conclude that 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. Solution

© 2010 Pearson Prentice Hall. All rights reserved 10-140 Objectives 1.Explain the logic of hypothesis testing 2.Test the hypotheses about a population mean with  known using the classical approach 3.Test hypotheses about a population mean with  known using P-values 4.Test hypotheses about a population mean with  known using confidence intervals 5.Distinguish between statistical significance and practical significance.

© 2010 Pearson Prentice Hall. All rights reserved 10-142 To test hypotheses regarding the population mean assuming the population standard deviation is known, two requirements must be satisfied: 1.A simple random sample is obtained. 2.The population from which the sample is drawn is normally distributed or the sample size is large (n≥30). If these requirements are met, the distribution of is normal with mean  and standard deviation.

© 2010 Pearson Prentice Hall. All rights reserved 10-143 Recall the researcher who believes that the mean length of a cell phone call has increased from its March, 2006 mean of 3.25 minutes. Suppose we take a simple random sample of 36 cell phone calls. Assume the standard deviation of the phone call lengths is known to be 0.78 minutes. What is the sampling distribution of the sample mean? Answer: is normally distributed with mean 3.25 and standard deviation.

© 2010 Pearson Prentice Hall. All rights reserved 10-144 Suppose the sample of 36 calls resulted in a sample mean of 3.56 minutes. Do the results of this sample suggest that the researcher is correct? In other words, would it be unusual to obtain a sample mean of 3.56 minutes from a population whose mean is 3.25 minutes? What is convincing or statistically significant evidence?

© 2010 Pearson Prentice Hall. All rights reserved 10-145 When observed results are unlikely under the assumption that the null hypothesis is true, we say the result is statistically significant. When results are found to be statistically significant, we reject the null hypothesis.

© 2010 Pearson Prentice Hall. All rights reserved 10-146 One criterion we may use for sufficient evidence for rejecting the null hypothesis is if the sample mean is too many standard deviations from the assumed (or status quo) population mean. For example, we may choose to reject the null hypothesis if our sample mean is more than 2 standard deviations above the population mean of 3.25 minutes. The Logic of the Classical Approach

© 2010 Pearson Prentice Hall. All rights reserved 10-147 Recall that our simple random sample of 36 calls resulted in a sample mean of 3.56 minutes with standard deviation of 0.13. Thus, the sample mean is standard deviations above the hypothesized mean of 3.25 minutes. Therefore, using our criterion, we would reject the null hypothesis and conclude that the mean cellular call length is greater than 3.25 minutes.

© 2010 Pearson Prentice Hall. All rights reserved 10-150 Because sample means greater than 3.51 are unusual if the population mean is 3.25, we are inclined to believe the population mean is greater than 3.25.

© 2010 Pearson Prentice Hall. All rights reserved 10-151 A second criterion we may use for sufficient evidence to support the alternative hypothesis is to compute how likely it is to obtain a sample mean at least as extreme as that observed from a population whose mean is equal to the value assumed by the null hypothesis. The Logic of the P-Value Approach

© 2010 Pearson Prentice Hall. All rights reserved 10-153 Recall So, we compute The probability of obtaining a sample mean of 3.56 minutes or more from a population whose mean is 3.25 minutes is 0.0087. This means that fewer than 1 sample in 100 will give us a mean as high or higher than 3.56 if the population mean really is 3.25 minutes. Since this outcome is so unusual, we take this as evidence against the null hypothesis.

© 2010 Pearson Prentice Hall. All rights reserved 10-154 Assuming that H 0 is true, if the probability of getting a sample mean as extreme or more extreme than the one obtained is small, we reject the null hypothesis. Premise of Testing a Hypothesis Using the P-value Approach

© 2010 Pearson Prentice Hall. All rights reserved 10-156 Testing Hypotheses Regarding the Population Mean with σ Known Using the Classical Approach To test hypotheses regarding the population mean with  known, we can use the steps that follow, provided that two requirements are satisfied: 1.The sample is obtained using simple random sampling. 2.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).

© 2010 Pearson Prentice Hall. All rights reserved 10-159 Step 3: Provided that the population from which the sample is drawn is normal or the sample size is large, and the population standard deviation, , is known, the distribution of the sample mean,, is normal with mean and standard deviation. Therefore, represents the number of standard deviations that the sample mean is from the assumed mean. This value is called the test statistic.

© 2010 Pearson Prentice Hall. All rights reserved 10-160 Step 4: The level of significance is used to determine the critical value. The critical region represents the maximum number of standard deviations that the sample mean can be from  0 before the null hypothesis is rejected. The critical region or rejection region is the set of all values such that the null hypothesis is rejected.

© 2010 Pearson Prentice Hall. All rights reserved 10-166 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.

© 2010 Pearson Prentice Hall. All rights reserved 10-167 Parallel Example 2: The Classical Approach to Hypothesis Testing A can of 7-Up states that the contents of the can are 355 ml. A quality control engineer is worried that the filling machine is miscalibrated. In other words, she wants to make sure the machine is not under- or over-filling the cans. She randomly selects 9 cans of 7-Up and measures the contents. She obtains the following data. 351360358356359 358355361352 Is there evidence at the  =0.05 level of significance to support the quality control engineer’s claim? Prior experience indicates that  =3.2ml. Source: Michael McCraith, Joliet Junior College

© 2010 Pearson Prentice Hall. All rights reserved 10-168 Solution The quality control engineer wants to know if the mean content is different from 355 ml. Since the sample size is small, we must verify that the data come from a population that is approximately normal with no outliers.

© 2010 Pearson Prentice Hall. All rights reserved 10-171 Solution Step 1: H 0 :  =355 versus H 1 :  ≠355 Step 2: The level of significance is  =0.05. Step 3: The sample mean is calculated to be 356.667. The test statistic is then The sample mean of 356.667 is 1.56 standard deviations above the assumed mean of 355 ml.

© 2010 Pearson Prentice Hall. All rights reserved 10-172 Solution Step 4: Since this is a two-tailed test, we determine the critical values at the  =0.05 level of significance to be -z 0.025 = -1.96 and z 0.025 =1.96 Step 5: Since the test statistic, z 0 =1.56, is less than the critical value 1.96, we fail to reject the null hypothesis. Step 6: There is insufficient evidence at the  =0.05 level of significance to conclude that the mean content differs from 355 ml.

© 2010 Pearson Prentice Hall. All rights reserved 10-174 A P-value is the probability of observing a sample statistic as extreme or more extreme than the one observed under the assumption that the null hypothesis is true.

© 2010 Pearson Prentice Hall. All rights reserved 10-175 Testing Hypotheses Regarding the Population Mean with σ Known Using P-values To test hypotheses regarding the population mean with  known, we can use the steps that follow to compute the P-value, provided that two requirements are satisfied: 1.The sample is obtained using simple random sampling. 2.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).

© 2010 Pearson Prentice Hall. All rights reserved 10-176 Step 1: A claim is made regarding the population mean. The claim is used to determine the null and alternative hypotheses. Again, the hypothesis can be structured in one of three ways:

© 2010 Pearson Prentice Hall. All rights reserved 10-183 Step 5: Reject the null hypothesis if the P-value is less than the level of significance, . The comparison of the P-value and the level of significance is called the decision rule. Step 6: State the conclusion.

© 2010 Pearson Prentice Hall. All rights reserved 10-184 Parallel Example 3: The P-Value Approach to Hypothesis Testing: Left-Tailed, Large Sample The volume of a stock is the number of shares traded in the stock in a day. The mean volume of Apple stock in 2007 was 35.14 million shares with a standard deviation of 15.07 million shares. A stock analyst believes that the volume of Apple stock has increased since then. He randomly selects 40 trading days in 2008 and determines the sample mean volume to be 41.06 million shares. Test the analyst’s claim at the  =0.10 level of significance using P-values.

© 2010 Pearson Prentice Hall. All rights reserved 10-185 Solution Step 1: The analyst wants to know if the stock volume has increased. This is a right-tailed test with H 0 :  =35.14 versus H 1 :  >35.14. We want to know the probability of obtaining a sample mean of 41.06 or more from a population where the mean is assumed to be 35.14.

© 2010 Pearson Prentice Hall. All rights reserved 10-187 Solution Step 4: P(Z > z 0 )=P(Z > 2.48)=0.0066. The probability of obtaining a sample mean of 41.06 or more from a population whose mean is 35.14 is 0.0066.

© 2010 Pearson Prentice Hall. All rights reserved 10-188 Solution Step 5: Since the P-value= 0.0066 is less than the level of significance, 0.10, we reject the null hypothesis. Step 6: There is sufficient evidence to reject the null hypothesis and to conclude that the mean volume of Apple stock is greater than 35.14 million shares.

© 2010 Pearson Prentice Hall. All rights reserved 10-189 Parallel Example 4: The P-Value Approach of Hypothesis Testing: Two-Tailed, Small Sample A can of 7-Up states that the contents of the can are 355 ml. A quality control engineer is worried that the filling machine is miscalibrated. In other words, she wants to make sure the machine is not under- or over-filling the cans. She randomly selects 9 cans of 7-Up and measures the contents. She obtains the following data. 351360358356359 358 355 361 352 Use the P-value approach to determine if there is evidence at the  =0.05 level of significance to support the quality control engineer’s claim. Prior experience indicates that  =3.2ml. Source: Michael McCraith, Joliet Junior College

© 2010 Pearson Prentice Hall. All rights reserved 10-190 Solution The quality control engineer wants to know if the mean content is different from 355 ml. Since we have already verified that the data come from a population that is approximately normal with no outliers, we will continue with step 1.

© 2010 Pearson Prentice Hall. All rights reserved 10-191 Solution Step 1: The quality control engineer wants to know if the content has changed. This is a two-tailed test with H 0 :  =355 versus H 1 :  ≠355.

© 2010 Pearson Prentice Hall. All rights reserved 10-193 Solution Step 4: Since this is a two-tailed test, P-value = P(Z 1.56) = 2*(0.0594)=0.1188. The probability of obtaining a sample mean that is more than 1.56 standard deviations away from the assumed mean of 355 ml is 0.1188.

© 2010 Pearson Prentice Hall. All rights reserved 10-194 Solution Step 5: Since the P-value= 0.1188 is greater than the level of significance, 0.05, we fail to reject the null hypothesis. Step 6: There is insufficient evidence to conclude that the mean content of 7-Up cans differs from 355 ml.

© 2010 Pearson Prentice Hall. All rights reserved 10-195 One advantage of using P-values over the classical approach in hypothesis testing is that P-values provide information regarding the strength of the evidence. Another is that P- values are interpreted the same way regardless of the type of hypothesis test being performed. the lower the P-value, the stronger the evidence against the statement in the null hypothesis.

© 2010 Pearson Prentice Hall. All rights reserved 10-197 When testing H 0 :  =  0 versus H 1 :  ≠  0, if a (1-  )·100% confidence interval contains  0, we do not reject the null hypothesis. However, if the confidence interval does not contain  0, we have sufficient evidence that supports the statement in the alternative hypothesis and conclude that  ≠  0 at the level of significance, .

© 2010 Pearson Prentice Hall. All rights reserved 10-198 Parallel Example 6: Testing Hypotheses about a Population Mean Using a Confidence Interval Test the hypotheses presented in Parallel Examples 2 and 4 at the  =0.05 level of significance by constructing a 95% confidence interval about , the population mean can content.

© 2010 Pearson Prentice Hall. All rights reserved 10-199 Solution Lower bound: Upper bound: We are 95% confident that the mean can content is between 354.6 ml and 358.8 ml. Since the mean stated in the null hypothesis is in this interval, there is insufficient evidence to reject the hypothesis that the mean can content is 355 ml.

© 2010 Pearson Prentice Hall. All rights reserved 10-201 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.

© 2010 Pearson Prentice Hall. All rights reserved 10-202 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.

© 2010 Pearson Prentice Hall. All rights reserved 10-203 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. Assuming a standard deviation of 4.8, determine whether the mean age has increased using a significance level of  =0.05.

© 2010 Pearson Prentice Hall. All rights reserved 10-204 Solution Step 1: To determine whether the mean age has increased, this is a right-tailed test with H 0 :  =25.2 versus H 1 :  >25.2. Step 2: The level of significance is  =0.05. Step 3: Recall that the sample mean is 25.5. The test statistic is then

© 2010 Pearson Prentice Hall. All rights reserved 10-205 Solution Step 4: Since this is a right-tailed test, P-value = P(Z > 2.17) = 0.015. The probability of obtaining a sample mean that is more than 2.17 standard deviations above the assumed mean of 25.2 is 0.015. Step 5: Because the P-value is less than the level of significance, 0.05, we reject the null hypothesis.

© 2010 Pearson Prentice Hall. All rights reserved 10-206 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). 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.

© 2010 Pearson Prentice Hall. All rights reserved 10-209 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  with s, follows Student’s t-distribution with n-1 degrees of freedom.

© 2010 Pearson Prentice Hall. All rights reserved 10-211 1.The t-distribution is different for different degrees of freedom. 2.The t-distribution is centered at 0 and is symmetric about 0. Properties of the t-Distribution

© 2010 Pearson Prentice Hall. All rights reserved 10-212 1.The t-distribution is different for different degrees of freedom. 2.The t-distribution is centered at 0 and is symmetric about 0. 3.The area under the curve is 1. Because of the symmetry, the area under the curve to the right of 0 equals the area under the curve to the left of 0 equals 1/2. Properties of the t-Distribution

© 2010 Pearson Prentice Hall. All rights reserved 10-214 4.As t increases (or decreases) without bound, the graph approaches, but never equals, 0. 5.The area in the tails of the t-distribution is a little greater than the area in the tails of the standard normal distribution because using s as an estimate of  introduces more variability to the t-statistic. Properties of the t-Distribution

© 2010 Pearson Prentice Hall. All rights reserved 10-215 6.As the sample size n increases, the density curve of t gets closer to the standard normal density curve. This result occurs because as the sample size increases, the values of s get closer to the values of  by the Law of Large Numbers. Properties of the t-Distribution

© 2010 Pearson Prentice Hall. All rights reserved 10-217 Testing Hypotheses Regarding a Population Mean with σ Unknown To test hypotheses regarding the population mean with  unknown, we use the following steps, provided that: 1.The sample is obtained using simple random sampling. 2.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).

© 2010 Pearson Prentice Hall. All rights reserved 10-232 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  =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?

© 2010 Pearson Prentice Hall. All rights reserved 10-233 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, we follow the steps for testing hypotheses about a population mean for large samples.

© 2010 Pearson Prentice Hall. All rights reserved 10-234 Solution Step 1: H 0 :  =5710 versus H 1 :  ≠5710 Step 2: The level of significance is  =0.05. Step 3: The sample mean is = 5708.07 and the sample standard deviation is s=992.05. The test statistic is

© 2010 Pearson Prentice Hall. All rights reserved 10-235 Solution: Classical Approach Step 4: Since this is a two-tailed test, we determine the critical values at the  =0.05 level of significance with n-1=45-1=44 degrees of freedom to be approximately -t 0.025 = -2.021 and t 0.025 =2.021. Step 5: Since the test statistic, t 0 =-0.013, is between the critical values, we fail to reject the null hypothesis.

© 2010 Pearson Prentice Hall. All rights reserved 10-236 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 -t 0.025 = -0.013 and to the right of t 0.025 =0.013. That is, P-value = 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.

© 2010 Pearson Prentice Hall. All rights reserved 10-237 Solution Step 6: There is insufficient evidence at the  =0.05 level of significance to conclude that the mean RMR of males listening to calm classical music differs from 5710 kJ/day.

© 2010 Pearson Prentice Hall. All rights reserved 10-238 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  =0.05 level of significance, is there evidence to conclude that state quarters have a weight different than 5.67 grams?

© 2010 Pearson Prentice Hall. All rights reserved 10-239 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.

© 2010 Pearson Prentice Hall. All rights reserved 10-242 Solution Step 1: H 0 :  =5.67 versus H 1 :  ≠5.67 Step 2: The level of significance is  =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

© 2010 Pearson Prentice Hall. All rights reserved 10-243 Solution: Classical Approach Step 4: Since this is a two-tailed test, we determine the critical values at the  =0.05 level of significance with n-1=18-1=17 degrees of freedom to be -t 0.025 = -2.11 and t 0.025 =2.11. Step 5: Since the test statistic, t 0 =2.75, is greater than the critical value 2.11, we reject the null hypothesis.

© 2010 Pearson Prentice Hall. All rights reserved 10-244 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 -t 0.025 = -2.75 and to the right of t 0.025 =2.75. That is, P-value = 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.

© 2010 Pearson Prentice Hall. All rights reserved 10-245 Solution Step 6: There is sufficient evidence at the  =0.05 level of significance to conclude that the mean weight of the state quarters differs from 5.67 grams.

© 2010 Pearson Prentice Hall. All rights reserved 10-246 Summary Which test to use? Provided that the population from which the sample is drawn is normal or that the sample size is large, if  is known, use the Z-test procedures from Section 10.2 if  is unknown, use the t-test procedures from this Section.

© 2010 Pearson Prentice Hall. All rights reserved 10-248 Objectives 1.Test the hypotheses about a population proportion 2.Test hypotheses about a population proportion using the binomial probability distribution.

© 2010 Pearson Prentice Hall. All rights reserved 10-250 Recall: The best point estimate of p, the proportion of the population with a certain characteristic, is given by where x is the number of individuals in the sample with the specified characteristic and n is the sample size.

© 2010 Pearson Prentice Hall. All rights reserved 10-251 Recall: The sampling distribution of is approximately normal, with mean and standard deviation provided that the following requirements are satisfied: 1.The sample is a simple random sample. 2. np(1-p) ≥ 10. 3.The sampled values are independent of each other.

© 2010 Pearson Prentice Hall. All rights reserved 10-252 Testing Hypotheses Regarding a Population Proportion, p To test hypotheses regarding the population proportion, we can use the steps that follow, provided that: 1.The sample is obtained by simple random sampling. 2. np 0 (1-p 0 ) ≥ 10. 3.The sampled values are independent of each other.

© 2010 Pearson Prentice Hall. All rights reserved 10-255 Step 3: Compute the test statistic Note: We use p 0 in computing the standard error rather than. This is because, when we test a hypothesis, the null hypothesis is always assumed true.

© 2010 Pearson Prentice Hall. All rights reserved 10-267 Parallel Example 1: Testing a Hypothesis about a Population Proportion: Large Sample Size In 1997, 46% of Americans said they did not trust the media “when it comes to reporting the news fully, accurately and fairly”. In a 2007 poll of 1010 adults nationwide, 525 stated they did not trust the media. At the  =0.05 level of significance, is there evidence to support the claim that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997? Source: Gallup Poll

© 2010 Pearson Prentice Hall. All rights reserved 10-268 Solution We want to know if p>0.46. First, we must verify the requirements to perform the hypothesis test: 1.This is a simple random sample. 2.np 0 (1-p 0 )=1010(0.46)(1-0.46)=250.8>10 3.Since the sample size is less than 5% of the population size, the assumption of independence is met.

© 2010 Pearson Prentice Hall. All rights reserved 10-269 Solution Step 1: H 0 : p=0.46 versus H 1 : p>0.46 Step 2: The level of significance is  =0.05. Step 3: The sample proportion is. The test statistic is then

© 2010 Pearson Prentice Hall. All rights reserved 10-270 Solution: Classical Approach Step 4: Since this is a right-tailed test, we determine the critical value at the  =0.05 level of significance to be z 0.05 = 1.645. Step 5: Since the test statistic, z 0 =3.83, is greater than the critical value 1.645, we reject the null hypothesis.

© 2010 Pearson Prentice Hall. All rights reserved 10-271 Solution: P-Value Approach Step 4: Since this is a right-tailed test, the P- value is the area under the standard normal distribution to the right of the test statistic z 0 =3.83. That is, P-value = P(Z > 3.83)≈0. Step 5: Since the P-value is less than the level of significance, we reject the null hypothesis.

© 2010 Pearson Prentice Hall. All rights reserved 10-272 Solution Step 6: There is sufficient evidence at the  =0.05 level of significance to conclude that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997.

© 2010 Pearson Prentice Hall. All rights reserved 10-275 Parallel Example 4: Hypothesis Test for a Population Proportion: Small Sample Size In 2006, 10.5% of all live births in the United States were to mothers under 20 years of age. A sociologist claims that births to mothers under 20 years of age is decreasing. She conducts a simple random sample of 34 births and finds that 3 of them were to mothers under 20 years of age. Test the sociologist’s claim at the  = 0.01 level of significance.

© 2010 Pearson Prentice Hall. All rights reserved 10-276 Parallel Example 4: Hypothesis Test for a Population Proportion: Small Sample Size Approach: Step 1: Determine the null and alternative hypotheses Step 2: Check whether np 0 (1-p 0 ) is greater than or equal to 10, where p 0 is the proportion stated in the null hypothesis. If it is, then the sampling distribution of is approximately normal and we can use the steps for a large sample size. Otherwise we use the following Steps 3 and 4.

© 2010 Pearson Prentice Hall. All rights reserved 10-277 Parallel Example 4: Hypothesis Test for a Population Proportion: Small Sample Size Approach: Step 3: Compute the P-value. For right-tailed tests, the P-value is the probability of obtaining x or more successes. For left-tailed tests, the P- value is the probability of obtaining x or fewer successes. The P-value is always computed with the proportion given in the null hypothesis. Step 4: If the P-value is less than the level of significance, , we reject the null hypothesis.

© 2010 Pearson Prentice Hall. All rights reserved 10-278 Solution Step 1: H 0 : p=0.105 versus H 1 : p<0.105 Step 2: From the null hypothesis, we have p 0 =0.105. There were 34 mothers sampled, so np 0 (1-p 0 )=3.57<10. Thus, the sampling distribution of is not approximately normal.

© 2010 Pearson Prentice Hall. All rights reserved 10-279 Solution Step 3: Let X represent the number of live births in the United States to mothers under 20 years of age. We have x=3 successes in n=34 trials so = 3/34= 0.088. We want to determine whether this result is unusual if the population mean is truly 0.105. Thus, P-value = P(X ≤ 3 assuming p=0.105 ) = P(X = 0) + P(X =1) + P(X =2) + P(X = 3) = 0.51

© 2010 Pearson Prentice Hall. All rights reserved 10-280 Solution Step 4: The P-value = 0.51 is greater than the level of significance so we do not reject H 0. There is insufficient evidence to conclude that the percentage of live births in the United States to mothers under the age of 20 has decreased below the 2006 level of 10.5%.

© 2010 Pearson Prentice Hall. All rights reserved 10-283 Chi-Square Distribution If a simple random sample of size n is obtained from a normally distributed population with mean  and standard deviation , then where s 2 is a sample variance has a chi-square distribution with n-1 degrees of freedom.

© 2010 Pearson Prentice Hall. All rights reserved 10-284 Characteristics of the Chi-Square Distribution 1.It is not symmetric. 2.The shape of the chi-square distribution depends on the degrees of freedom, just as with Student’s t-distribution. 3.As the number of degrees of freedom increases, the chi-square distribution becomes more nearly symmetric. 4.The values of  2 are nonnegative, i.e., the values of  2 are greater than or equal to 0.

© 2010 Pearson Prentice Hall. All rights reserved 10-287 Testing Hypotheses about a Population Variance or Standard Deviation To test hypotheses about the population variance or standard deviation, we can use the following steps, provided that 1.The sample is obtained using simple random sampling. 2.The population is normally distributed.

© 2010 Pearson Prentice Hall. All rights reserved 10-296 Step 4: Use Table VII to determine the approximate P-value for a left-tailed or right-tailed test by determining the area under the chi- square distribution with n-1 degrees of freedom to the left (for a left-tailed test) or right (for a right-tailed test) of the test statistic. For two-tailed tests, we recommend the use of technology or confidence intervals. P-Value Approach

© 2010 Pearson Prentice Hall. All rights reserved 10-302 Parallel Example 1: Testing a Hypothesis about a Population Standard Deviation A can of 7-Up states that the contents of the can are 355 ml. A quality control engineer is worried that the filling machine is miscalibrated. In other words, she wants to make sure the machine is not under- or over-filling the cans. She randomly selects 9 cans of 7-Up and measures the contents. She obtains the following data. 351360358356359 358355361352 In section 9.2, we assumed the population standard deviation was 3.2. Test the claim that the population standard deviation, , is greater than 3.2 ml at the  =0.05 level of significance.

© 2010 Pearson Prentice Hall. All rights reserved 10-303 Solution Step 1: H 0 :  =3.2 versus H 1 :  >3.2. This is a right-tailed test. Step 2: The level of significance is  =0.05. Step 3: From the data, the sample standard deviation is computed to be s=3.464. The test statistic is

© 2010 Pearson Prentice Hall. All rights reserved 10-304 Solution: Classical Approach Step 4: Since this is a right-tailed test, we determine the critical value at the  =0.05 level of significance with 9-1=8 degrees of freedom to be  2 0.05 = 15.507. Step 5: Since the test statistic,, is less than the critical value, 15.507, we fail to reject the null hypothesis.

© 2010 Pearson Prentice Hall. All rights reserved 10-305 Solution: P-Value Approach Step 4: Since this is a right-tailed test, the P-value is the area under the  2 -distribution with 9-1=8 to the right of the test statistic. That is, P-value = P(  2 > 9.37)>0.10. Step 5: Since the P-value is greater than the level of significance, we fail to reject the null hypothesis.

© 2010 Pearson Prentice Hall. All rights reserved 10-306 Solution: Step 6: There is insufficient evidence at the  =0.05 level of significance to conclude that the standard deviation of the can content of 7-Up is greater than 3.2 ml.

© 2010 Pearson Prentice Hall. All rights reserved 10-320 Parallel Example 1: Computing the Probability of a Type II Error In Section 10.2, we tested the hypothesis that the mean trade volume of Apple stock was greater than 35.14 million shares, H 0 :  =35.14 versus H 1 :  >35.14, based upon a random sample of size n=40 with the population standard deviation, , assumed to be 15.07 million shares at the  =0.1 level of significance. Compute the probability of a type II error given that the population mean is  =40.62.

© 2010 Pearson Prentice Hall. All rights reserved 10-321 Solution Step 1: Since z 0.9 =1.28, we let Z=1.28,  0 =35.14,  =15.07, and n=40 to find the sample mean that separates the rejection region from the nonrejection region: For any sample mean less than 38.19, we do not reject the null hypothesis.

© 2010 Pearson Prentice Hall. All rights reserved 10-326 The complement of , 1- , is the probability of rejecting the null hypothesis when the alternative hypothesis is true. The value of 1-  is referred to as the power of the test. The higher the power of the test, the more likely the test will reject the null when the alternative hypothesis is true.

© 2010 Pearson Prentice Hall. All rights reserved Chapter Hypothesis Tests Regarding a Parameter 10.

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