Homework ReviewIntroductionHypothesis Testing for the MeanThe HypothesesThe Level of SignificanceThe Test StatisticThe Value of the Test StatisticThe p-ValueThe DecisionThe ConclusionAnother ExampleThe TI-83AssignmentHypothesis Testing for MeansLecture 33Sections 10.1-10.2Robb T. KoetherHampden-Sydney CollegeMon, Mar 22, 2010Robb T. Koether (Hampden-Sydney College) Hypothesis Testing for Means Mon, Mar 22, 2010 1 / 50Outline1Homework Review2Introduction3Hypothesis Testing for the MeanThe HypothesesThe Level of SignificanceThe Test StatisticThe Value of the Test StatisticThe p-ValueThe DecisionThe Conclusion4Another Example5The TI-836AssignmentRobb T. Koether (Hampden-Sydney College) Hypothesis Testing for Means Mon, Mar 22, 2010 2 / 50Outline1Homework Review2Introduction3Hypothesis Testing for the MeanThe HypothesesThe Level of SignificanceThe Test StatisticThe Value of the Test StatisticThe p-ValueThe DecisionThe Conclusion4Another Example5The TI-836AssignmentRobb T. Koether (Hampden-Sydney College) Hypothesis Testing for Means Mon, Mar 22, 2010 3 / 50Homework ReviewExercise 9.25, page 595.A new method for detecting a type of cancer has been developed.Among 80 adults who have this type of cancer, this method failed todetect the cancer in five of the adults. Provide a 92% confidenceinterval estimate for the failure rate for this method.The sample proportion (point estimate) for the failure rate isˆp =580= 0.0625.For a 92% C.I., α = 0.08, so α/2 = 0.04.The coefficient is z0.04= invNorm(0.04) = −1.751.So, the confidence interval isˆp ± z0.04rˆp(1 −ˆp)n= 0.0625 ± 1.751r(0.0625)(0.9375)80= 0.625 ± (1.751)(0.0271)= 0.0625 ± 0.0474.Robb T. Koether (Hampden-Sydney College) Hypothesis Testing for Means Mon, Mar 22, 2010 4 / 50Homework ReviewExercise 9.25, page 595.A new method for detecting a type of cancer has been developed.Among 80 adults who have this type of cancer, this method failed todetect the cancer in five of the adults. Provide a 92% confidenceinterval estimate for the failure rate for this method.The sample proportion (point estimate) for the failure rate isˆp =580= 0.0625.For a 92% C.I., α = 0.08, so α/2 = 0.04.The coefficient is z0.04= invNorm(0.04) = −1.751.So, the confidence interval isˆp ± z0.04rˆp(1 −ˆp)n= 0.0625 ± 1.751r(0.0625)(0.9375)80= 0.625 ± (1.751)(0.0271)= 0.0625 ± 0.0474.Robb T. Koether (Hampden-Sydney College) Hypothesis Testing for Means Mon, Mar 22, 2010 4 / 50Homework ReviewExercise 9.25, page 595.A new method for detecting a type of cancer has been developed.Among 80 adults who have this type of cancer, this method failed todetect the cancer in five of the adults. Provide a 92% confidenceinterval estimate for the failure rate for this method.The sample proportion (point estimate) for the failure rate isˆp =580= 0.0625.For a 92% C.I., α = 0.08, so α/2 = 0.04.The coefficient is z0.04= invNorm(0.04) = −1.751.So, the confidence interval isˆp ± z0.04rˆp(1 −ˆp)n= 0.0625 ± 1.751r(0.0625)(0.9375)80= 0.625 ± (1.751)(0.0271)= 0.0625 ± 0.0474.Robb T. Koether (Hampden-Sydney College) Hypothesis Testing for Means Mon, Mar 22, 2010 4 / 50Homework ReviewExercise 9.25, page 595.A new method for detecting a type of cancer has been developed.Among 80 adults who have this type of cancer, this method failed todetect the cancer in five of the adults. Provide a 92% confidenceinterval estimate for the failure rate for this method.The sample proportion (point estimate) for the failure rate isˆp =580= 0.0625.For a 92% C.I., α = 0.08, so α/2 = 0.04.The coefficient is z0.04= invNorm(0.04) = −1.751.So, the confidence interval isˆp ± z0.04rˆp(1 −ˆp)n= 0.0625 ± 1.751r(0.0625)(0.9375)80= 0.625 ± (1.751)(0.0271)= 0.0625 ± 0.0474.Robb T. Koether (Hampden-Sydney College) Hypothesis Testing for Means Mon, Mar 22, 2010 4 / 50Homework ReviewExercise 9.25, page 595.A new method for detecting a type of cancer has been developed.Among 80 adults who have this type of cancer, this method failed todetect the cancer in five of the adults. Provide a 92% confidenceinterval estimate for the failure rate for this method.The sample proportion (point estimate) for the failure rate isˆp =580= 0.0625.For a 92% C.I., α = 0.08, so α/2 = 0.04.The coefficient is z0.04= invNorm(0.04) = −1.751.So, the confidence interval isˆp ± z0.04rˆp(1 −ˆp)n= 0.0625 ± 1.751r(0.0625)(0.9375)80= 0.625 ± (1.751)(0.0271)= 0.0625 ± 0.0474.Robb T. Koether (Hampden-Sydney College) Hypothesis Testing for Means Mon, Mar 22, 2010 4 / 50Homework ReviewSolutionOr you could use the TI-83 function 1-PropZInt.EnterIx = 5.In = 80.IC-Level = 0.92.The calculator reports the interval (.01512,.10988).Robb T. Koether (Hampden-Sydney College) Hypothesis Testing for Means Mon, Mar 22, 2010 5 / 50Outline1Homework Review2Introduction3Hypothesis Testing for the MeanThe HypothesesThe Level of SignificanceThe Test StatisticThe Value of the Test StatisticThe p-ValueThe DecisionThe Conclusion4Another Example5The TI-836AssignmentRobb T. Koether (Hampden-Sydney College) Hypothesis Testing for Means Mon, Mar 22, 2010 6 / 50IntroductionWe now ask the two basic questions, this time about the mean.IIs a given hypothesis concerning µ true?IWhat is the value of µ?In most ways, Chapter 10 will be like Chapter 9.In one way, it will be very different.Robb T. Koether (Hampden-Sydney College) Hypothesis Testing for Means Mon, Mar 22, 2010 7 / 50IntroductionWe now ask the two basic questions, this time about the mean.IIs a given hypothesis concerning µ true?IWhat is the value of µ?In most ways, Chapter 10 will be like Chapter 9.In one way, it will be very different.Robb T. Koether (Hampden-Sydney College) Hypothesis Testing for Means Mon, Mar 22, 2010 7 / 50IntroductionWe now ask the two basic questions, this time about the mean.IIs a given hypothesis concerning µ true?IWhat is the value of µ?In most ways, Chapter 10 will be like Chapter 9.In one way, it will be very different.Robb T. Koether (Hampden-Sydney College) Hypothesis Testing for Means Mon, Mar 22, 2010 7 / 50IntroductionWe now ask the two basic questions, this time about the mean.IIs a given hypothesis concerning µ true?IWhat is the value of µ?In most ways, Chapter 10 will be like Chapter 9.In one way, it will be very different.Robb T. Koether (Hampden-Sydney College) Hypothesis Testing for Means Mon, Mar 22, 2010 7 / 50IntroductionWe now ask the two basic questions, this time about the mean.IIs a given hypothesis concerning µ
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