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Slide 1Practice ProblemsAdditional Reading and ExamplesTest 5Slide 5Inferences on the Population ProportionSampling DistributionSampling DistributionsAssumptionsCentral Limit TheoremSampling Distribution of pSampling Distribution of pExample 66/84Example 66/84Example 66/84Example 66/84Example 66/84Example 66/84Example 66/84Example 66/84Example 66/84Example 66/84Example 66/84Example 66/84Example 66/84Example 66/84Example 67/85Example 67/85Example 67/85Example 67/85Example 67/85Example 67/85Example 67/85Example 67/85Example 67/85Example 67/85Example 67/85Example 67/85ProbabilityProbabilityProbabilityExample 68/86Example 69/87Example 69/87Example 69/87Example 69/87Example 70/88Example 70/88Example 70/88Motivating ExampleMotivating ExampleSlide 52Motivating Example SolutionStatistical InferenceConfidence IntervalsExampleSlide 57Point EstimateConfidence Interval for pSampling Distribution of pSampling Distribution of pSampling Distribution of pZ-Score TransformationConfidence IntervalConfidence IntervalConfidence IntervalConfidence IntervalConfidence Interval – Z*Slide 69Confidence Interval – Z*Confidence Interval for πConfidence Interval for πConfidence IntervalsConfidence Interval for πTI-83/84 CalculatorExample 71/89Example 71/89Example 71/89Example 71/89Example 71/89Example 71/89Example 71/89Example 71/89Example 72/90Example 72/90Example 72/90Example 72/90Example 72/90Example 72/90Slide 90Example 72/90Example 72/90Example 72/90Example 73/91Example 73/91Example 73/91Example 73/91Example 73/91Example 73/91Example 73/91Example 73/91Example 73/91Example 73/91Example 73/91Example 73/91Slide 106Motivating ExampleMotivating ExampleMotivating ExampleMotivating Example SolutionSee you Wednesday!STAT 210Lecture 27Sampling Distribution of the Sample Proportion ANDConfidence Intervals for Population ProportionsOctober 31, 2016Practice ProblemsSailboat: Pages 252 through 257Relevant problems: IX.1 through IX.7Recommended problems: IX.1, IX.3 and IX.7Hummingbird: Pages 210 through 215Relevant problems: VIII.1 through VIII.7Recommended problems: VIII.1, VIII.3 and VIII.7Additional Reading and ExamplesSailboat: Read pages 248 through 251Pay particular attention to pages 248 – 249Hummingbird: Read pages 206 through 209Pay particular attention to pages 206 - 207Test 5Monday, November 7Questions for the first 10 minutes, then test – papers due promptly at the end of class!Covers chapters 7 & 8 (in Hummingbird book) or chapters 7 & 9 (in Sailboat book)Combination of multiple choice questions and written/short answer questions and problems.Formulas provided; Bring a calculator!Practice Tests and Formula Sheet on Blackboard.ClickerInferences on the Population ProportionStatistical inference involves using statistics computed from sample data to make statements about unknown population parameters.In this chapter we will learn how to use the sample proportion p to make statistical inferences about the population proportion p.Sampling DistributionA sampling distribution of a statistic is the distribution of values taken by the statistic in a large number of simple random samples of the same size n taken from the same population.Sampling DistributionsTo theoretically describe a sampling distributionwe must describe the (1) shape, (2) center, (3) spread, and (4) unusual features of the distribution.Assumptions1. The data being used to make inferences must be a simple random sample from the population.2. The population distribution must be known to be normal, or the sample size must be “large enough” for the Central Limit Theorem to apply.Since a 0-1 random variable will NEVER have a normal population shape, then we require that the sample size is large enough for the Central Limit Theorem to apply.Central Limit TheoremSince the population is not normal, we must use the Central Limit Theorem. If the sample size is “large enough” then the distribution of the sample proportion p is approximately normal.The sample will be “large enough” if BOTH the expected number of successes np and the expected number of failures n(1 - p) are greater than or equal to 10.Sampling Distribution of pIf the population proportion p is known and if both assumptions are satisfied, then the sampling distribution of the sample proportion p can be described as follows.1. Center: the mean of p is mp = p2. Spread: the standard deviation of p is sp = p (1-p) n3. Shape: the shape will be normal4. Unusual features: there will be no unusual features.Sampling Distribution of pWe can summarize the sampling distribution by writing:p ~ N(p, p (1-p) ) nThis is read “the sampling distribution of the sample proportion p is distributed normal with mean p and standard deviation p (1-p) . nExample 66/84“success” = ???“success” = student passes the statistics coursep = proportion of population that are successes = ???Example 66/84“success” = student passes the statistics coursep = proportion of population that are successes = .80Example 66/84“success” = student passes the statistics coursep =.80n = 64Example 66/84“success” = student passes the course on their first attemptp =.80n = 64Do we have a simple random sample? ClickerExample 66/84“success” = student passes the course on their first attemptp =.80n = 64Do we have a simple random sample? YesExample 66/84“success” = student passes the course on their first attemptp =.80n = 64Do we have a simple random sample? YesIs the sample size large enough for the CLT to apply? ClickerExample 66/84“success” = student passes the statistics coursep =.80n = 64np = 64(.80) = 51.2 > 10n(1 - p) = 64(1 - .80) = 12.8 > 10Hence the Central Limit Theorem applies.Example 66/84“success” = student passes the statistics coursep =.80n = 64Mean: mp = ???Example 66/84“success” = student passes the statistics coursep =.80n = 64Mean: mp = p = .80Example 66/84“success” = student passes the statistics coursep =.80n = 64Mean: mp = p = .80Standard Deviation: sp = ???Example 66/84“success” = student passes the statistics coursep =.80n = 64Mean: mp = p = .80Standard Deviation: sp = p(1 - p) = (.80)(1 - .80) n 64 = .0025 = .05Example 66/84“success” = student passes the statistics coursep =.80n = 64Mean: mp = p = .80Standard Deviation: sp = p(1 - p) = (.80)(1 - .80) n 64 = .0025 = .05Shape: NormalExample 66/84“success” = student passes the statistics coursep =.80n = 64Mean: mp = p = .80Standard Deviation: sp = p(1
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