Slide 1Test 5Practice ProblemsAdditional Reading and ExamplesStatistical InferenceSlide 6Point EstimateSampling Distribution of pConfidence Interval for pAssumptionsConfidence IntervalPropertiesSignificance Test for pSignificance Test for pSlide 15Card ExerciseCard ExerciseCard ExerciseCard ExerciseCard ExerciseCard ExerciseCard ExerciseCard ExerciseCard ExerciseCard ExerciseCard Exercise Confidence IntervalSTAT 210Lecture 29Inferences on Population ProportionsNovember 3, 2017Test 5Monday, November 6Covers chapter on proportions and concepts of chapter VII.Combination of multiple choice/fill in the blank questions and problems/written questions.Formulas and tables provided, please bring calculator and writing instrument.Practice ProblemsSailboat: Pages 252 through 257Relevant problems: IX.1 through IX.18Recommended problem: IX.15Hummingbird: Pages 210 through 215Relevant problems: VIII.1 through VIII.18Recommended problem: VIII.15Additional Reading and ExamplesSailboat: Read pages 248 through 251Hummingbird: Read pages 206 through 209Statistical InferenceStatistical inference involves using statistics computed from a sample data to make statements about some parameter of the population.This chapter we have made inferences about the population proportion p.Top Hat 2Point EstimateThe point estimate of the population proportion p is the sample proportion p.p = number of successes in the sample sample size nSampling Distribution of pAssumptions:1. Simple random sample from the population2. A large enough sample so that the central limit theorem applies. The sample is large if both np and n(1 - p) are greater than or equal to 10.Then the sample proportion p is distributed approximately normalwith mean m p = p and standard deviation sp = p(1 - p) . np ~ N( p, p(1-p)/n )Confidence Interval for pGoal: Estimate the unknown population proportion pPoint Estimate: Sample proportion pSituation: While p should be close to p, it is very unlikely that p will equal p exactly.Therefore, to the point estimate we subtract and add a margin of error to create a 100(C)% confidence interval estimate for p.AssumptionsSimple random sampleBoth the number of successes in the sample np and the number of failures in the sample n(1-p) are both greater than or equal to 10.Confidence IntervalThen a 100 C% confidence interval for p is:p + z* p(1 - p)/nThe z* values are found in the table on page 340.Properties1. Increase the degree of confidence, then the margin of error and hence the width of the interval increase. Decrease the degree of confidence, then the margin of error and hence the width of the interval decrease.2. Increase the sample size, then the margin of error and hence the width of the interval decrease. Decrease the sample size, then the margin of error and hence the width of the interval increase.Significance Test for pWe hypothesize that the population proportion p equals some specified value p0 and we want to use the data in a sample to test whether this null hypothesis is appropriate or whether we should reject the null hypothesis in favor of some alternative hypothesis.Null hypothesis H0: p = p 0Ha: p > p 0Alternative hypothesis Ha: p < p 0Ha: p = p 0Significance Test for pAssumptions:Simple random sampleBoth the hypothesized number of successes np0 and the hypothesized number of failures n(1 - p0 ) are both greater than or equal to 10.Z = p - p0 p0(1 - p0)/nTop HatCard Exercise1. Half the cards in the deck are red and half the cards in the deck are black, so proportion of cards that are red is .50.Card Exercise1. Half the cards in the deck are red and half the cards in the deck are black, so proportion of cards that are red is .50.2. H0: p = .50 versus HA: p = .50Card Exercise1. Half the cards in the deck are red and half the cards in the deck are black, so proportion of cards that are red is .50.2. H0: p = .50 versus HA: p = .503. np = 25(.50) = 12.5 red cards and likewise n(1-p) = 25(1-.50) = 12.5 black cards.Card Exercise4. Since we have a simple random sample and both np and n(1-p) are greater than 10, then the assumptions are satisfied.mp = p = .50sp = p(1-p) = .50(1-.50) = .10 n 25Shape is approximately normal.Card ExerciseNumber of red cards chosen = Number of black cards chosen = p = _____ = . 25Card Exercise6. Does this data provide some evidence that the alternative hypothesis HA: p = .50 is true? Top HatCard Exercise7. (1) H0: p = .50 versus HA: p = .50, test at a = .10 (2) Assumptions satisfied (question 4) p = . Z = . - .50 = . = .50(1-.50) .10 25 (3) p-value = 2P(Z > | |) = 2P(Z > ) = 2[1-P(Z < )] = 2(1 - ) = 2( ) =Card Exercise (4) a = .10 Since p-value <>.10, we (5) There is sufficient/insufficient evidence that the proportion of cards in the deck that are red is different from .50.Card ExerciseWe concluded that there is [not] sufficient evidence that the proportion of cards in the deck that are red is different from .50. Is this the correct conclusion? Top Hat 2Card ExerciseNumber of red cards chosen = Number of black cards chosen = p = _____ = . 25If appropriate, use these results to calculate and interpret a 90% confidence interval for the proportion of red cards in this deck.Card Exercise Confidence Intervaln = 25, p = /25 = We have a simple random sample, and np = and n(1-p) = are greater than 10. For a 90% confidence interval, z* = 1.645.. + 1.645 (. )(1-. ) = . + . = (., .) 25 We have 90% confidence that the proportion of all cards in the deck that are red is between . and .
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