Sectio 9.2 Tests About a Populatio Proportio P H A N T O M S Parameters Hypothesis Assess Coditios Name the Test Test Statistic (Calculate) Obtai P value Make a decisio State coclusio Sectio 9.2 Tests About a Populatio Proportio After this sectio, you should be able to Parameters & Hypothesis Parameter: p = the actual proportio of free throws the shooter makes i the log ru. CHECK coditios for carryig out a test about a populatio proportio. CONDUCT a sigificace test about a populatio proportio. CONSTRUCT a cofidece iterval to draw a coclusio about for a two sided test about a populatio proportio. Hypothesis: H 0 : p = 0.80 H a : p < 0.80 Recall our basketball player who claimed to be a 80% free throw shooter. I a SRS of 50 freethrows, he made 32. His sample proportio of made shots, 32/50 = 0.64, is much lower tha what he claimed. Does it provide covicig evidece agaist his claim? Assess Coditios: Radom, Normal & Idepedet. Radom We ca view this set of 50 shots as a simple radom sample from the populatio of all possible shots that the player takes. Normal Assumig H 0 is true, p = 0.80. the p = (50)(0.80) = 40 ad (1 p) = (50)(0.20) = 10 are both at least 10, so the ormal coditio is met. Idepedet I our simulatio, the outcome of each shot does is determied by a radom umber geerator, so idividual observatios are idepedet. 1
Name the Test, Test Statistic (Calculate) & Obtai P value Name Test: Oe sample z test p ˆ p 0.80 ad stadard deviatio p ˆ test statistic = p(1 p) statistic - parameter stadard deviatio of statistic (0.8)(0.2) 50 0.0566 Make a Decisio & State the Coclusio Make a Decisio: P value is 0.0023. Sice the p value is so small we reject the ull hypothesis. State the Coclusio: We have covicig evidece that the basketball player does ot make 80% of his free throws. z Z score: 2.83 P value: 0.0023 0.64 0.80 2.83 0.0566 Name the Test TINspire: Meu, 6: Statistics, 7: Stats Tests, 5: 1 Prop z Test Notes o Test Statistic ad P value A sigificace test uses sample data to measure the stregth of evidece agaist H 0. Here are some priciples that apply to most tests: The test compares a statistic calculated from sample data with the value of the parameter stated by the ull hypothesis. Values of the statistic far from the ull parameter value i the directio specified by the alterative hypothesis give evidece agaist H 0. A test statistic measures how far a sample statistic diverges from what we would expect if the ull hypothesis H 0 were true, i stadardized uits. Test Statistic (Calculate) & Obtai P value The Oe Sample z Test for a Proportio The z statistic has approximately the stadard Normal distributio whe H 0 is true. Oe Sample z Test for a Proportio Choose a SRS of size from a large populatio that cotais a ukow proportio p of successes. To test the hypothesis H 0 : p =, compute the z statistic p ˆ p z (1 ) Fid the P-value by calculatig the probability of gettig a z statistic this large or larger i the directio specified by the alterative hypothesis H a : 2
Example: Oe Potato, Two Potato A potato chip producer has just received a truckload of potatoes from its mai supplier. If the producer determies that more tha 8% of the potatoes i the shipmet have blemishes, the truck will be set away to get aother load from the supplier. A supervisor selects a radom sample of 500 potatoes from the truck. A ispectio reveals that 47 of the potatoes have blemishes. Carry out a sigificace test at the α= 0.10 sigificace level. What should the producer coclude? Example: Oe Potato, Two Potato Name the Test Example: Oe Potato, Two Potato State Parameter & State Hypothesis Example: Oe Potato, Two Potato Test Statistic (Calculate) ad Obtai P value α = 0.10 sigificace level Parameter: p is the actual proportio of potatoes i this shipmet with blemishes. Hypothesis: H 0 : p = 0.08 H a : p > 0.08 Example: Oe Potato, Two Potato Assess Check Coditios Radom The supervisor took a radom sample of 500 potatoes from the shipmet. Normal Assumig H 0 : p = 0.08 is true, the expected umbers of blemished ad ublemished potatoes are = 500(0.08) = 40 ad (1 ) = 500(0.92) = 460, respectively. Because both of these values are at least 10, we should be safe doig Normal calculatios. Idepedet Because we are samplig without replacemet, we eed to check the 10% coditio. It seems reasoable to assume that there are at least 10(500) = 5000 potatoes i the shipmet. Example: Oe Potato, Two Potato Make a Decisio ad State Coclusio Make a Decisio: Sice our P value, 0.1251, is greater tha the chose sigificace level of α = 0.10, we fail to reject H 0. State Coclusio: There is ot sufficiet evidece to coclude that the shipmet cotais more tha 8% blemished potatoes. The producer will use this truckload of potatoes to make potato chips. 3
Smokig i High School Accordig to the Ceters for Disease Cotrol ad Prevetio (CDC) Web site, 50% of high school studets have ever smoked a cigarette. Taeyeo woders whether this atioal result holds true i his large, urba high school. For his AP Statistics class project, Taeyeo surveys a SRS of 150 studets from his school. He gets resposes from all 150 studets, ad 90 say that they have ever smoked a cigarette. What should Taeyeo coclude? Give appropriate evidece to support your aswer. Name the Test, Test Statistic (Calculate) & Obtai P value Name: Oe Proportio Z Test P0: 0.50 x: 90 : 150 Test Statistic: z = 2.449 Obtai p value: p = 0.0143 State Parameter & State Hypothesis Perform at test at the α = 0.05 sigificace level Parameter: p is the actual proportio of studets i Taeyeo s school who would say they have ever smoked cigarettes Hypothesis: H 0 : p = 0.50 H a : p 0.50 Make a Decisio & State Coclusio Make Decisio: Sice our P value, 0.0143, is less tha the chose sigificace level of α = 0.05. we have sufficiet evidece to reject H 0. State Coclusio: We have covicig evidece to coclude that the proportio of studets at Taeyeo s school who say they have ever smoked differs from the atioal result of 0.50. Assess Coditios Radom Taeyeo surveyed a SRS of 150 studets from his school. Normal Assumig H 0 : p = 0.50 is true, the expected umbers of smokers ad osmokers i the sample are = 150(0.50) = 75 ad (1 ) = 150(0.50) = 75. Because both of these values are at least 10, we should be safe doig Normal calculatios. Idepedet We are samplig without replacemet, we eed to check the 10% coditio. It seems reasoable to assume that there are at least 10(150) = 1500 studets a large high school. Why Cofidece Itervals Give More Iformatio The result of a sigificace test is basically a decisio to reject H 0 or fail to reject H 0. Whe we reject H 0, we re left woderig what the actual proportio p might be. A cofidece iterval might shed some light o this issue. Taeyeo foud that 90 of a SRS of 150 studets said that they had ever smoked a cigarette. Before we costruct a cofidece iterval for the populatio proportio p, we should check that both the umber of successes ad failures are at least 10. The umber of successes ad the umber of failures i the sample are 90 ad 60, respectively, so we ca proceed with calculatios. Our 95% cofidece iterval is: p ˆ z * p ˆ (1 p ˆ ) 0.60 1.96 0.60(0.40) 0.60 0.078 (0.522,0.678) 150 We are 95% cofidet that the iterval from 0.522 to 0.678 captures the true proportio of studets at Taeyeo s high school who would say that they have ever smoked a cigarette. 4
Cofidece Itervals ad Two Sided Tests There is a lik betwee cofidece itervals ad two sided tests. The 95% cofidece iterval gives a approximate rage of s that would ot be rejected by a two sided test at the α = 0.05 sigificace level. The lik is t perfect because the stadard error used for the cofidece iterval is based o the sample proportio, while the deomiator of the test statistic is based o the value from the ull hypothesis. A two sided test at sigificace level α (say, α = 0.05) ad a 100(1 α)% cofidece iterval (a 95% cofidece iterval if α = 0.05) give similar iformatio about the populatio parameter. However, if the sample proportio falls i the reject H 0 regio, the resultig 95% cofidece iterval would ot iclude. I that case, both the sigificace test ad the cofidece iterval would provide evidece that is ot the parameter value. Basketball Step 3b: Calculate Test Statistic p ˆ p 0.80 ad stadard deviatio p ˆ test statistic = z p(1 p) statistic - parameter stadard deviatio of statistic 0.64 0.80 2.83 0.0566 (0.8)(0.2) 0.0566 50 The, Usig Table A, we fid that the P value is P(z 2.83) = 0.0023. Example: Oe Potato, Two Potato Step 3: Calculatios The sample proportio of blemished potatoes is pˆ 47 / 500 0.094. Oe Proportio Z Test by Had Test statistic z p ˆ (1 ) 0.094 0.08 1.15 0.08(0.92) 500 P value Usig Table A the desired P value is P(z 1.15) = 1 0.8749 = 0.1251 Basketball Step 3a: Calculate Mea & Stadard Deviatio If the ull hypothesis H 0 : p = 0.80 is true, the the player s sample proportio of made free throws i a SRS of 50 shots would vary accordig to a approximately Normal samplig distributio with mea High School Smokig The sample proportio is Test statistic z ˆ p 60/150 0.60. p ˆ 0.60 0.50 2.45 (1 ) 0.50(0.50) 150 p ˆ p 0.80 ad stadard deviatio p ˆ p(1 p) (0.8)(0.2) 0.0566 50 P value To compute this P value, we fid the area i oe tail ad double it. Usig Table A or ormalcdf(2.45, 100) yields P(z 2.45) = 0.0071 (the righttail area). So the desired P value is2(0.0071) = 0.0142. 5
Sectio 9.2 Tests About a Populatio Proportio Summary I this sectio, we leared that As with cofidece itervals, you should verify that the three coditios Radom, Normal, ad Idepedet are met before you carry out a sigificace test. Sigificace tests for H 0 : p = are based o the test statistic p ˆ p z 0 (1 ) with P values calculated from the stadard Normal distributio. The oe sample z test for a proportio is approximately correct whe (1) the data were produced by radom samplig or radom assigmet; (2) the populatio is at least 10 times as large as the sample; ad (3) the sample is large eough to satisfy 10 ad (1 ) 10 (that is, the expected umbers of successes ad failures are both at least 10). Summary Sectio 9.2 Tests About a Populatio Proportio I this sectio, we leared that Follow the four step process whe you carry out a sigificace test: STATE: What hypotheses do you wat to test, ad at what sigificace level? Defie ay parameters you use. PLAN: Choose the appropriate iferece method. Check coditios. DO: If the coditios are met, perform calculatios. Compute the test statistic. Fid the P value. CONCLUDE: Iterpret the results of your test i the cotext of the problem. Cofidece itervals provide additioal iformatio that sigificace tests do ot amely, a rage of plausible values for the true populatio parameter p. A two sided test of H 0 : p = at sigificace level α gives roughly the same coclusio as a 100(1 α)% cofidece iterval. 6