Inference for a Single ProportionLarge-sample Confidence IntervalSignificance Test: One Sample Z-testTwo Examples of Significance TestSample Size CalculationComparing Two ProportionsExamples of Comparing Two ProportionsLarge-sample Confidence Interval for a difference in proportions (1-2)Significance Test for a Difference in ProportionsChapter 8-Inference for ProportionsAnh DaoJuly 27th, 2009Chapter 8-Inference for ProportionsInference for a Single ProportionLarge-sample Confidence IntervalThe same principles used for the confidence interval for the mean areused for the confidence interval of the population proportion.Here we want to obtain a plausible range of values for the populationproportion, π.Keep in mind, π, should have a value between 0 and 1Previously, we used p as an estimate of π, so that initially we mightconsider p when trying to construct confidence intervals for π.However, using p can lead to confidence intervals which contain valuesoutside of 0 and 1Why would this be a problem?Chapter 8-Inference for ProportionsInference for a Single ProportionLarge-sample Confidence IntervalRecall that sample proportion is:p =Xn,where X is the counts and n is the sample size. The standard error of pis:SEp=rp (1 − p )nand the margin of error of confidence level C is: m = z∗SEpand anapproximate level C confidence interval for p is: p ± m.Thus, we can rewrite the confidence interval for a population proportionexplicitly in this form:p ± z∗rp (1 − p )nThis formula should be used when both np ≥ 15 and n(1 − p) ≥ 15.Chapter 8-Inference for ProportionsInference for a Single ProportionLarge-sample Confidence IntervalRadio ExampleA recent fire in a warehouse that contained 100,000 radios damaged anunknown number of the radios. A freight broker who purchasesdamaged goods offers to purchase the entire contents from theinsurance company that provides coverage for the warehouse. Thefreight broker will eventually sort through all the radios and sell thosethat are not damaged. Before the broker makes an offer to theinsurance company, he would like to know what proportion of the100,000 radios are damaged and cannot be sold. (from Graybill, Iyerand Burdick, Applied Statistics, 1998).Find a 99% confidence interval for the proportion of the 100,000 radioswhich are damaged.Chapter 8-Inference for ProportionsInference for a Single ProportionLarge-sample Confidence IntervalRadio ExampleSuppose a random sample of 200 radios is taken from the warehouse and 34of them were damaged.We first obtain p: p =Xn=34200= 0.17.Next we use the formulap ± z∗rp (1 − p )n= 0.17 ± 2.576r0.17(1 − 0.17)200= (0.101, 0.239)This is a 99% Confidence Interval for the true proportion.Chapter 8-Inference for ProportionsInference for a Single ProportionSignificance Test: One Sample Z -testFor one single proportion, we want to test H0: π = π0, v.s. Ha: π 6= π0(or Ha: π > π0or Ha: π < π0).The first thing recommended is to check the following rules of thumb:nπ0≥ 10n(1 − π0) ≥ 10where n is the sample size and π0is the hypothesized proportion.Chapter 8-Inference for ProportionsInference for a Single ProportionSignificance Test: One Sample Z -test1State the null hypothesis: H0: π = π02State the alternative hypothesis: Ha: π 6= π0or π > π0or π < π03State the level of significance.RECALL: we assume α = 0.05 unless otherwise stated.4Calculate the test statistic (z-statistics):z =p − π0qπ0(1−π0)nChapter 8-Inference for ProportionsInference for a Single ProportionSignificance Test: One Sample Z -test5Find the p -valueFor a Two-sided Test: Ha: π 6= π0p − value = P(Z ≥ |z| or Z ≤ |z|) = 2P(Z ≥ |z|)For an One-sided Test: Ha: π > π0p − value = P(Z ≥ z)For an One-sided Test: Ha: π < π0p − value = P(Z ≤ z)Chapter 8-Inference for ProportionsInference for a Single ProportionSignificance Test: One Sample Z -testChapter 8-Inference for ProportionsInference for a Single ProportionSignificance Test: One Sample Z -test6Reject or fail to reject H0based on the p-value.If the p-value is less than or equal to α, reject H0.It the p-value is greater than α, fail to reject H0.7State your conclusion.Your conclusion should reflect your original statement of thehypotheses.Furthermore, your conclusion should be stated in terms of thealternative hypotheses, e.g.If H0is rejected, “there is significant statistical evidence thatthe population proportion is different than π0.”If H0is not rejected, “there is not significant statisticalevidence that the population mean is different than π0.”Chapter 8-Inference for ProportionsInference for a Single ProportionTwo Examples of Significance TestOrange Trees ExampleThe owner of an orange grove wants to determine if the proportion ofdiseased trees in the grove is more than 10%. He will use thisinformation to determine if it will be cost effective to spray the entiregrove. The owner would like to know the exact value of π, but herealizes that he cannot know the exact value unless he examines everyone of the 6,010 trees, which would be too expensive. He decides totake a simple random sample of 150 trees and examine them for thedisease. He finds that 12 of the 150 trees are diseased. (adapted fromGraybill, Iyer and Burdick, Applied Statistics, 1998).Chapter 8-Inference for ProportionsInference for a Single ProportionTwo Examples of Significance TestOrange Trees ExampleInformation we have:n = 150, X = 12, π0= 0.10p =Xn=12150= 0.08.We first check the rules of thumb.nπ0= 150(0.10) = 15 > 10n(1 − π0) = 150(1 − 0.10) = 135 > 10The assumptions for the test are met.Chapter 8-Inference for ProportionsInference for a Single ProportionTwo Examples of Significance TestOrange Trees ExampleState the null hypothesis: H0: π ≤ 0.10State the alternative hypothesis: Ha: π > 0.10State the level of significance: assume α = 0.05.Calculate the test statisticz =p − π0qπ0(1−π0)n=0.08 − 0.10q0.10(1−0.10)150= −0.82Chapter 8-Inference for ProportionsInference for a Single ProportionTwo Examples of Significance TestOrange Trees ExampleFind the p-valuep − value = P(Z ≥ z) = P(Z ≥ −0.82) = 1 − P(Z < −0.82) = 0.7939Do we reject or fail to reject H0based on the p -value?p -value = 0.7939 is greater than α = 0.05.State the conclusion.We fail to reject H0and conclude that: “There is not significant statisticalevidence that the true proportion of diseased orange trees is greaterthan 10%.”Chapter 8-Inference for
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