Statistics 303Section 8.1: Confidence Interval for ProportionsConfidence Interval for ProportionsConfidence Interval for ProportionsSlide 5Sample SizeConfidence Interval ExampleSlide 8Slide 9Tests of Significance: One ProportionTests of Significance: ProportionsSlide 12Tests of SignificanceSlide 14Two Examples of Significance TestsSlide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Section 8.2: Comparing Two ProportionsComparing Two ProportionsSlide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Statistics 303Chapter 8Inference for ProportionsSection 8.1: Confidence Interval for Proportions•The same principles used for the confidence interval for the mean are used for the confidence interval of the population proportion.•Here we want to obtain a plausible range of values for the population proportion, π–Keep in mind, π, should have a value between 0 and 1Confidence Interval for Proportions•Previously, we used p as an estimate of π, so that initially we might consider p when trying to construct confidence intervals for π•However, using p can lead to confidence intervals which contain values outside of 0 and 1–Why would this be a problem?Confidence Interval for Proportions –The confidence interval has the following form:)( merrormargin of estimate p))((*pSEznXp z* is the same as beforenppSEp)1( Confidence Interval for Proportions•Thus, we can rewrite the confidence interval for a population proportion explicitly in this form.nppzp)1(*This formula should be used when n is larger than 5.Sample Size•To get a desired margin of error (m) by adjusting the sample size n we use the following:–Determine the desired margin of error (m).–Use the following formula: **2*1 ppmznwhere p* is a guessed value for the proportion of successes in the future sample.Confidence Interval Example•“A recent fire in a warehouse that contained 100,000 radios damaged an unknown number of the radios. A freight broker who purchases damaged goods offers to purchase the entire contents from the insurance company that provides coverage for the warehouse. The freight broker will eventually sort through all the radios and sell those that are not damaged. Before the broker makes an offer to the insurance company, he would like to know what proportion of the 100,000 radios are damaged and cannot be sold” (from Graybill, Iyer and Burdick, Applied Statistics, 1998).•Find a 99% confidence interval for the proportion of the 100,000 radios which are damaged.Confidence Interval Example•Suppose a random sample of 200 radios is taken from the warehouse and 34 of them were damaged.nppzp)1(*:obtain first We pnXp 2003417.0Next we use the formula:200)17.01(17.0576.217.02.576 (99% C.I.)200)0266.0)(576.2(17.0 0687.017.0 )239.0 , 101.0(99% Confidence Interval for the true proportion:Confidence Interval Example•A confidence interval of (0.101, 0.239) is not very narrow.•How large of a sample size would be needed to have a margin of error (m) of 0.01? **2*1 ppmzn2.5760.01We can use 0.17 as a good guess since it is estimated from the data. 17.0117.001.0576.2208.9363The freight broker should use a sample of size 9,364 to achieve a margin of error of 0.01.Tests of Significance: One Proportion•Steps for Testing a Population Proportion (Section 8.1)–Before using the techniques of the following slides, it is recommended that we check the following rule of thumb:•nπ0 > 10 and•n(1- π0) > 10•where n is the sample size and π0 is the hypothesized proportion as shown in the following slides.Tests of Significance: Proportions•Steps for Testing a Population Proportion–1. State the null hypothesis:–2. State the alternative hypothesis:–3. State the level of significance.•RECALL: we assume = 0.05 unless otherwise stated–4. Calculate the test statistic00:H0a:H):Hor :H be also (could0a0anpz)1(000):Hor :H be also (could0000Tests of Significance: Proportions•Steps for Testing a Population Proportion–5. Find the P-value:•For a two-sided test:•For a one-sided test:•For a one-sided test: zZzZzZP 2Pr or Pr value-a 0H :p p>a 0H :p p< zZP Pr value- zZP Pr value-0a:HTests of SignificanceLarge-Sample Significance Test for a Population Proportion•Draw an SRS of size n from a large population with unknown proportion π of successes. To test the hypothesis H0: π = π0, compute the z statistic•In terms of a standard normal random variable Z, the approximate P-value for a test of H0 against Ha: π > π0 is P(Z > z)Ha: π < π0 is P(Z < z)Ha: π ≠ π0 is 2P(Z > |z|)npz)1(000Tests of Significance: Proportions•Steps for Testing a Population Proportion–6. Reject or fail to reject H0 based on the P-value.•If the P-value is less than or equal to , reject H0.•It the P-value is greater than , fail to reject H0.–7. State your conclusion.•Your conclusion should reflect your original statement of the hypotheses.•Furthermore, your conclusion should be stated in terms of the alternative hypotheses•For example, if Ha: π ≠ π0 as stated previously–If H0 is rejected, “There is significant statistical evidence that the population proportion is different than π0.”–If H0 is not rejected, “There is not significant statistical evidence that the population mean is different than π0.”Two Examples of Significance Tests•Example 1: Orange Trees–The owner of an orange grove wants to determine if the proportion of diseased trees in the grove is more than 10%. He will use this information to determine if it will be cost effective to spray the entire grove. The owner would like to know the exact value of p, but he realizes that he cannot know the exact value unless he examines every one of the 6,010 trees, which would be too expensive. He decides to take a simple random sample of 150 trees and examine them for the disease. He finds that 12 of the 150 trees are diseased. (adapted from Graybill, Iyer and Burdick, Applied Statistics, 1998).Two Examples of Significance Tests•Orange Trees Example–Information
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