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CH8 SPRING 2010

# CH8 SPRING 2010 - Chapter 8 PROPORTIONS Many statistical...

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Unformatted text preview: Chapter 8 PROPORTIONS Many statistical studies produce counts rather than measurements. Example: Did you vote in the last election? The response would be either a “Yes” or a “No”. If I did a survey, 1 would accumulate the count of “Yes” responses and describe this count as a proportion of the total. Proportion that voted = g = 338/ 1500: .225 Example: What academic year are you in at Purdue. The response would be either “Freshman”, “Sophomore”, “Junior”, or “Senior”. Again, I could accumulate the count of each and describe each as a proportion of the total. Proportion of freshmen = .2045 Proportion of sophomores = .2688 Proportion of juniors = .3215 Proportion of seniors = .2052 Population and Sample proportions: In statistical sampling we often want to estimate the proportion, p, of “successes” in a population. “Success” is when the categorical variable takes on one particular value of interest. p = count of successes in population size of population 2 X / N = a fixed but unknown value between 0 and 1.0 When we take a sample of our population; our estimator is the sample proportion of successes: A p = count of successes in sample size of sample Page 1 /\ = X/n = a known value but another sample would give a different value. Example: 1. You ﬂip a coin 20 times and record whether a head (success) or a tail is tossed. In this sample, a head is recorded 11 times. What is the sample proportion of heads? 06‘ = 11/20 = .55 Inference for a Single Proportion: So far we have only looked at making a statistical inference on population means. Now we will look at making a statistical inference on a population proportion. Examples: 0 A newspaper runs a poll to determine the popularity of a congressman in the district. The poll results in a value P = .52. In this example we are interested in estimating the unknown proportion p in the population of all voters in the district. The point estimate of that populationparameter Q is the sample proportion Q, a statistic. Sampling Distribution of a Sample Proportion: If you select a SRS of size n from a large population, and determine the sample proportion in favor of a certain issue, [A9 = X/n, you will find that: 0 The sample proportion is a statistic and ﬂuctuates, 0 The sampling distribution of [3 is approximately normal provided the sample size is big enough to produce at least 15 success and 15 failures, and does not exceed 1% of the population. 0 The mean of the sampling distribution is p, the population proportion. 0 The standard deviation of the sampling distribution is Page 2 A ' . Large-Sample Conﬁdence Iﬁerval for aPopulation ﬁoportion: Choose an SRS of size n from a large population with unknown proportion p of successes. The sample proportion is: [3 = X/n The level C conﬁdence interval for p is approximately: [Aai Z*SEA where /\ w = standard error of f n It is standard error rather than standard deviation because we do not know the value of the population proportion p, and we use the sample proportion in its place. The margin of error is: m=z*SEA P Use this interval when the number of successes and the number of failures in the sample are both at least 15 and the sample size does not exceed 1% of the population. The value of z* depends on the confidence level desired: Confidence Level 2* 90% l .645 95% 1.960 99% 2.576 Page 3 Example: 1. When trying to hire managers and executives, companies sometimes verify the academic credentials described by the applicants. One company that performs these checks summarized its findings for a six— month period. Of the 84 applicants whose credentials were checked, 15 lied about having a degree. 6 ﬁg wﬁyf 7M) #1 (fa! a Give the estimate of the proportion of applicants who lie bout having a degree and give the estimate for the standard error of /\ p. A /¢ EEH Mmmmg W1 w Wema‘m mm mm SE §£W§m1J7gﬁngWW® WWW ”W 1,1151%” aWWN me*m 3 ,09/738W b. Consider these data to be a random sample of credentials from a large collection of similar applicants. Give a 95% confidence interval for the true proportion of applicants who lie about having a degree. Page 4 Large-Sample Signiﬁcance Test for a Population Proportion: 1. State the Null and Alternative hypothesis. H0319 : p 0 H a 2 p > [90, one side right, or H a 2 p < [90, one side left, or Haipipo, two Slde 2. Find the test statistic: Draw a SRS of size n from a large population with unknown proportion p of successes. To test the hypothesis, compute the z statistic: E" P0 [900— pg) 71 3. Calculate the p-value. In terms of a standard normal random variable Z, the approximate P—value for a test of H 0 against Ha I p > p0 is P(Z 2 z) one side right Ha 2 p < p0 1s P(Z S z) one Side left Ha :p i [90 is 2P(Z 2| zl) two side 4. State the conclusions in terms of the problem. Choose a significance level such as or = 0.05, then compare the P—value to the or level. If P—value S Ct, then reject H 0 If P-value > or, then fail to reject H 0 Use the Large—Sample Significance Test for a Population Proportion if the expected number of successes (npo) in the sample and the expected number of failures (n(1— p0» in the sample are both greater than 10. Also, the population size must be at least 20 times larger than the sample. Page 5 Example: 1. Shereka, a starting player for a major college basketball team, made 60% of her free throws in her last three seasons. During the summer she worked hard on improving her free-throw accuracy. In the first nine games of this season Shereka made 48 free throws in 67 attempts. Let p be her population probability of success in making each free throw she shoots. a. Does this data indicate that her population proportion of success is higher now? State the null and alternate hypothesis to test this. Ht: ’7? m 5 Hail: M? \$53.33? A b. Calculate the test statistic, z. W H 0\ 4: N :3? méx’gﬁ __ Z 1: I7ié‘fl “53 :_ [lgiwim :é(iﬁﬁ} 47. c. Determine the P Value. If or = 0.05 what decision is indicated? f3V33t m i ”,3?33 3 . as? d. Give a 90% confidence interval for Shereka’s free— throw success probability for the new season. Are you convinced that she 1s now a better free- thrower shooter? ‘w33?11435 .3133L4iu7mwm t W m - é 7 ,7IQL92 : 362333 32333 865?92 e. What ssumptions are needed forth validity of the test and confidence interval calculations that you performed? 43 Faé53333 3 (3 13\$?th 3» m?" 5;??? .132??? f??? 035 f3 3. H \$3??333?3 Page 6 Sample Size for Desired Margin of Error: The level C confidence interval for a proportion [9 will have a margin of error approximately equal to a specified value m when the sample size satisfies * 2 Z A] p*(1—p*> m Here z* is the critical value for confidence C, and p* is an estimated value for the proportion of successes in the future sample. The estimated value can be either based on a previous pilot study or it can be assumed to be 0.5 if we have no prior knowledge. If we use 0.5 for p*, we are guaranteed that the margin of error will not exceed In no matter what the sample proportion is in our new sample. Example: 1. You want to estimate the proportion of students at your college or university who are employed for 10 or more hours per week while classes are in session. You plan to present your results using a 95% confidence interval. Using the estimated value, p* = .25, find the sample size required if the interval is to have a margin of error, m, not exceeding .06. ﬂ :3 .250 “’25)<~Li§Q->z= Zawga :75 am Repeat the calculation using p*=0.5. m : ,S"(I‘".S) (#{f—iz 26,6]? "53> 2C7 Repeat the calculation using p*=.75 n = 350775) {/1220 >2 1 200(023 22> 20/ Page 7 Section 8.2 Comparing Two Proportions Assumptions for comparing two Proportions: 0 The data consist of the two independent SRS’s 0 The two SRS’s are large enough that we get at least 10 successes and at least 10 failures in both samples. Typically we want to compare two proportions by giving a confidence interval for the difference, p1 — p2, or by testing the hypothesis of no difference, HO 2 [91- p2 = 0. Confidence Intervals for Comparing Proportions in Two Populations. Choose an SRS of size 111 from population 1 having proportion p1 of successes; choose a SRS of size n2 from population 2 having proportion p2 of success. An approximate level C confidence interval for 191 - [92 is A _ A >l< (p1 p2) i Z S E D ‘Where A X A X 1912—1 and p2=-2 ”i ”2 are the point estimates of the population proportions, and the standard error of the difference is p1(1_p1) + p2(1—I92) n1 "2 SE and z* is the value for the standard Normal density curve with area C between —z* and z*. The margin of error is: m: 2*SED Page 8 Use this method when the number of successes and the number of failures in both sample sizes are at least 10. Example 1. Is lying about credentials by job applicants changing? From a previous example, one company that performs these checks summarized its findings for the first six—month period. Of the 84 applicants whose credentials were checked, 15 lied about having a degree. The company performed the same checks for a second period, and the results for both periods are shown below. a) Find the 95% confidence interval for the difference between the population proportions for the two time periods (raga?! imagiﬁ "x MSIGZ I: vista; is: (twat?) 172%) b) Based on this confidence interval, what can you say about the change over time of the population proportion of applicants that lied on their application for the two periods? 77)::th 1.3 a egéifgerertgea ﬁefween \M [@eméa’ a ma Paws! / [706.3 nervnc/Ucfe CD Page 9 Signiﬁcance Tests for Comparing Two Proportions 1. State the Null and Alternative hypothesis. Hozpl—pzzO fl 3‘sz Haipl—p2>0,one side right or a?! :w {5:32. H a 2 p1 - p2 <0, one side left or “\$33!; \$1 {3% H :p —p ¢O,twoside a 1 2 7f 3!: 2. Find the test statistic: I “102. Choose an SRS of size mi from population 1 having proportion p1 of successes; choose a SRS of size 712 from population 2 having proportion p2 of success. To test the hypothesis, compute the z statistic: zzpl—Pz SED p where p1=X1/n1 and pZZXz/nz and where the pooled standard error is andwhereE=<X1+X2)/(n1+n2) 9 777g: F‘ﬁél/ﬁd’ F Page 10 3. Calculate the p-value. In terms of a standard normal random variable Z, the approximate P- value for a test of H 0 against Ha 2p1—p2 >0 is P(Z 2 z) one side right Ha 2191—192 <0 is P(Z S z) one side left Ha 2p1~p2 i0 is 2P(Z 2| zl) two side 4. State the conclusions in terms of the problem. Choose a significance level such as 0t 2 0.05, then compare the P—value to the 0L level. If P-value S on, then reject H0 If P-value > 0L, then fail to reject H 0 Examples 1. Data on the proportion of applicants who lied about having a degree in two consecutive six-month periods are given in the previous example as: __—— 50 a. Does this data indicate that the population proportion has increased between period 1 and period 2? Write the appropriate null and alternative hypotheses for this question. 51¢ : ”in; M if; :3. at: a? .7432} V i g a g {Eﬁwngj’gwﬁ J95: Z firm 2 0 3 E, f, 3; ‘ 1 am: «L. L) ' “‘2 lg Zé éQ Zé (W 3% Page 11 0. Determine the P Value. F5!” :22 “Qaﬁiﬂ Rt»??? “my: PM: ’5 )“.?53§'; ,éé‘EQi k .22 d. If a = .05 what decision is indicated? *2 ‘ 2 2‘ 2“ :2 a: 22 2? 5’52: 5*??? ﬁg?) , ,:,_.:::::2232 ~62 : 22: a»? 2,25 22 : 77:: 5’ ﬂy :2 2:22 fg :22 225,222; i “2 m: «£1222 ’22 22:? i e: H: 52;, :1 2:2?22 f2: W??? 22222 2, Pilaf??? 23"? ,O/éfél {‘2 am“ A??? Paw" i i 2" 5 :2? i :27” f” 32?: ¢:.:~:;'2 3L7??? 22“?”‘213 5 C? A? \$5: \«Vg’ % ‘2 ‘ 5f 5?; 2““? 222? ‘ 25’- ‘ 2*? i“ if” h g 2-22 2 1 “23 f ’3? “‘2 22 (”:3 :3? i 2:. 2E E ‘2‘" ,5” 5i 2? Q < 5‘, j “:93 ’2“ 2 221322 5w 2" R5232 1’ 2 v} 3"} if f _ ‘ I m," E}; g I if 2’ :2 t ., 22?? 2‘ is Mgéiwig? (:3fo §?\$g if? {hfé‘jﬂgﬁe Q i Page 12 ...
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