9 Master Presentation 8.1 Example • A survey of 436 workers showed that 192 of them said that it was seriously unethical to monitor employee e-mail. When 121 senior- level bosses were surveyed, 40 said that it was seriously unethical to monitor employee e-mail. • Let and be the population proportion of workers and bosses that feel it‟s unethical to pW pB monitor e-mail. We might want to obtain a CI for pW. pB We would first need an estimate of this difference. It should seem reasonable that an estimate be pW pB ˆ ˆ 192 / 436 40 /121 0.1097 The standard error of ˆ pBˆ pW is estimated by pW (1 pW ) pB (1 pB ) ˆ ˆ ˆ ˆ s.e.( pW pB ) ˆ ˆ nW nB and just thinking intuitively, this means a CI for pW is pB pW pB z s.e.( pW pB ) ˆ ˆ ˆ * ˆ • To compute a CI for B pWwepneed andˆ B p ˆ pW which are 192/436= 0.4403 and 40/121=0.3305 respectively. This gives a standard error of 0.4403 (1 0.4403 ) 0.3305 (1 0.3305 ) 0.0489 436 121 • Now, if we want to obtain an 80% CI for pW pB have ˆ ˆ we 0.1097 1.282 (0.0489 ) (0.047 ,0.172 ) • Suppose we want to test the claim that the a larger percentage of workers feel that it‟s unethical to monitor email. That is H1 : pW pB H1 : pW pB 0 Again, it should seem intuitive that the test statistic will be of the form pW pB ˆ ˆ pW (1 pW ) pB (1 pB ) nW nB but under H0, pW and pB are equal. So, in the denominator, we can simply replace this with p. pW pB ˆ ˆ p (1 p ) p (1 p) nW nB An estimate for p is (192+40)/(436+40) =0.4165. This gives the test statistic as 2.1656. Similar to the one sample tests, we can make a decision by • comparing the test statistic to the critical value. If α = 0.05, then the critical value is 1.645. Since TS > CV, reject H0. • or we can compare the p-value to α. The p- value is found as P(Z > 2.1656) =0.015. Since this value is less than α, we reject H0. Another example A major court case on the health effects of drinking contaminated water took place in the town of Woburn, Massachusetts. A town well was contaminated with industrial chemicals. During the period when the well was open, 16 birth defects out of 414 births. When this particular well was shut off from and water was supplied from other wells, 3 out of 228 birth defects were reported. The plaintiffs suing the firm responsible for contaminating the well claim that the rate of birth defects is higher when the contaminated well was in use. Denote the contaminated well as „C‟ and the other uncontaminated wells as „U‟ and p be the proportion of birth defects. What exactly are the plaintiffs wanting to test? • Obtain a 98% confidence interval for the difference in the rate of birth defects for when the well was on compared to when it was shut off. • What is the test statistic? • What‟s the critical value if we use α=0.01? • What‟s the conclusion? Should the plaintiffs be favored here? Confidence Interval for p Reasonable Range of Values for True Population Proportion p Confidence Interval for p • The goal is to take a sample and be able to make intelligent guesses about the true value of the proportion p in the population. • A valuable tool is the confidence interval: the range of values for p in the population that could reasonably have produced the sample p-hat we observed. CI Formula • A confidence interval for the population p is given by: p (1 p ) ˆ ˆ pZ ˆ * n CI Formula • A 95 percent confidence interval for the population p is given by: p (1 p ) ˆ ˆ p 1.96 ˆ n Example • Suppose we cure p-hat = .9 of n=1000 heartworm infected dogs. What is the reasonable range for the cure rate p of our new . CI for treatment? Do 95%9(.1) p. .9 1.96 1000 .9 1.96(.009487) .9 .0185 (.8815,.9185) Example • Reasonable range for p (.88, .92) is same range argued in previous section on sampling distributions for p-hat. • The only reasonable values for p are those that could produce p-hats only a couple of standard deviations removed from the truth. Reeses Pieces Example • What is the proportion of orange candies, p? • To study this unknown, but very important value p, we will construct confidence intervals for p from samples of candies. • Each bag represents a random sample of size n from the population of these candies. • From each bag your group should: find n, Reeses Pieces Example • On whiteboard place your information in tabular form: Grou N P-hat CI p 1 2 3 4 5 6 Reeses Pieces Example • A histogram of p-hat values should result in a representation of the sampling distribution of p-hat. • The center of this histogram should be p. What do you think p is? Reeses Pieces Example • From the CI‟s, what do you think the true p is? • Is an evenly distributed color distribution p=1/3, a reasonable hypothesis based on our data? Why or why not? • Pay attention to the written conclusion I provide on the board ! Vietnam Veterans Divorce Rate • N=2101 veterans interviewed found p-hat=777/2101 = .3698 had been divorced at least once. • What is reasonable range of values for true divorce proportion p? .3698 1.96(.01053) .3698 .02064 (.349,.390) Vietnam Vets Divorces • Do you think true divorce proportion is greater than .5? • Ans: No. The reasonable range of values for the true p is (.349, .390). This range is entirely below p=.5, so we have strong evidence that the true divorce proportion is BELOW .5 not above it. Vietnam Vets Divorces • Do you think the true divorce proportion could be .37? • Ans: Yes, a proportion like .37 is a reasonable value for the true p according to our range of reasonable values, so the truth could reasonably be .37. Domestic Violence • For those women who had experienced some abuse before age 18, the sample proportion that had experienced some abuse in the past 12 months was p-hat = 236/569 = .4147 • CI for p: (.374, .455). • Suppose the true proportion currently abused for those not abuse before age 18 was .11. • Is there evidence the true population proportion in our study is greater than .11? Ask Marilyn – Let‟s Make a Deal • In 1991 a reader wrote to Marilyn Vos Savant (highest documented IQ) and asked whether a player should switch doors when playing Let‟s Make a Deal. • There are 3 doors, two with goats and one with a car. You pick a door. The host, Monty Hall shows you a door you have not picked and there is a goat behind it. You are then asked if you wish to switch doors. Should you switch? Let‟s Make a Deal • Marilyn said yes, you should switch doors. • There was a storm of angry letters from bad colleges with bad statistics professors. • “you are the goat”, “take my intro class”, “it is clearly 50-50 with no advantage to switching”. • The next week stats professors from elite universities like Harvard, Stanford, UMM wrote in and said that Marilyn was correct, but her reasoning was wrong. Let‟s Make a Deal • Let‟s play the game on the computer simulation, be sure to play the strategy of switching doors after a goat is shown to you. Keep track of how many times you win divided by the number of plays. Compute p-hat. • Who is right? Marilyn or the bad professors? • Do a 95% CI for p, the proportion of Level of Confidence • A CI for p includes a statement of a confidence level, usually 95%. • You should know how to compute confidence intervals for any level of confidence, but particularly for 80%, 90%, 95%, 98%, 99%. • The formula is the same for each, but the Z multiplier changes. Z Multiplier • For any confidence level, the Z multiplier is obtained by drawing a standard normal curve and then placing symmetric boundaries around the mean zero. • For a 95% interval these boundaries should contain 95% of the observations within these bounds. That means there is 2.5% of the observations outside these bounds in each tail to add to the remaining 5%. Finding Z* Z-Multiplier • This means that the upper boundary is at the 97.5 percentile, and the lower boundary is at the 2.5 percentile. • Use your normal table and look up in the middle for .975 (97.5%), go to the edges to observe that the z-value corresponding to this point is 1.96. That is why we have used 1.96 for the 95% CI multiplier. Other Z-Multipliers • You should be able to verify that the correct multipliers for other confidence levels are: 1.28, 1.64, 2.33, 2.57. • Do you know how these were obtained? What Does 95% Confidence Mean Anyway? • A 95% CI means that the method used to construct the interval will produce intervals containing the true p in about 95% of the intervals constructed. • This means that if the 95% CI method was used in 100 samples, we should expect that about 95 of the intervals will contain the true p, and about 5 intervals should miss the true p. Diagram of Confidence 95% of intervals Contain true p, but Some do not. About 5% miss truth. p CI Meaning • We never know if our CI has contained the true p or not, but we know the method we used has the property that it catches the truth 90% of the time (for a 90% CI), so it probably has done well in our study, or at least is not far from the truth. Butterfly Net • A confidence interval is like a butterfly net for catching the true p within its boundaries. • Take a swing at the butterfly (p) with your net (CI), you have a known reliability of catching the butterfly (p), say 90%, but you will never know if your net caught the butterfly or not, just that it is typically a good method for catching butterflies, and so it was probably good for you too! Percent Confidence • The percent confidence refers to the reliability of the CI method to produce intervals that contain the true p. • Why not do a 100% confidence interval? Then we would be completely sure that the interval has contained the true p. 100 % CI for p • A 100% CI for p is (0, 1), this interval is sure to contain the true p. • However this is not very useful. This illustrates the trade-off between %confidence and the usefulness of the interval to simplify the world. • We usually choose 90, 95, or 99 percent confidence levels. CI Cautions ! • Don‟t suggest that the parameter varies: There is a 95% chance the true proportion is between .37 and .42. YUCK!! It sounds like the true proportion is wandering around like an intoxicated (blank) fan. (Fill in your most hated sports team in the blank). The true p is fixed, not random. • Don‟t claim that other samples will agree with yours: 95% of samples will have proportions supporting proposal X between .37 and .42. NOPE!! This range is not about sample proportions as this statement implies. CI Cautions ! (Continued) • Don‟t be certain about the parameter: The cure rate is between 37 and 42 percent. UGG !! This makes it seem like the true p could never be outside this range. We are not sure of this, just sorta-kinda-sure. • Don‟t forget: It‟s the parameter (not the statistic): Never, ever say that we are 95% sure the sample proportion is between .37 and .42. DUH ! There is NO uncertainty in this, it HAS to be true. • Don‟t claim to know too much. • Do take responsibility (for the uncertainty). CI Cautions ! (Continued) • Don‟t claim to know too much: “I‟m 95% confident that between 37 and 42 percent of people in the universe are lunkheads.” Well your population really wasn‟t the whole universe, just Podunk State U. • Do take responsibility (for the uncertainty): You are the one who is uncertain, not the parameter p. You must accept that only 95% of CI‟s will contain the true value of p. Usefulness of CI‟s • There is a trade-off between reliability (confidence) and the width of the interval. • Increasing confidence means the interval width becomes greater (wider). By increasing the sample size, n, the interval becomes narrower. • How big should the sample size be to get useful, precise information about the population p? CI Behavior Margin of Error • The margin of error (m) of a confidence interval is the plus and minus part of the confidence interval, m=Z se(p-hat) • P-hat +/- Z se(p-hat) • P-hat +/- m • A confidence interval that has a margin of error of plus or minus 3 percentage points means that the margin of error m=.03. Margin of Error • From the formula m=Z se (p-hat), you can see that the margin of error depends on the confidence level (Z multiplier) and through the sample size n inside the expression for se(p-hat). • A common problem in statistics is to figure out what sample size will be needed to obtain the desired accuracy (margin of error m). Sample Size Formula • The sample size n needed to get desired margin of error m is given by, 2 Z * * n p (1 p ) * m Sample Size • The margin of error desired m, is usually provided in the problem. The value Z* is determined by the level of confidence that is desired. If no level is given, just assume 95% confidence. • The p* value is a bit of a chicken and egg problem. P* is your best guess about the value of the true p. Sample Size • Mmmm, let‟s see, we are trying to do a study to estimate p, but we need to know p (p*) to compute the needed sample size. This seems impossible! • Quit whining and do the best you can. Give the best or most current state of knowledge about p as p*. Usually there is some information about what p might be. If you know absolutely nothing, then use p*=.5. Why use p*=.5? • Here is a graph of p*(1-p*) for values of p*: p*(1-p*) .25 p*=0 .5 p* 1 Why use p*=.5 • The graph shows that p*(1-p*) will be largest when p*=.5. This means the sample size will be largest when p*=.5. This means that the sample size will be at least as big as actually needed. • This is called being conservative because you are using more data than would actually be needed to achieve the margin of error desired. Sample Size Example • NBA Games: I had a basketball viewing orgy at my house. I watched n=30 NBA games from my big blue chair, drank beverages of God, ate lots of popcorn. I found that X=18 games were won by the home team. This means p-hat = 18/30 = .6. • What is a 95% CI for true home court win proportion p? NBA Games Example .6(.4) .6 1.96 30 .6 .1753 (.4246,.7753) NBA Games Example • Plausible range of values for true home court winning proportion was (.42, .78). This is not very helpful, I knew this even before the first popcorn kernel popped. • Why was the procedure not more helpful? • Problem was the margin of error. It was huge ! It was about m=.17, .18. The sample size was too small to make our inference more precise. We need a bigger sample size. How big? NBA Sample Size • Suppose we wish to obtain a margin of error of m=.02 in a 95% CI for p. What sample size is needed? • n=(1.96/.02)^2 .6(1-.6) = 2304.96 • Round up to n=2305 games. Oh Joy! What a fiesta ! • Note that our best knowledge was the small study done at my house, there p-hat =.6 so it is our best knowledge of the true p, so p*=.6. Vietnam Vets Example • If you go back a few slides you will find that in the Vietnam Vets divorce rate example, the margin of error was about .02. Notice this is a small value for m, and it was obtained because the sample size was huge for that problem. Sample size was over 2000 subjects! Relationship between m and n m n Graph Computation • When p*=.5, m=.05, n=385 • When m=.03, n=1068 • When m=.02, n=2401 • etc Relationship between m and n • Notice that as the sample size increases initially, there is a big drop in the margin of error. It drops substantially early on. • However, for larger sample sizes there is almost no additional reduction in margin of error for increasing the sample size. • Most big surveys are below 2000 – 3000 subjects. Do you see why? Poor, Ignorant Phil ! Right Eye Dominance • Hold a piece of paper with small hole in middle out in front of you with both hands. Focus on an object across the room to be visible in the hole with both eyes open. • Now shut one eye, if the object is still visible, the open eye is the dominant eye. • Do a 95% CI for the proportion of the population that is right eye dominant, p. A Recent Poll (Gallup) Poll Details • Certainly, one of the challenges for the winner of this year's election will be to bring a divided nation together again. Survey Methods • These results are based on telephone interviews with a randomly selected national sample of 1,013 adults, aged 18 and older, conducted Oct. 14-16. For results based on this sample, one can say with 95% confidence that the maximum error attributable to sampling and other random effects is ±3 percentage points. In addition to sampling error, question wording and practical difficulties in conducting surveys can introduce error or bias into the findings of public opinion polls. Hypothesis Tests for p Decisions About the True Population Proportion p Hypothesis Test for p • You have seen previously the method for producing a confidence interval or reasonable range for parameter p. • Hypothesis tests can also be performed with one sample proportion to learn about the population proportion of interest. Hypothesis Test Formula H 0 : p p0 H a : p p0 , p0 , p0 p p0 ˆ Z p0 (1 p0 ) n P Value P( Z Zobs), P( Z Zobs), 2 * P( Z | Zobs |) Ask Marilyn Example H 0 : p .5 H a : p .5 p p0 ˆ .689 .5 Z 2.54 p0 (1 p0 ) .5(1 .5) n 45 P Value P ( Z 2.54) .0055 Ask Marilyn Example • Data is unlikely under Ho, data is inconsistent with Ho. • We have evidence to doubt Ho. • We have evidence to support Ha. • We have evidence the proportion of wins by switching doors, p, is greater than .5. We have evidence that Marilyn is right, we should switch doors. Reeses Pieces Example • What is the proportion of orange candies, p? • I believe our data were something like p- hat=.52 for n=60 candies. Do appropriate hypothesis test. Right Eye Dominance • Hold a piece of paper with small hole in middle out in front of you with both hands. Focus on an object across the room to be visible in the hole with both eyes open. • Now shut one eye, if the object is still visible, the open eye is the dominant eye. • Do a hypothesis test that the proportion of the population that is right eye dominant, p is not equal to .5. Spinning Pennies • We wish to test the hypothesis that the proportion of spins that will turn heads is different than .5. • Some students perform an experiment and find that 17 heads were obtained from 40 spins. This means p-hat=17/40 = .425. Spinning Pennies H 0 : p .5 H a : p .5 p p0 ˆ .425 .5 Z .95 p0 (1 p0 ) .5(1 .5) n 40 P Value 2 * P( Z .95) 2 * P( Z .95) P Value 2 * (.171) .342 Spinning Pennies Conclusion • The data is consistent with Ho. • There is no evidence to doubt the Ho. • There is no evidence to support the Ha. • There is no evidence to suggest the proportion of spins that are heads is anything other than .5. Spinning Pennies • Let‟s do the experiment ourselves. Inference for Two Population Proportions Tests and CI‟s about two population proportions. Data Situation • We now have two populations, and we wish to compare the proportions of these populations. • Population 1 Data: n_1 and p-hat_1. • Population 2 Data: n_2 and p-hat_2. Data Situation Data : X1 Sample _ 1 : n1 , p1 ˆ n1 X2 Sample _ 2 : n2 , p2 ˆ n2 Hypothesis Test Formula H 0 : p1 p2 0 H a : p1 p2 0, 0, 0 p1 p2 0 ˆ ˆ Z , where 1 1 p(1 p) ˆ ˆ n1 n2 X1 X 2 p ˆ n1 n2 P Value P( Z Zobs), P( Z Zobs), 2 * P( Z | Zobs |) Hypothesis Test Formula • Notice the p-hat with no subscript in the denominator of the Z statistic. This is called the pooled proportion. • Under the Ho we hypothesize that both populations have the same proportion, so the natural thing to do is use all the data to estimate the common proportion. Simply add all events and divide by the total sample size. Red Dye #2 Example • 2 samples conducted on lab animals. One group was given a typical animal diet with 44 animals. Four developed tumors. Thus, p-hat=.091 • In a group given red dye # 2, there were 14 animals developing tumors out of 44. Thus p-hat=.318. Red Dye Hypothesis Test H 0 : pR pC 0 H a : pR pC 0 p1 p2 0 ˆ ˆ .318 .091 Z 1 1 1 1 p (1 p ) ˆ ˆ .205(1 .205) n1 n2 44 44 Z 2.64 X1 X 2 4 14 p ˆ .205 n1 n2 44 44 P Value P( Z 2.64) .0041 Red Dye #2 Conclusion • The data are unusual if the Ho is true. The data are inconsistent with the Ho. • There is evidence to doubt the Ho. • There is evidence to support the Ha. • There is evidence that p_r > p_c, and this means there is evidence the red dye #2 group has a higher proportion of animals with cancerous tumors than the control diet. This is evidence that RD#2 is a carcinogen. Red Dye #2 Historical Note • All red color food disappeared for a while. No Jello, no red M&M‟s, no Hawaiian Punch, etc, poor young Jon . • Eventually another red dye was approved for sale. Jon‟s favorite mass-produced junk items returned . Saracco Study (Italy) • Study of heterosexual couples where one member of the couple was HIV infected. • First group used condoms regularly, 171 couples. Of these 3 subsequently became infected. P-hat = 3/171=.0175 • Second group did not use condoms regularly. There were 55 such couples, and 8 subsequently became infected, p- hat = 8/55 = .14545. Saracco Hypothesis Test H 0 : pR p N 0 H a : pR p N 0 pR p N 0 ˆ ˆ .0175 .14545 Z 1 1 1 1 p(1 p) ˆ ˆ .04867(1 .04867) n1 n2 171 55 Z 3.84 X1 X 2 38 p ˆ .04867 n1 n2 171 55 P Value P( Z 3.84) .0002 Saracco Conclusion • The data are unusual under Ho, so data are inconsistent with Ho. • There is evidence to doubt the Ho. • There is evidence to support the Ha. • There is evidence that p_r<p_n, this means evidence that HIV infection proportion is less in group that used condoms regularly. Saracco Historical Note • This was the study that prompted world health officials to proclaim that regular condom use was “effective” in preventing HIV infection. • This does not mean that using condoms is risk-free, all it means is that the infection proportion was statistically less than not using them. Confidence Interval Formula p1 (1 p1 ) p2 (1 p2 ) ˆ ˆ ˆ ˆ p1 p2 Z ˆ ˆ * n1 n2 Confidence Intervals • The crucial value used to evaluate these intervals is zero. If all values are above zero, it implies that proportion p_1 is greater than p_2. • If the interval is all negative, there is evidence p_1<p_2. • If the interval contains zero, it means no difference is a plausible/reasonable statement, and thus no evidence to say that the proportions differ. Woburn Mass CI • In Woburn Massachusetts there were public wells that provided the city‟s water supply. • When the questionable water was being consumed there were 16 adverse birth outcomes out of 414 births. P-hat = 16/414=.039. • When the water was not being consumed, there were 3 adverse birth outcomes out Woburn Confidence Interval p y (1 p y ) pn (1 pn ) ˆ ˆ ˆ ˆ p y pn Z ˆ ˆ * ny nn .039(1 .039) .013(1 .013) .039 .013 1.96 414 228 .026 1.96(.012) .026 .024 (.002,.05) Woburn Water Conclusion • The plausible range of value for p_y-p_n is (.002, .05). • The entire plausible range is positive. • This means there is evidence the p_y > p_n, and that the proportion of adverse birth events with the water on is greater than when the water was not used. There is evidence the water is responsible for an increase in adverse birth events in Woburn. Woburn Water Note • Entertainment note: Hollywood film, A Civil Action, starring John Travolta and Robert DuVall is based on this problem situation. • I believe the parents shown in the video clip are part of the plot of the movie. Propranolol Study • Potential usefulness of propranolol for recent heart attack victims. Population proportion p_c=proportion death within 2 years. Population proportion p_p= proportion death within two years. Propranolol Confidence Interval pc (1 pc ) p p (1 p p ) ˆ ˆ ˆ ˆ pc p p Z ˆ ˆ * nc np .0954 .0704 1.645 * .0954(1 .0954) .0704(1 .0704) 1919 1918 .025 1.645(.00889) .025 .0147 (.0103,.0397) Propranolol Conclusion • Note was 90% CI. The plausible range of values for p_c – p_p is (.01, .04). • This range includes only positive values. • This implies p_c > p_p, and that there is a higher death proportion under usual care, and that two year death rates are reduced when using propranolol. • Is this a big deal? Personal Ads Data Personal Ads Comparisons • Compute four confidence intervals – one for each of the four attributes. • Compute 95% CI‟s for p_male – p_female. • Write complete conclusion for each interval. Large-sample Confidence Interval for a Population Proportion •A confidence interval for a population characteristic is an interval of plausible values for the characteristic. It is constructed so that, with a chosen degree of confidence, the value of the characteristic will be captured inside the interval. Confidence Level •The confidence level associated with a confidence interval estimate is the success rate of the method used to construct the interval. Recall For the sampling distribution of p, p(1 p) mp = p, p and for large* n n The sampling distribution of p is approximately normal. Specifically when n is large*, the statistic p has a sampling distribution that is approximately normal with mean p and standard deviation p(1 p) . n * np 10 and np(1-p) 10 Some considerations Approximately 95% of all large samples will result in a value of p that is within p(1 p) of the true population 1.96p 1.96 n proportion p. Some considerations Equivalently, this means that for 95% of all possible samples, p will be in the interval p(1 p) p(1 p) p 1.96 to p 1.96 n n Since p is unknown and n is large, we estimate p(1 p) p(1 p) with n n This interval can be used as long as np 10 and np(1-p) 10 The 95% Confidence Interval When n is large, a 95% confidence interval for p is p(1 p) p(1 p) p 1.96 , p 1.96 n n The endpoints of the interval are often abbreviated by p(1 p) p 1.96 n where - gives the lower endpoint and + the upper endpoint. Example •For a project, a student randomly sampled 182 other students at a large university to determine if the majority of students were in favor of a proposal to build a field house. He found that 75 were in favor of the proposal. •Let p = the true proportion of students that favor the proposal. Example - continued 75 p 0.4121 182 So np = 182(0.4121) = 75 >10 and n(1-p)=182(0.5879) = 107 >10 we can use the formulas given on the previous slide to find a 95% confidence interval for p. p(1 p) 0.4121(0.5879) p 1.96 0.4121 1.96 n 182 0.4121 0.07151 The 95% confidence interval for p is (0.341, 0.484). The General Confidence Interval The general formula for a confidence interval for a population proportion p when 1. p is the sample proportion from a random sample , and 2. The sample size n is large (np 10 and np(1-p) 10) is given by p(1 p) p z critical value n Finding a z Critical Value •Finding a z critical value for a 98% confidence interval. 2.33 Looking up the cumulative area or 0.9900 in the body of the table we find z = 2.33 Some Common Critical Values Confidence z critical level value 80% 1.28 90% 1.645 95% 1.96 98% 2.33 99% 2.58 99.8% 3.09 99.9% 3.29 Terminology The standard error of a statistic is the estimated standard deviation of the statistic. For sample proportions, the standard deviation is p(1 p) n This means that the standard error of the sample proportion is p(1 p) n Terminology The bound on error of estimation, B, associated with a 95% confidence interval is (1.96)·(standard error of the statistic). The bound on error of estimation, B, associated with a confidence interval is (z critical value)·(standard error of the statistic). Sample Size The sample size required to estimate a population proportion p to within an amount B with 95% confidence is 2 1.96 n p(1 p) B The value of p may be estimated by prior information. If no prior information is available, use p = 0.5 in the formula to obtain a conservatively large value for n. Generally one rounds the result up to the nearest integer. Sample Size Calculation Example •If a TV executive would like to find a 95% confidence interval estimate within 0.03 for the proportion of all households that watch NYPD Blue regularly. How large a sample is needed if a prior estimate for p was 0.15. We have B = 0.03 and the prior estimate of p = 0.15 2 2 1.96 1.96 n p(1 p) (0.15)(0.85) 544.2 B 0.03 A sample of 545 or more would be needed. Sample Size Calculation Example revisited •Suppose a TV executive would like to find a 95% confidence interval estimate within 0.03 for the proportion of all households that watch NYPD Blue regularly. How large = 0.03 and should use p = 0.5 in We have B a sample is needed if we have no reasonable prior estimate for p. the formula. 2 2 1.96 1.96 n p(1 p) (0.5)(0.5) 1067.1 B 0.03 The required sample size is now 1068. Notice, a reasonable ball park estimate for p can lower the needed sample size. Another Example •A college professor wants to estimate the proportion of students at a large university who favor building a field house with a 99% confidence interval accurate to 0.02. If one of his students performed a B = 0.02, a prior estimate estimated p to We havepreliminary study and p = 0.412 and we be 0.412, how large a sample should use the z critical value 2.58 (for a 99% confidence interval) take. should he 2 2 2.58 2.58 n p(1 p) (0.412)(0.588) 4031.4 B 0.02 The required sample size is 4032. Large Sample Hypothesis Test for a Single Proportion To test the hypothesis H0: p = hypothesized proportion, compute the z statistic p hypothesized value z hypothesized value(1-hypothesized value) n In terms of a standard normal random variable z, the approximate P-value for this test depends on the alternate hypothesis and is given for each of the possible alternate hypotheses on the next 3 slides. Hypothesis Test Large Sample Test of Population Proportion p hypothesized value P-value P z hypothesized value(1-hypothesized value) n Hypothesis Test Large Sample Test of Population Proportion p hypothesized value P-value P z hypothesized value(1-hypothesized value) n Hypothesis Test Large Sample Test of Population Proportion p hypothesized value P-value 2P z hypothesized value(1-hypothesized value) n Hypothesis Test Example Large-Sample Test for a Population Proportion •An insurance company states that the proportion of its claims that are settled within 30 days is 0.9. A consumer group thinks that the company drags its feet and takes longer to settle claims. To check these hypotheses, a simple random sample of 200 of the company‟s claims was obtained and it was found that 160 of the Example 2 Single Proportion p = proportion of the company’s claims that are continued settled within 30 days H0: p = 0.9 HA: p 0.9 160 The sample proportion is p 0.8 200 0.8 0.9 0.8 0.9 z 4.71 (0.9)(1 0.9) 0.9(0.1) 200 200 P-value P(z 4.71) 0 2 Single Proportion continued The probability of getting a result as strongly or more strongly in favor of the consumer group's claim (the alternate hypothesis Ha) if the company’s claim (H0) was true is essentially 0. Clearly, this gives strong evidence in support of the alternate hypothesis (against the null hypothesis). Example 2 Single Proportion We would say continued support for that we have strong the claim that the proportion of the insurance company’s claims that are settled within 30 days is less than 0.9. Some people would state that we have shown that the true proportion of the insurance company’s claims that are settled within 30 days is statistically significantly less than 0.9. Hypothesis Test Example Single Proportion •A county judge has agreed that he will give up his county judgeship and run for a state judgeship unless there is evidence at the 0.10 level that more then 25% of his party is in opposition. A SRS of 800 party members included 217 who opposed him. Please advise this judge. Hypothesis Test Example Single Proportion continued p = proportion of his party that is in opposition H0: p = 0.25 HA: p > 0.25 = 0.10 Note: hypothesized value = 0.25 217 n 800, p 0.27125 800 0.27125 0.25 z 1.39 0.25(0.75) 800 Hypothesis Test Example Single Proportion continued P-value=P(z 1.39) 1 0.9177 0.0823 •At a level of significance of 0.10, there is sufficient evidence to support the claim that the true percentage of the party members that oppose him is more than 25%. •Under these circumstances, I Large-Sample Inferences Difference of Two Population (Treatment) Proportions Some notation: Population Sample Sample Proportion of Proportion of Size Successes Successes Population or treatment 1 n1 p1 p1 Population or treatment 2 n2 p2 p2 Properties: Sampling Distribution of p1- p2 If two random samples are selected independently of one another, the following properties hold: 1. m p p p1 p2 1 2 p1 (1 p1 ) p2 (1 p2 ) 2. 2 2 2 p1 p 2 p1 p2 and n1 n2 p1 (1 p1 ) p2 (1 p2 ) p p 1 2 n1 n2 3. If both n1 and n2 are large [n1 p1 10, n1(1- p1) 10, n2p2 10, n2(1- p2) 10], then p1 and p2 each have a sampling distribution that is approximately normal Large-Sample z Tests for p1 – p2 = 0 The combined estimate of the common population proportion is n1p1 n 2 p 2 pc n1 n 2 total number of successes in two samples total sample size Large-Sample z Tests for p1 – p2 = 0 Null hypothesis: H0: p1 – p2 = 0 Test statistic: p1 p 2 z p c (1 p c ) p c (1 p c ) n1 n2 Assumptions: 1. The samples are independently chosen random samples OR treatments are assigned at random to individuals or objects (or vice versa). 2. Both sample sizes are large: n1 p1 10, n1(1- p1) 10, n2p2 10, n2(1- p2) 10 Large-Sample z Tests for p1 – p2 = 0 Alternate hypothesis and finding the P-value: 1. Ha: p1 - p2 > 0 P-value = Area under the z curve to the right of the calculated z 2. Ha: p1 - p2 < 0 P-value = Area under the z curve to the left of the calculated z 3. Ha: p1 - p2 0 i. 2•(area to the right of z) if z is positive ii. 2•(area to the left of z) if z is negative Example - Student Retention A group of college students were asked what they thought the “issue of the day”. Without a pause the class almost to a person said “student retention”. The class then went out and obtained a random sample (questionable) and asked the question, “Do you plan on returning next year?” The responses along with the gender of the person responding are summarized in the following table. Response Yes No Maybe Male 211 45 19 Gender Female 141 32 9 Test to see if the proportion of students planning on returning is the same for both genders at the 0.05 level of significance? Example - Student Retention p1 = true proportion of males who plan on returning p2 = true proportion of females who plan on returning n1 = number of males surveyed n2 = number of females surveyed p1 = x1/n1 = sample proportion of males who plan on returning p2 = x2/n2 = sample proportion of females who plan on returning Null hypothesis: H0: p1 – p2 = 0 Alternate hypothesis: Ha: p1 – p2 0 Example - Student Retention Significance level: = 0.05 Test statistic: p1 p 2 z p c (1 p c ) p c (1 p c ) n1 n2 Assumptions: The two samples are independently chosen random samples. Furthermore, the sample sizes are large enough since n1 p1 = 211 10, n1(1- p1) = 64 10 n2p2 = 141 10, n2(1- p2) = 41 10 Example - Student Retention Calculations: n1p1 n 2 p 2 211 141 352 pc 0.7702 n1 n 2 275 182 457 p1 p 2 z p c (1 p c ) p c (1 p c ) 275 182 0.76727 0.77473 0.77024(1 0.77024) 0.77024(1 0.77024) 275 182 -0.0074525 -0.19 0.040198 Example - Student Retention P-value: The P-value for this test is 2 times the area under the z curve to the left of the computed z = -0.19. P-value = 2(0.4247) = 0.8494 Conclusion: Since P-value = 0.849 > 0.05 = , the hypothesis H0 is not rejected at significance level 0.05. There is no evidence that the return rate is different for males and females.. Example A consumer agency spokesman stated that he thought that the proportion of households having a washing machine was higher for suburban households then for urban households. To test to see if that statement was correct at the 0.05 level of significance, a reporter randomly selected a number of households in both suburban and urban environments and obtained the following data. Number Proportion having having Number washing washing surveyed machines machines Suburban 300 243 0.810 Urban 250 181 0.724 Example p1 = proportion of suburban households having washing machines p2 = proportion of urban households having washing machines p1 - p2 is the difference between the proportions of suburban households and urban households that have washing machines. H0: p1 - p2 = 0 Ha: p1 - p2 > 0 Example Significance level: = 0.05 Test statistic: p1 p 2 z p c (1 p c ) p c (1 p c ) n1 n2 Assumptions: The two samples are independently chosen random samples. Furthermore, the sample sizes are large enough since n1 p1 = 243 10, n1(1- p1) = 57 10 n2p2 = 181 10, n2(1- p2) = 69 10 Example Calculations: n1p1 n 2 p 2 243 181 424 pc 0.7709 n1 n 2 300 250 550 p1 p2 z pc (1 pc ) pc (1 pc ) n1 n2 0.810 0.742 1 1 0.7709(1 0.7709) 300 250 2.390 Example P-value: The P-value for this test is the area under the z curve to the right of the computed z = 2.39. The P-value = 1 - 0.9916 = 0.0084 Conclusion: Since P-value = 0.0084 < 0.05 = , the hypothesis H0 is rejected at significance level 0.05. There is sufficient evidence at the 0.05 level of significance that the proportion of suburban households that have washers is more that the proportion of urban households that have washers. Large-Sample Confidence Interval for p1 – p2 When 1. The samples are independently selected random samples OR treatments that were assigned at random to individuals or objects (or vice versa), and 2. Both sample sizes are large: n1 p1 10, n1(1- p1) 10, n2p2 10, n2(1- p2) 10 A large-sample confidence interval for p1 – p2 is p1 (1 p1 ) p 2 (1 p 2 ) (p1 p2 ) z critical value n1 n2 Example A student assignment called for the students to survey both male and female students (independently and randomly chosen) to see if the proportions that approve of the College’s new drug and alcohol policy. A student went and randomly selected 200 male students and 100 female students and obtained the data summarized below. Number Number that Proportion surveyed approve that approve Female 100 43 0.430 Male 200 61 0.305 Use this data to obtain a 90% confidence interval estimate for the difference of the proportions of female and male students that approve of the new policy. Example For a 90% confidence interval the z value to use is 1.645. This value is obtained from the bottom row of the table of t critical values (Table III). We use p1 to be the female’s sample approval proportion and p2 as the male’s sample approval proportion. 0.430(1 0.430) 0.305(1 0.305) (0.430 0.305) 1.645 100 200 (0.125) 0.097 or (0.028,0.222) Based on the observed sample, we believe that the proportion of females that approve of the policy exceeds the proportion of males that approve of the policy by somewhere between 0.028 and 0.222.
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