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Confidence Interval (CI) for a Proportion http://www.rossmanchance.com/iscam2/applets/BinomDi st/BinomDist.html http://www.shodor.org/interactivate/activities/Adju stableSpinner/?version=1.5.0_06&browser=MSIE&vendor =Sun_Microsystems_Inc
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Confidence Interval (CI) for a Proportion. http://www.rossmanchance.com/iscam2/applets/BinomDist/BinomDist.html http://www.shodor.org/interactivate/activities/AdjustableSpinner/?version=1.5.0_06&browser=MSIE&vendor=Sun_Microsystems_Inc. Critical Values (95% confidence). - PowerPoint PPT Presentation

### Transcript of Confidence Interval (CI) for a Proportion

Populations

Confidence Interval (CI)for a Proportionhttp://www.rossmanchance.com/iscam2/applets/BinomDist/BinomDist.html

1Critical Values (95% confidence)

2Critical Values (95% confidence)

3

Critical Values (95% confidence)Look up 0.975 (area to the left) to get the positive z score; 0.025 for the negative.0.0250.0250.954

5Mean / S D of a Sample ProportionThe sample proportion (a statistic) is the count X divided by the sample size n. The sample proportion

If both the expected count of Successes and Failures are at least 10 (np and n(1 p) both 10), has approximate Normal distribution.

6C% Confidence Interval

With approximately C% = (1 )100% probabilityGiven results from an appropriately obtained sampleWith approximately C% confidence

7Confidence Interval for a (Population) Proportion p

This is the sample proportion.

The Z value comes from the confidence C%.8C% Confidence Interval for pConditions for use of this methodRandom sample from a categorical population 2 categories (S / F)If sample w/o replacement: Population at least 20 times the sample sizeAt least 10 Successes and Failures in the sample (this ensures that the Normal is appropriate)When we collect data from 1 random sample and compute the sample proportion, the interval of values

is a C% confidence interval (CI) for p.** which, in this type of application, is unknown.9ExampleWhat proportion of students smoke?p = ??N = ??p = probability a student smokes= proportion of all students who smokeA simple random sample of 368 students is surveyed. n = 368X = # of sampled students who smokevaries depending on the sample10p = ??n = 368X = # of sampled students whoN = ?? smoke. Varies.The survey is conducted: 79 of the 368 are smokers.X = 79 is observed for one sample. (Other samples would yield (somewhat) different values.)n X = 368 79 = 28979/368 = 0.215 =Not p (nearly impossible its exactly p). = 0.215 is the statistic estimating the parameter p.the sample (observed) proportion the point estimate of p

11A simple random sample of 368 students finds that 79 smoke. Obtain a 95% confidence interval for the proportion of all students who smoke.The book would saytrue proportion of students who smoke.Each person in the population is a Success (smoker) or Failure (nonsmoker); the sample is randomThe population is huge (much bigger than 20(368))There are 79 smokers; 289 nonsmokers. Both are well above 5.The confidence interval based on the Normal distribution can be used.12

130.215 0.042 (within 0.042 of 0.215)0.215 0.042 = 0.1730.215 + 0.042 = 0.2570.173 < p < 0.257Between 0.173 and 0.257.InterpretationWe are approximately 95% confident that the proportion of all* students who smoke is between 0.173 and 0.257.*Its important to say all or population proportion14Confidence Interval ExampleProper formatting of CIs:0.215 0.042(0.173, 0.257)0.173 to 0.2570.173 p 0.257For the last three: The low value is written first.All CIs should be accompanied by a statement interpreting them, including the confidence level (here 95%) and an indication that you are making a statement about an unknown parameter p = population proportion.15Confidence Interval for pTo use this formula we needRandom sample from a categorical population 2 categories (S / F)If sampling w/o replacement: Population at least 20 times the sample sizeAt least 5 Successes and Failures in the sample (this ensures that the Normal is appropriate; 10 is better)If one of these is violated: The confidence is not really the value of C used in the formula.If the sample is not random the confidence associated with this method could be anything but is likely to be much lower than C.

16Example 2A marketer works for an electronics store. He wishes to estimate the percent of coupons that will be redeemed at the stores. 927 customers are randomly sampled and sent coupons; 27 of them redeem their coupon.Obtain a 90% CI for the proportion of all customers that redeem this coupon.Check conditions:17Example 2Of the 927 coupons, 27 are redeemed

Dont overround. Keep at least 3 significant figures in intermediate results.1.64 or 1.65either is fine18Example 2Error margin

19Example 2Error margin

0.9709 = (1 0.0291) proportion not redeemed20Example 2Error margin

Be careful. These can be small. Keep at least 3 significant figures.21Example 2Error margin

22Example 2Of the 927 coupons, 27 are redeemed

We are (approximately) 90% confident that between 2.00% and 3.82% of all coupons will be redeemed.23Example 3What proportion of voters currently approve of the Presidents handling of the economic situation?p = _____________________________p = probability voter approves= proportion of all voters who approveA random sample of 1000 likely voters is taken, using random digit dialing. n = 1000X = # of sampled voters approve24p = ??n = 1000X = # of sampled voters who approve. Varies.The survey is conducted: 557 of the sampled voters approve.Compute X/n = __________ .Which of the following is correct?

Fill in the blanks with the appropriate terms 0.557 is the statistic estimating the parameter p.

Example 30.557

25p = ??n = 100X = # of sampled voters who approve. Varies.The survey is conducted: 557 of the sampled voters approve.Compute X/n = __________ .Which of the following is correct?

Fill in the blanks 0.557 is the statistic estimating the parameter p.

Example 30.557

26Example 3Obtain a 95% confidence interval for the proportion of all people who approve.1st: Check ConditionsRandom sample from a categorical population 2 categories (S / F)Population at least 20 times the sample sizeAt least 5 Successes and Failures in the sample (this ensures that the Normal is appropriate; 10 is better)

The number of Successes and Failures are 557 and 443 respectively, both well above 5. The confidence interval based on the Normal distribution can be used.27Summary of Information

28Summary of Information

29Final NumbersWithin 0.031of 0.5570.557 0.031 = 0.5260.557 + 0.031 = 0.588Between 0.526 and 0.588. Any of these suffices0.557 0.031(0.527, 0.588)0.527 to 0.5880.527 p 0.588

30Assessing the Error MarginThe error margin covers random sampling errors. It does not account for errors due to improper sampling, or inaccurate data collection.Is the sample drawn from a collection of units that may not be representative of the entire population?If so, perhaps the interval is appropriate for the population defined by that collection.That is: Define a new (reduced) population.Is any judgment required in categorizing the units as Success and Failure?

31What Confidence MeansImagine a population for which 45% of the population approves of the states governor.Consider all samples of size n = 1000 from this population. For each sample a 90% CI is obtained.Before any sampling is donebefore any data is collected:The probability of a randomly chosen sample giving a CI that covers the parameter of p = 0.45 is 0.90.32

What Confidence MeansThe probability of a random sample giving a CI that covers the parameter of p = 0.45 is 0.90.

90% of black intervals cover the blue line at p.90% of all 90% CIs cover the estimated parameter.33

What Confidence MeansA histogram of the black dots would be Normal, with mean 0.45. Approximately 10% of the time, the black dot would be far enough from 0.45 so that the interval (roughly 0.026) would not cover 0.45.34What Confidence MeansExample 4: A random sample of 2136 adults was asked, Do you favor or oppose abolishing the penny? 59% answered oppose.0.59 2136 = 1260.241260 answered oppose.1260/2136 = 0.5899 = 0.590 to 3 significant digits.

59.0% 2.1% is a 95% confidence interval.

35What Confidence MeansWe are 95% confident that between 56.9% and 61.1% of all Americans oppose abolishing the penny. (p represents this unknown proportion.)In a real study: Exactly one random sample is chosen. Once the data is recorded there is nothing random (certainly p is not random). The location of the blue line is unknown. It exists: We just dont know where. We dont know whether or not p is covered (the probability is either 0 or 1).We use the word confidence after the random sample is drawn. We dont use the word probability (unless we are explaining what confidence is).

36QuizA sample of jokes from The Daily Show found that 83 of 252 were of a political nature. Assume this was a random sample from all jokes. Then a 95% confidence interval is (0.271, 0.387).Answer true or false.37QuizA sample of jokes from The Daily Show found that 83 of 252 were of a political nature. Assume this was a random sample from all jokes. Then a 95% confidence interval is (0.271, 0.387).Answer true or false.95% of jokes are political in nature.FALSE. 95% is the confidence we have in the result, it has nothing to do with the prevalence (in the sample or for the entire population) of political jokes on The Daily Show).38QuizA sample of jokes from The Daily Show found that 83 of 252 were of a political nature. Assume this was a random sample from all jokes. Then a 95% confidence interval is (0.271, 0.387).Answer true or false.We are 95% confident that between 27.1% and 38.7% of the sampled jokes were political in nature.39QuizA sample of jokes from The Daily Show found that 83 of 252 were of a political nature. Assume this was a random sample from all jokes. Then a 95% confidence interval is (0.271, 0.387).Answer true or false.We are 95% confident that between 27.1% and 38.7% of the sampled jokes were political in nature.FALSE. 83 / 252 = 0.329. The probability is 100% that the sample proportion lies within the bounds of the interval it centers the interval and always falls within the bounds.40QuizA sample of jokes from The Daily Show found that 83 of 252 were of a political nature. Assume this was a random sample from all jokes. Then a 95% confidence interval is (0.271, 0.387).Answer true or false.The confidence is 0.95 that another random sample of jokes would have between 0.271 and 0.387 of the jokes political in nature.FALSE. Confidence intervals are not intended to predict what will happen with other random samples. They estimate a parameter (in this case, p).41QuizA sample of jokes from The Daily Show found that 83 of 252 were of a political nature. Assume this was a random sample from all jokes. Then a 95% confidence interval is (0.271, 0.387).Answer true or false.The probability is 0.95 that between 0.271 and 0.387 of all jokes on The Daily Show are political in nature.FALSE. The probability is either 0 or 1 we just dont know what p is. Probability refers to an outcome that has uncertainty due to randomness. The uncertainty here is due to ignorance.42QuizA sample of jokes from The Daily Show found that 83 of 252 were of a political nature. Assume this was a random sample from all jokes. Then a 95% confidence interval is (0.271, 0.387).Answer true or false.95% of all samples of The Daily Show jokes give an interval that cover p = the proportion of all jokes that are political in nature. Our 1 sample, randomly drawn, gives (0.271, 0.387). We dont know if p is in there or not, but we are 95% confident it is.Thats it! TRUE!43Polls apart: Why polls vary on presidential raceThe groups pollsters randomly choose to interview are bound to differ from each other, and sometimes do significantly.Every poll has a margin of sampling error, usually around 3 percentage points for 1,000 people.* That means the results of a poll of 1,000 people should fall within 3 points of the results you would expect had the pollster instead interviewed the entire population of the U.S. But and this is important the results are expected to be that accurate only 95 percent of the time. That means that one time in 20, pollsters expect to interview a group whose views are not that close** to the overall population's views.* Using p = 0.5 at 95% confidence gives n 1068** not within the error margin

^44ExampleSuppose we randomly sample people for a telephone poll on the issue of Presidential approval.Well sample 1000 people, using 95% confidence.People of different political leanings have systematically different behaviors.Refusing telephone surveys is one such behavior.45ExampleSuppose (to oversimplify) that in the population 88 million people approve of the President and 72 million disapprove. So the Presidents approval rating is p = 88/160 = 0.55.ButThe people that approve of the President are crankier than those that do not. They are less likely to put up with an intruding phone call. In fact, 40% of the approvers will not respond (thats 35.2 million people). The disapprovers are more willing to take the call: only 10% of them will refuse (thats 7.2 million people). 46ExampleAmong everyone the approval rate is 55%.Among responders, the approval rate is52.8 / 117.6 = 45%The CI formed from the data estimates 45% (not 55%).ApproveDisapproveTotalRespond52.864.8117.6Refuse35.27.242.4Total88.072.0160.047How Confident Are We?