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Point and Confidence Interval Estimation of a Population Proportion, p. We are frequently interested in estimating the proportion of a population with a characteristic of interest, such as: the proportion of smokers the proportion of cancer patients who survive at least 5 years

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## Point and Confidence Interval Estimation of a Population Proportion, p

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**Point and Confidence Interval Estimation of a Population**Proportion, p**We are frequently interested in estimating the proportion of**a population with a characteristic of interest, such as: • the proportion of smokers • the proportion of cancer patients who survive at least 5 years • the proportion of new HIV patients who are female**If we take a random sample from a population**• observe the number of subjects with the characteristic of interest (# of “successes”) • we are observing a binomial random variable. • Now, however, we will focus on • estimating the true proportion , p, in the population • rather than focusing on the count.**Again, one way to deal with this type of data is to define a**random variable X that can take two values: • X = 1, if characteristic is present – a “success” • X = 0, if characteristic is absent – a “failure” • Then • if we sum all values in a population, • we are summing zeros and ones – • this will give a count of the number of individuals in the population with the characteristic:**The population mean is the Proportion of individuals in**population with the characteristic: The sample proportion is then: Therefore, p is the estimator of p, the proportion with a characteristic of interest.**By the Central Limit Theorem, we know, for n large**even when X is not normally distributed. When X is a 0,1 variable, for n large we know from the central limit theorem.**What is the variance, s2, for a 0,1 variable?**We know By use of algebra, and the fact that xs2 = xs. for a 0,1 variable, we can show that**For those who want the algebra:**expand x2 = x, for 0,1 sum over constant**Hence,**The standard error of the sample proportion is Standard error of P:**We also know, by the central limit theorem, that for large**n, P is approximately normally distributed: For Estimation of the population proportion, p: Point Estimate:Confidence Interval Estimate:**Example: Suppose that a sample of 1000 voters is taken to**determine presidential preference. In this sample, 585 persons indicated that they would vote for candidate A. Construct a 95% confidence interval estimate for the true proportion, p, in the population planning to vote for candidate A. The confidence interval forptakes the form:**The point estimate of the proportion is: p= (585/1000) =**.585 • The 95% confidence interval estimator of p is • However we don’t know p, so we will use p in it’s place to estimate the standard error:**The 95% CI on the proportion preferring Candidate A is**(.554, .616). This does not include the value .50: Either we obtained an unusually large sample mean (such that the interval estimate did not overlap µ=0.5) if µ really is .5, or the population mean is not .5, suggesting that candidate A will win the election.**When is the sample large enough to use the normal**approximation to the binomial? • When (n)(π)5, and (n)(1-π)5 • That is, • when both the expected number of successesand the expected number of failures is greater than 5.**Aside: improve to the normal Appoximation for a Binomial**• The Binomial distribution is discrete, while the normal distribution is continuous. When the true proportion,π, is known, we can match the binomial distribution better to a normal distribution by including a correction. The correction is called the ‘continuity correction’. • For example, when π = .5, and n = 10, to approximate We use instead the normal approximation for the probability**Example of ‘Continuity Correction' to the Normal**Approximation to the Binomial. Suppose π = .5 and n = 16. Compare the exact normal approximation and continuity corrected values of P(.4375 ≤ P ≤ .5). • From Binomial Table: • Using Normal Approximation, no correction • Using Correction:**Using P in place of p to estimate the standard errorsp:**• 1.If (n)(π)5 and (n)(1-π)5, use P: • 2.Otherwise, a) Assume π=.5,or b) use an ‘exact ’method for the CI • We do this to avoid underestimating the variance, p(1– p) which is at a maximum when p=.5 • Don’t use Student’s t with proportions since the assumption of normality of the underlying population elements is not satisfied by a 0,1 variable.**What do we use when the normal approximation is not**appropriate? • Exact Binomial Confidence Intervals for p can be computed: • Solve for x in the following and then substitute into p= x/n: • Lower Limit: • Upper Limit: • Clearly, exact binomial CI is not simple to compute**Go to Minitab or other software**Stat Basic Statistics 1 Proportion Leave blank for Binomial CI; Check for Normal approx. n x**EXACT Binomial:**Test and CI for One Proportion Test of p = 0.5 vs p not = 0.5 Exact Sample X N Sample p 95.0% CI P-Value 1 585 1000 0.585 (0.553748, 0.615750) 0.000 Normal Approximation: Test and CI for One Proportion Test of p = 0.5 vs p not = 0.5 Sample X N Sample p 95.0% CI Z-Value P-Value 1 585 1000 0.585 (0.554461, 0.615539) 5.38 0.000**Sample Size Estimation when the goal is**Estimating a Population Proportion, p • The pattern is the same as when goal is estimation of a mean: • If we know • the desired precision (width of interval) • confidence level • “guess” of the proportion to get std error • we can estimate the sample size, n.**The width of a confidence interval for P is:**w = 2[z1-a/2 (sP)] , where sP is the standard error of P w ) ( P P – z1-a/2(sP) P + z1-a/2(sP) Using we have**Solving for n gives us**• Note: • this requires information about p, which is our goal! • However, p(1–p) is at a maximum when p=.5 • To be conservative • (over- rather than under-estimate sample size) • use (.5) in place of p**Substituting in .5 for p gives a conservative sample size**estimator of:**Example:**• For an election poll, how many voters should be surveyed to estimate the proportion, to within 5%, in favor of re-electing the current mayor, with 95% confidence? • We have a confidence level, 1–a = .95 z.975 = 1.96 • We have a desired width of 5% = .05, w = .10 • Conservative: n = (z1-a/2)2/w2 = (1.96)2/(.10)2 = 384.16 • We should poll 385 voters to achieve a 95% CI of 5%**What if we have some information on p?**• A previous poll tells us that the current office-holder had ~ 75% of the voter support. • Assuming p = .75: • n = 4p(1–p)(z1-a/2)2/w2 • = 4(.75)(.25)(1.96)2/(.10)2 = 288.12 • Using available information • we get a sample size estimate of 289 voters • which can save us considerable time and expense, compared to the more conservative estimate.**Confidence Interval Calculation for the**Difference between two proportions, p1 – p2, Two independent groups • We are often interested in comparing proportions from 2 populations: • Is the incidence of disease A the same in two populations? • Patients are treated with either drug D, or with placebo. Is the proportion “improved” the same in both groups?**Suppose we take independent, random samples from two groups,**and estimate a proportion in each. For large enough sample size, we know: Then the standard error of the difference between the sample proportions is the square root of the sum of the variances:**Or, since we don’t know the true proportions, the sample**estimate of the standard error: Thus, for n large, the (1-a) confidence interval estimator is:**Example:**In a clinical trial for a new drug to treat hypertension, 50 patients were randomly assigned to receive the new drug, and 50 patients to receive a placebo. 34 of the patients receiving the drug showed improvement, while 15 of those receiving placebo showed improvement. Compute a 95% confidence interval estimate for the difference between proportions improved.**Point Estimate of (p1 – p2):**• p1 = 34/50 = .68 • p2 = 15/50 = .30 (p1 – p2)= .68 – .30 = .38 • Since we have n1 = n2 = 50, our sample size is large enough to use the sample estimate of standard error:**Confidence coefficient: For 1 – a = .95, z1-a/2 =**z.975 = 1.96 • Confidence Interval Estimate: • The 95% CI estimate is: • (.199 , .561) or (19.9% , 56.1%) • The difference between proportions improved is bounded away from zero – it seems that the proportion improved by the drug is clearly greater than the proportion by placebo.**Using Minitab: Stat Basic Statistics 2 Proportions**Enter sample sizes n1 and n2 Enter # of successes x1 and x2**Test and Confidence Interval for Two Proportions**Sample X N Sample p 1 34 50 0.680000 2 15 50 0.300000 Estimate for p(1) - p(2): 0.38 95% CI for p(1) - p(2): (0.198748, 0.561252)**The same cautions apply here, as for estimates for a single**proportion • the sample size should be large enough in each group, so that the normal approximation will hold: • nπ5 and n(1-π)5 for each sample • Otherwise: a) use .5 in place of π when estimating the variance for the confidence interval.b) use some other method. • Minitab offers the option to compute a pooled estimate of the standard error**And in summary:**• Confidence interval estimates provide • a range of likely values • an associated probability, or confidence level. • The width of the confidence interval depends upon: • The underlying variability in the population • The sample size • The confidence level**It is important to keep track of assumptions that we must**make about the data: • Samples should be selected randomly • selection of any element is independent of selection of any others • For many cases, we must assume that the underlying population follows a normal distribution • without this assumption, probabilities computed using the • t-distribution • c2–distribution • F-distribution may not be correct.**When we speak of “knowing” the population variance, s2,**• we really mean that we have an outside source of information • previous research, census data, etc. • the key is that we are not using the sample estimate, s2, based upon the current sample.**The key to confidence interval estimation is to know**• what parameter you are estimating • the point estimate of the parameter • the confidence level • what distributional assumptions are required • the associated distribution for computing probabilities. • I have started a summary table for you below – completing this table will be a good review exercise.

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