Last few slides from last time…Example 3: What is the probability that p’ will fall in a certain range, given p? • Flip a coin 50 times. If the coin is fair (p=0.5), what is the probability of getting an estimate, p’, greater than or equal to 0.7 (=35 heads). • E(P’) = 0.5 • Std. error(P’) = sqrt((.5)(.5)/50) = .0707 • z = (0.7-0.5)/0.0707 ≈ 2.83 • P(z > 2.83) ≈ (1-0.9953)/2 = 0.0024 = P(p’ > 0.7) – Pretty unlikely to get such high estimates of pMore examples on finding the mean and standard deviation of a r.v. •x~N(µx, σx), y~N(µy, σy) • Z = x + 4y + 2 – E(Z) = E(x) + E(4y) + E(2) = µx+ 4µy+ 2 – Var(Z) = var(x) + var(4y) + var(2) 2= σx2 + 16 σy •Z = (2x1+ 2x2-y)/5 – E(Z) = (E(2x) + E(2x) - E(y))/5 = 4/5 µx –1/5 µy – Var(Z) = var(2x/5) + var(2x/5) + var(y/5) = 8/25 σx2 + σy2/25Confidence intervals 9.07 2/26/2004From last time… • Sampling theory – Start with a large population. – Take a sample of N units from the population, and compute a statistic on that sample. • E.G. statistic = sample mean = estimate of the population mean. – We imagine doing this many times, and make deductions about how the sample statistic (estimator) is distributed.The purpose of confidence intervals • To estimate an unknown population parameter with an indication of how accurate the estimate is and of how confident we are that our estimate correctly captures the true value of the parameter.Relationship of confidence intervals to sampling theory • Confidence intervals – Start with one sample of N units from a population. – We make inferences from this one sample about the parameters of the underlying population.Deduction: Reasoning from a hypothesis to a conclusion. Induction: Reasoning from a set of observations to a reasonable hypothesis. Deduce properties of the sample statistic from knowledge of the population. samplepopulation Infer properties of the population parameter from knowledge of the sample.Probability that the statistic takes on a particular range of values. samplepopulationWhy confidence vs. probability? The population parameter has a true value, which either is or is not in the given range. Confidence that the population parameter falls within a particular range of values.Archery & confidence intervals • Suppose you have an archer who can hit the 10 cm radius bull’s eye 95% of the time. • One arrow in 20 misses. 10cmArchery & confidence intervals • You look at the target from the back. The archer shoots a single arrow. • Knowing the archer’s ability, what can you say about where you expect the center of the bull’s eye to be?Archery & confidence intervals • Well, 95% of the time, that arrow is going to be within 10cm of the center of the bull’s eye. • Thus, 95% of the time, the center of the bull’s eye is going to be within 10cm of the arrow.Archery & confidence intervals • Draw a 10cm circle centered around the arrow. • If you did this manytimes, your circleswould include (“cover”) the center ofthe target 95% of thetime. • You are 95% confident that the center of the target lies in this circle. true centerExample: A Poll • A pollster draws a random sample of 1000 voters and asks what they think of a candidate • 550 voters favor the candidate • What is the true proportion, p, of the population who favor the candidate? • From our single sample, – n=1000, p’=.55A Poll • Recall the sampling distribution of P’ – The distribution of P’ estimates, given the size of the sample, n, and the true population proportion, p. – Approximately normal (especially for n=1000). – E(P’) = p σP' = sqrt(p(1-p)/ n)A Poll • From last time: what is the range of proportions such that we expect 95% of our p’ estimates to fall within this range? • From the z-tables, and some algebra: 0.95 = P(-1.96 ≤ z ≤ 1.96) 0.95 = P(-1.96 ≤ (p’-p)/ σP' ≤ 1.96) 0.95 = P(p-1.96* σP' ≤ p’ ≤ p+1.96* σP') 95% of the p’ “arrows” land between p-1.96* σP' and p+1.96* σP'A Poll • More algebra: 0.95 = P(p-1.96* σP' ≤ p’ ≤ p+1.96* σP') 0.95 = P(p’-1.96* σP' ≤ p ≤ p’+1.96* σP') • This says that 95% of the time, the true value of the population parameter, p, lies within p’-1.96* σP' and p’+1.96* σP' • There’s just one catch – we don’t know p, so we don’t know σP' =sqrt(p(1-p)/n).A Poll • We don’t know σP'. • So, we fudge a bit, and approximate p by p’. Then σP' ≈ sqrt(p’(1-p’)/n) • Back to the poll: – p’ = 0.55, σP' ≈ sqrt(.55*.45/1000) = .0157 – We are 95% confident that p, the true % of the voters that favor the candidate, is within the rangep’± 1.96 σP' = 0.550 ± (1.96)(0.0157) = .550 ± 0.031 • Polls will say, “55% favored the candidate, with a margin of error of 3%.” (Polls typically use a 95% confidence level, just like we did here.)95% confidence intervals. Computer simulation of 20 samples of size n=1000. Red dots = estimates of p’. Arrows indicate All intervals include the actual value of p, except this one.About 5% (1 in 20)of our confidence intervals will not cover the true value of p. Sample Distribution of P'pFigure by MIT OCW.Note a subtle error on that last slide • The picture of the 20 confidence intervals on the last slide was adapted from a book. • It actually doesn’t quite match the procedure we just followed. •We estimated σP' ≈ sqrt(p’(1-p’)/n) • Confidence interval = ± 1.96 σP' • If we estimate σP' we expect the width of the confidence interval to vary with our estimate of p’.A comment on the symmetry of sampling theory & confidence intervals • 95% of the p’ “arrows” land between p-1.96 σP' and p+1.96 σP' • 95% of the time, the true value of the population parameter, p, lies within p’-1.96 σP' and p’+1.96 σP' • We get this symmetry because the normal distribution is symmetric about the mean (the parameter we are estimating).Consider what would happen if parameter estimates had an asymmetric distribution • 95% of estimates of µ fall within an asymmetric region. µ-l µµ+r • Suppose you observe estimate m.Consider what would happen if parameter estimates had an asymmetric distribution • Suppose you observe estimate m. • If that estimate is < µ (i.e. to the left), then it must be pretty close to µ. – Within l. – m within l of the left of µ, => µ is