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Estimation • Read Sections 7.1, 7.2, and 7.3 of Devore. • The problem with which we are concerned is estimating a parameter educated guess • That is we want to form an of the parameter . at the value on the basis of available data. • That is, on the basis of observations. • An estimate of the value, in the form of a single number, (a single “best guess”) is called a point estimate . • In Stat 2593 we don’t sweat the problem of how to calculate point estimates. • Read Chapter 6 of Devore to find out a bit about what this problem involves. Estimation (Cont’d.) point estimates • In the context of this course, come by. are easy to • You estimate the population quantity by the corresponding sample quantity. • E.g. you estimate the population mean mean . • Likewise you estimate the population sample by the sample proportion . 1 proportion by the • But point estimates are guesses and, as they said in Casablanca, a guess is just a guess. • What we REALLY WANT is some measure of such a guess is. how good Interval Estimation • The usual way in which the “goodness” of a point estimate is expressed, is through a confidence • Such an interval is also called an interval. interval estimate . • The “goodness” being measured depends on how much variability or “volatility” there is in the observations and how MANY observations there are. • Less variability and/or more observations makes for a narrower or “tighter” confidence interval. • That is, for a more precise estimate. Technical Definition of a Confidence Interval • A 100(1 − α)% confidence interval for the parameter in which we are interested is an interval whose endpoints are random variables A and B such that P (A ≤ parameter ≤ B) = 1−α • The random variables A and B are calculated from (i.e. are functions of ) the available data. 2 The (fairly) General Form of a Confidence Interval • In the majority of contexts (in all contexts with which we are concerned) the forgoing formalism translates into: A confidence interval is given by point estimate ± (table value) × (st. dev. of point estimate) “t-tables” • The table value comes from the • It depends on the desired confidence level i.e. upon . α . • The “usual” or “standard” values of α are 0.10, 0.05, and 0.01 with 0.05 being the most “standard of all”. • For α = 0.05 we are talking about 95% confidence intervals. Degrees of Freedom • The table value also depends on how precisely the standard deviation of the population, i.e. σ has been estimated. • This dependence is specified by the so-called degrees of freedom . • Large degrees of freedom mean that σ is well estimated. • Small degrees of freedom mean that σ is not so well estimated. 3 • If σ is known — so that we don’t have to estimate it at all — then the degrees of freedom are infinity, i.e. ∞. perfectly • I.e. σ is “estimated”. Degrees of Freedom (Cont’d.) • When σ must actually be estimated (i.e. in real life (!), the degrees of sample size n freedom depend on the . • In the sort of problem with which we are initially concerned, namely one sample problems, the degrees of freedom are equal to n−1 n−k • More generally, the degrees of freedom are equal to where k is the number of parameters that you had to estimate BEFORE you could start to estimate σ . WARNING “classical” • This whole procedure for finding confidence intervals is valid ONLY WHEN EITHER the distribution of the data is large enough OR n is are “effectively” ∞. normal so that the degrees of freedom • What is going on is that we need n large enough so that (1) the Central Limit Theorem kicks in, and (2) the sample standard deviation is such a good estimate of σ that we can pretend that it IS σ . 4 . Confidence Intervals for Means • Setting: We have a sample X1 , X2 , . . . , Xn from a (normal) population with mean µ and variance σ 2 . • We want a 100(1 − α)% • The point estimate of µ is X̄. confidence interval for µ. √ σ/ n • The standard deviation of X̄ is . • When σ is unknown, we √ estimate it by the sample standard deviation S, so we actually use S/ n. • To calculate S we must first calculate X̄ which is of course an estimate of µ. • That is we have to estimate 1 parameter namely µ before we can start estimating σ. • Hence k=1 and so the degrees of freedom are Using the t-Tables n−1 • The degrees of freedom tell you what line or row of the t-tables to look at in order to find your table value . • If the degrees of freedom are larger than 30 most people (including me) take them to be equal to ∞. • I.e. if the degrees of freedom are larger than 30 we might as well pretend that σ is known. • This is just a pretence, but it makes no practical difference to the answer. • The column of the t-tables to look at is the 5 α/2 column. . • This is because we are putting half of the “missing confidence” at each end. • E.g. if α = 0.05 and n = 17, the table value is 16 and column 0.025 of the t-tables. 2.120 from row SUMMARY normal • If the data are , or if n is large, a 100(1 − α)% confidence interval for the population mean is given S X̄ ± (table value) × √ n • In the unlikely event that σ is known, replace S by foregoing. σ • The table value comes from the (degrees of freedom) = row and the α/2 column of the t-tables. in the n−1 • If σ is known or if n is large take the degrees of freedom to be Interpretation of Confidence Intervals ∞ . • Roughly speaking a confidence interval is an interval such that you true value of the are “pretty damned sure” that the parameter that you are estimating lies in the interval. • But does “sure” mean here? It SHOULD be interpreted in terms of probability. • Remember that “theoretically” a confidence interval [A, B] is a pair of random variables A and B such that P (A ≤ parameter ≤ B) = 1 − α. 6 observed • But when we calculate a confidence interval we have our data and so A and B turn magically from random variables into numbers. • This makes the foregoing tricky to think about. probability statement Confidence and Probability • E.g. suppose your (95%) confidence interval for µ turns out to be [42, 56] . It doesn’t really MEAN anything to say “P (42 ≤ µ ≤ 56) = 0.95”. constant • Since µ is not a random variable, but rather a statement 42 ≤ µ ≤ 56 is either true or it isn’t. Theres’s no probability involved. , the • You should be aware that there are some strange people called “Bayesians” who try to get around this problem by thinking of µ as a random variable. • But this just messes everything up and has to rest on a concept called “personal” probability which is just plain silly. Thought Experiments • One rational way of thinking about the meaning of the probability statement is through a “gedanken experiment” . • This experiment consists in considering what would happen if if a “large number” of people were to collect data (independently) and were each to calculate (e.g.) a 95% confidence interval estimated. 7 for the parameter being • The meaning of the probability statement is that in such a situation 95% of those confidence intervals would contain or “cover” the true value of the parameter. (about) • And about 5% would miss it. Believeable Values • An informal but very useful way of thinking about confidence plausible intervals is to think of them as being “a range of believeable values” for the parameter being estimated. or • That is you can “believe” a value for the parameter if that value lies in the confidence interval. • You don’t/can’t/shouldn’t believe values that the confidence interval. don’t lie in • Of course whether you “believe” a value or not depends on the level of confidence that you are imposing. Example of a Confidence Interval for a Mean • Problem 13, page 297 Devore 6th ed. page 293 5th ed. • Summary of information: n = 69, x̄ = 0.163 1.028 , and s = . • Since n is large we needn’t worry about whether the data are normal. • Also, since n is large, the degrees of freedom are 8 ∞ . • Hence a 95% confidence interval for µ is given by s x̄ ± (table value) × √ n 0.163 = 1.028 ± 1.96 × √ 69 = 1.028 ± 0.038 Example (Cont’d.) • Notice in the foregoing that since our sample has been observed, the random variables numbers x̄ and s. X̄ and S have (Poof!) turned into the • The foregoing calculation gives the interval [0.990, 1.066] . • Thus (at the 95% confidence level) a value of (e.g.) 1 is believable for the population mean dye layer density. • However a value of (e.g.) 1.1 is not (quite) believable. Another Example • Problem 32, page 306 Devore, 6th ed., page 302 5th ed. • Summary of info: n = 8, x̄ = 30.2 ,s= 3.1 . • Here we have a difficulty — n is small so our methods yield a valid confidence interval only if the data are normal. • There’s no way to tell if the data really ARE normal • The problem says to assume normality, so we will. 9 . • Under the assumption of normal data, a 95% confidence interval for µ is given by s x̄ ± (table value) × √ n 3.1 = 30.2 ± 2.365 × √ 8 2.59 = 30.2 ± Another Example (Cont’d) [26.61, 32.79] • The foregoing gives the interval . 28 • Hence (e.g.) is a plausible value for the population mean “interfacial shear yield stress”. • At least this is true if it is true that the data are normal . 25 is not a plausible value, subject again to the • But (e.g.) assumption of the normality of the data. One Last Example • Problem 5, page 290 Devore 6th ed., page 286 5th ed. • We are told to assume normal data, and that the population standard deviation is σ = 0.75 . 4.85 • Part (a): Info — n = 20, x̄ = . • A 95% confidence interval for µ is given by σ x̄ ± (table value) × √ n 0.75 = 4.85 ± 1.96 × √ 20 = 4.85 ± 10 0.328 . • This gives the interval [4.522, 5.178] . One Last Example (Cont’d.) 4.56 • Part (b): Info — n = 16, x̄ = • A 98% . confidence interval for µ is given by σ x̄ ± (table value) × √ n 0.75 = 4.56 ± 2.326 × √ 16 = 4.56 ± • This gives the interval 0.436 [4.124, 4.996] . One Last Example (Cont’d.) • Part (c): To get the we’d need width 0.75 1.96 × √ n of a 95% CI to be (at most) 0.40 ≤ 0.20 • This says that 54.02 n ≥ (1.96 × 0.75/0.2)2 = • So we could get away with n = 55 11 . One Last Example (Cont’d.) • Part (d): To get the half -width for a we’d need 0.75 2.576 × √ n 99% CI to be (at most) 0.2 ≤ 0.2 • This says that n ≥ (2.576 × 0.75/0.2)2 = • Hence we’d need n = 94 93.3 . How Big a Sample Do I Need? • The foregoing examples have illustrated how to deal with this issue. • To answer the question, one needs to specify the answers to three preliminary questions: 1. How “confident” 2. How precise 3. How variable do I want to be? do I want my estimate to be? will the observations be? • The answer to (1) is almost always 95%. • The answer to (2) is expressed as “to within ±d”. • The value of d depends on the practicalities of the area of study. • In a course such as Stat 2593 the required level of precision will always be supplied as part of the question, so you don’t need to worry about this issue. 12 Guessing at the Variability real difficulty • Finding an answer to (3) is the . • The problem is that you want to know, but you don’t know, and never will know the value of σ. guess • What you have to do is find some way of making a and then use this guess where you would like to use the true value of σ. • In effect, you pretend that your guess is the you know this is not really the case. truth at σ, even though • Having answered the three preliminary questions you calculate the sample size from n≥ (table value)2 × ( best guess at σ 2 ) d2 Getting the Table Value — df = ∞ at least as big “round high” . • Note that this says that n must be right hand side, so you always as the • The table value comes from the ∞ line of the t-tables. • This is because: • (a) You don’t know n so where else could you go? • (b) It will (always?) turn out that n is so big that the degrees of freedom might as well be ∞. • (c) You are pretending that σ is degrees of freedom should be ∞. 13 “known” anyhow, so the How to Guess at the Variability σ • Some ways of getting a guess at are: sample standard deviation from an • (i) Use the existing sample (which turned out to give inadequate precision). (This is the usual scenario for questions in introductory statistics courses.) • (ii) Find the results of “similar” research in the literature. • (iii) Do a pilot study. • (iv) Guess at the maximum and at the minimum values that will be observed. (People almost always have a better intuitive feeling for these than they have for σ≈ σ .) Then use max − min 4 = “range” 4 Confidence Intervals for Proportions • Suppose we are trying to estimate p which is the of successes in a certain population. proportion probability • We may also describe p as being the of getting a “success” when we draw a random item from the population. • We draw a sample of size n from the population, and on that basis we wish to form an (interval) estimate of p. • The general principal applies; such a confidence interval is given by point estimate±(table value)× (st. dev. of point estimate) 14 The Sample Size Must be Large • The methods we shall use for CIs for valid when n is LARGE proportions are only . degrees of freedom calculating a CI for a proportion are always ∞ . • Hence the associated to use when Point Estimates and St. Devs. for Proportions • The point estimate of p is the p̂ = X/n • This is the sample. sample proportion . where X is the number of “successes” in • The only problem left is to determine the standard deviation of p̂. • But we know that X ∼ Bin(n, p) . • Hence V (X) = np(1 − p). • Hence V (p̂) = V (X/n) = V (X)/n2 = np(1 − p)/n2 = p(1 − p)/n But We Don’t Know p! • Since we don’t know p — that’s what we’re trying to estimate for crying out loud! — we do the best we can under the circumstances. • That is, we replace p by our point estimate 15 of p, i.e. by p̂. • SUMMARY: A confidence interval for a proportion is given by p̂ ± (table value) × r p̂(1 − p̂) n • Devore does something a little more complicated — he solves a quadratic to get the CI. • See expressions (7.10), page 291 5th ed., page 295, 6th ed. • PLEASE DON’T use (7.10)!!! Use the simpler expression given above (equivalently expression (7.11) in Devore). Degrees of Freedom; Point Estimate • The table value to be used when calculating confidence intervals for proportions comes from the ∞ row of the t-tables. • This stuff only works if the sample size is large! • And if the sample size is large, the are (effectively) infinite. degrees of freedom • The point estimate in this setting is the sample proportion. p̂ = X/n • That is, it is where X is the number of “successes” in the sample and n is the sample size. Example • Problem 21, page 298 Devore 6th ed., page 294 5th ed. • Summary of info: n = 539, x = 133. Hence 16 p̂ = 133/539 . • The required confidence level is 99% so α = 0.01 whence α/2 = 0.005. 2.576 • The table value is therefore freedom = ∞ row of the tables. from the degrees of • The interval estimate of p is therefore given by r p̂(1 − p̂) p̂ ± (table value) × n r 133 1 133 406 = ± 2.576 × × × = 0.2468 ± 539 539 539 539 giving the interval 0.0478 [0.1990, 0.2946] How Big a Sample Do I Need? • The 0.05. half-width of the forgoing confidence interval is about • Suppose we wanted the estimate of p to be accurate to within ±0.03, with 99% confidence. • How big a sample should we take? • The procedure is very similar to what we do in the estimating a mean setting. We need r p(1 − p) 2.576 × n ≤ 0.03 • Now of course we don’t know p. • So we must come up with a “reasonable guess” at p . • This is analogous to the fact that in the “means” setting we needed a “reasonable guess” at σ . 17 Guessing at p “preliminary” • In the current situation we have a sample, i.e. the sample which gave us an (insufficiently precise) estimate of p. • We can use the p̂ from that preliminary sample, i.e. our “reasonable guess” at p. • This gives n≥ so we’d take 2.576 0.03 2 × 0.2468 × 0.7532 = 0.2468 as 1370.6 n = 1371 • But suppose we didn’t have a preliminary sample; what then? • We might have some sort of guess at p. The Worst That Can Happen “worst case scenario” • Failing that, we could use the • Note that proportions have an advantage over means in this regard. • For means there is no “worst case”; things can get arbitrarily bad . • For proportions the worst case is the value of p that makes p(1 − p) largest and hence makes n largest; that value is p = 1/2 • If we had to go with the worst case “guess” at p in the current example, i.e. with n≥ so we’d take p = 1/2 2.576 0.03 2 n = 1844 we’d get × 0.5 × 0.5 = 1843.3 . 18 . . Summary on Sample Size Calculations • For both means and proportions the sample size calculations boil down to: n≥ best guess at σ 2 (table value)2 × ( ) d2 where we want our estimate to be “accurate” to within ±d. • In the case of proportions, σ 2 becomes • The table value comes from the ∞ p(1 − p) . line of the t-tables. • In the case of proportions we can, if all else fails, make a guess at p by using the “worst case scenario” i.e. p = 0.5. • In the case of means, a good way of making a guess at σ is to guess instead at the max and min of the observations, set range = max − min and use σ ≈ range/4 . Using Minitab to Calculate Confidence Intervals • The calculations for confidence intervals can all be done with Minitab. • Click on Stat then on Basic Statistics , then on 1 sample t or 1 sample Z or 1 proportion . • The ‘‘1 sample Z’’ is used for (artificial) problems in which σ is known . • There’s a wee window in which you specify “Sigma”. 19 default • For all three possibilities the confidence level is 95%. • You can specify the confidence level (to be something other than 95%) by clicking on Options . Minitab and CIs for Means • You need only concern yourself about using the ‘‘1 sample t’’ and DATA ‘‘1 sample Z’’ utilities with . • I.e. use these utilities when you have the actual sample stored in a column of the Minitab worksheet. • Minitab 14 will do the calculations for you on the basis of the “summary statistics” namely n, x̄, and s (or σ in the “1 sample Z” case). • But the calculations are not arduous; you might as well do them by hand . Minitab and CIs for Proportions • For proportions, the only problems you will ever get will involve summary statistics (i.e. sample size and successes ). • Minitab will do such problems for you, if you insist. • But I’m not going to go there. • Minitab will also do proportions problems based on columns of data . 20 number of • I.e. it will determine the sample size (length of the column) and count the number of “successes” in the column. • And then do all the necessary calculations. • I won’t go there either! Prediction Intervals • As I have previously tried to emphasize, we often DO NOT CARE about the mean of the population. • We are more interested in individual members of the population, i.e. in future individual observations . • Confidence intervals tell us nothing directly about individual observations. • E.g. suppose that someone takes a sample of U.N.B. students and finds that a 95% confidence interval for the mean height of U.N.B. students is [168, 176] cm. • What does this tell you about the height of the next student to walk through the door? NADA! • To say something about the value of the next observation we need a prediction interval or PI for short. Normality is Crucial • The following technique for calculating prediction intervals is valid ONLY IF the data are normal . • There are various techniques for checking on whether data are normal. 21 goodness of fit tests • These include which we will touch on briefly later on in the course, and normal probability plots which we unfortunately don’t have time to cover. • You can read about probability plots on pages 188 to 197 of Devore 6th ed., pages 185 to 195 5th ed. • In any case you need draw any reasonably your data are normal. quite a lot of data before you can certain conclusions about whether Calculating Prediction Intervals for Normal Populations • Given a sample from a normal population we form a 100(1 − α)% confidence level prediction interval for “the next observation” as point estimate ± (table value) × st. dev. where the “point estimate” is the sample mean i.e. X̄ and where “st. dev.” is the appropriate standard deviation for prediction. • In fact “st. dev.” here is equal to the standard deviation of X − X̄ where X is predict. the new observation that you are trying to The “Appropriate” Standard Deviation • Explicitly “st. dev.” here is equal to S p 1 + 1/n where S is the sample standard deviation of your sample. • In the unlikely event that σ is known, you of course use this known value instead of the estimate S. 22 from X • The “1” inside the square root comes comes from X̄ and the “1/n” , • Actually the 1/n makes a negligible amount of difference and can be ignored for “large” n (e.g. n = 30 or 40 or more). • If you ignore 1/n then the prediction interval becomes x̄ ± (table value) × S . Example of a Prediction Interval • Problem 38 page 307 Devore 6th ed., problem 38, page 303 5th ed. • Summary of info: n = 25, s = 0.0065 x̄ = 0.0635 and . • Part (b) of the problem asks us to calculate a 95% prediction interval for the amount of warpage of a single piece of laminate. • This prediction interval is given by p 1 + 1/n p = 0.0635 ± 2.064 × 0.0065 × 1 + 1/25 x̄ ± (table value) × s = 0.0635 ± giving the interval 0.0137 [0.0498, 0.0772] . Example of a Prediction Interval (Cont’d.) • The table value comes from the row of the t-tables. “degrees of freedom” • In this current scenario, the degrees of freedom are n − 1 = 25 − 1 = 24 23 . • Since α = 0.05 we look in the α/2 = 0.025 column. • Compare the foregoing prediction interval with the much narrower confidence interval for the population mean warpage, explicitly [0.0608, 0.0662] . • For instance 0.0550 is “believable” as the value of an individual observation from this population. NOT believable • However it is population mean. 24 as the value of the