Questions and Answers #6 Confidence Intervals

Questions and Answers Sheet 6 Confidence Intervals Question #1 Calculate 𝑃[𝐴, 𝐵, 𝐶] from 𝑃[𝐴, 𝐵] and 𝑃[𝐵, 𝐶] Answer: Step 1 I think you can constrain 𝑃[𝐴, 𝐵, 𝐶] to an interval such as max(0, 𝑃[𝐴, 𝐵] + 𝑃[𝐴, 𝐶] − 𝑃[𝐴], 𝑃[𝐴, 𝐵] + 𝑃[𝐵, 𝐶] − 𝑃[𝐵], 𝑃[𝐴, 𝐶] + 𝑃[𝐵, 𝐶] − 𝑃[𝐶]) ≤ 𝑃[𝐴, 𝐵, 𝐶] ≤ min(𝑃[𝐴, 𝐵], 𝑃[𝐴, 𝐶], 𝑃[𝐵, 𝐶], 1 + 𝑃[𝐴, 𝐵] + 𝑃[𝐴, 𝐶] + 𝑃[𝐵, 𝐶] − 𝑃[𝐴] − 𝑃[𝐵] − 𝑃[𝐶]) 1 1 1 As an example, if 𝑃[𝐴] = 𝑃[𝐵] = 𝑃[𝐶] = and 𝑃[𝐴, 𝐵] = 𝑃[𝐴, 𝐶] = 𝑃[𝐵, 𝐶] = then 0 ≤ 𝑃[𝐴, 𝐵, 𝐶] ≤ 2 4 4 Question #2 Calculating confident intervals for the expected daily returns 𝑆𝑉 = 1 𝑛−1 ⋅ ∑𝑖 (𝐷𝑅𝑖 − 𝑆𝑀)2 Answer: Step 1 What you calculated as: 𝑆𝐸 = √ 𝑆𝑉 𝑛 is known as the standard-error of your estimate of the mean. It's basically the standard deviation of your estimator. Let's assume the true mean were 𝜇 . What's referred to the t-stat is: 𝑆𝑀−𝜇 𝑡= 𝑆𝐸 follows a t-distribution with 𝑛 − 1 degrees of freedom. By the CLT, the distribution of estimator SM converges to the normal distribution but there's a bit of extra complexity for the t-stat here in that we're dividing by the square root of our estimate of the variance, which is a random variable too. The tdistribution is basically the standard normal distribution with somewhat fatter tails. Where does the 1.96 etc... come from? A t distributed random variable with greater than approximately 1000 degrees of freedom will lie between -1.96 and 1.96 about 95 percent of the time. This can be seen using the inverse of the t-cdf. 𝐹𝑡−1 (0.975,1000) ≈ 1.9623 and 𝐹𝑡−1 (0.025, 1000) ≈ −1.9623 Question #3 Calculating the Confidence Interval It is found that in a random sample of 100 Science students, there are 48 studying statistics . To test whether the true proportion of students in statistics is 50% or not, suitable null and alternative hypotheses are: (a) 𝐻0: 𝑝 = 0.48; 𝐻1: 𝑝 = 0.48 (b) 𝐻0 ∶ 𝑝 = 0.5; 𝐻1 ∶ 𝑝 > 0.5 (c) 𝐻0 ∶ 𝑝 = 0.5; 𝐻1 ∶ 𝑝 < 0.5 (d) 𝐻0 ∶ 𝑝 = 0.5; 𝐻1 ∶ 𝑝 = 0.48 (e) 𝐻0 ∶ 𝑝 = 0.5; 𝐻1 ∶ 𝑝’ = 0.5 Find a 95% Confidence Interval for p in Q1. I have determined the answer to Q 1), which I believe is e. For question 2), I have been told the CI= estimator +/− (table value)(Standard error). The estimator is 0.48, and I have found the SE to be 0.05. However, I am unsure about how to calculate the "table value". Answer: You are correct to use 𝐻0 : 𝑝 = 0.5 versus 𝐻1 : 𝑝 ≠ 0.5 . for your hypothesis testing. In the US until several years ago, the traditional 95% ci for p is to use the point estimate 𝑝̂ = 𝑋/𝑛 = 48/100 . The standard error of 𝑝̂ is √𝑝(1 − 𝑝)/𝑛 , but because you don't know p, the traditional method is to estimate it by plugging in 𝑝̂ for p. Thus the traditional formula for the 95% CI is 𝑝̂ ± 1.96√ 𝑝̂(1−𝑝̂) 𝑛 , which computes to (0.382,0.578). This method uses the normal approximation to the binomial distribution, so 1.96 is the 'table value' you're looking for. However, this method has two sources of error: (a) using the normal approximation to the binomial distribution, and (b) estimating the standard error. Computational evidence is that the traditional method often does not actually have the 'promised' 95% coverage. Much of the error due to (b) can be eliminated by using 𝑛̃ = 𝑛 + 2 instead of n, and 𝑝̃ = (𝑋 + 2)/𝑛̃ = 50/104 instead of 𝑝̂ . The resulting adjusted interval, due to Agresti, is (0.385,0.577). (In your particular example, the adjustment is small.) The Wilson interval is also more accurate than the traditional one. You can google that. For the 95% confidence level, it is not much different from the more easily computed Agresti interval. The Wilson interval has been standard for many years in parts of Europe. All three intervals cover 𝑝 = 0.5 , so there is no reason to doubt that the true percentage of statistics majors is near 50%. Question #4 Can a prediction interval be interpreted as a probability? Suppose I find a 90% prediction interval for some data distribution. This implies that if I sample large enough data from this distribution, then 90% of such data will lie inside the prediction interval. Is it same as saying that any randomly sampled data point from the distribution will lie inside the prediction interval with 0.9 probability? Answer: Your interpretation of a prediction interval is incorrect. A 90% prediction interval will contain 90% of the probability of the true underlying distribution on average (not always nor "at a minimum"). What you are thinking about is a less-taught concept: a Tolerance Interval. A tolerance interval is specified by a confidence and a coverage: The coverage is the minimum probability that we want our interval to contain (as calculated using the true distribution), while the confidence is the probability that the interval will actually achieve its minimum coverage for a random sample from the population. In contrast, a 90% prediction interval is defined in reference only to the next point, not to all future points (like tolerance intervals). So, say you collect a sample S and calculate a 90%𝑃𝐼, then collect one more point, p. What the 90% tells us is that, 90% of the PI's constructed in this way from random samples S will contain the next point p. What is missing from PI's is the conditional argument: Let I be our prediction interval and 𝑝1 be the first "out of sample" point collected after we formed our interval. What we cannot say is: 𝑃(𝑝2 ∈ 𝐼|𝑝1 ∈ 𝐼) = 0.9; We also cannot say: P( p2  I | p1  I ) = 0.9 However, a tolerance interval is defined not in reference to the next point but to the entire population, so it retains its properties for all future "out of sample" observations (albeit with the given confidence). Question #5 90% Confidence Interval for Difference Between Two Proportions In random, independent samples of 225 adults and 250 teenagers who watched a certain television show, 99 adults and 125 teens indicated that they liked the show. Let 𝑝1 be the proportion of all adults watching the show who liked it, and let 𝑝2 be the proportion of all teens watching the show who liked it. Find a 90% confidence interval for 𝑝1 − 𝑝2 . 1. what is the lower limit of the 90% confidence interval 2. what is the upper limit of the 90% confidence interval Answer: (upper = 0.06, +1.64√ (lower = 0.06, −1.64√ p1(1−p1) n1 p1(1−p1) n1 0.44∗0.56 upper = .0.06, +1.64√ lower = .0.06, −1.64√ 225 + + + p2(1−p2) n2 p2(1−p2) n2 ) ) 0.5∗0.5 250 0.44 ∗ 0.56 0.5 ∗ 0.5 + 225 250 Thus, the interval is (−.015, .135) Notice the sign difference because I used Teens - Adults. It is the same answer as above, but make sure to interpret it correctly. R is of course a lot easier and supplies much more information Question #6 95% confidence interval for 𝜃2 − 𝜃1 from uniform (𝜃1 , 𝜃2 ) where 𝑋1 , … , 𝑋𝑛 are the random sample from 𝑈(𝜃1 , 𝜃2 ) , the uniform distribution with two parameters 𝜃1 , 𝜃2 . ̂2 = 𝑋(𝑛) , the I have the maximum likelihood estimators ̂ 𝜃1 = 𝑋(1) , the minimum order statistic, and 𝜃 maximum order statistic. But I don't know how to compute the confidence interval of 𝜃2 − 𝜃1 . Answer: Let 𝛿 ≔ 𝜃2 − 𝜃1 and 𝑍𝑖 ≔ 𝜃1 + 𝑋𝑖 . We have 𝑍𝑖 ∼ Unif(0, 𝛿) , and 𝑋(1) = 𝜃1 + 𝑍(1) and X_{(n)} = 𝜃1 + 𝑍(𝑛) . Then, 𝑅 ≔ 𝑋(𝑛) − 𝑋(1) = 𝑍(𝑛) − 𝑍(1) is a good estimate of 𝛿 . We note that [𝑍(𝑛) −𝑍(1) ] 𝛿 is a pivot, i.e. its distribution does not depend on 𝛿 and it is easy to find out. This can be used to construct the CI. 𝑍 Hint: Note that 𝑈𝑖 ≔ 𝑖 is uniform (0,1). You are basically computing the distribution of 𝑈(𝑛) − 𝑈(1) . You 𝛿 can either use the known joint distribution of (𝑈(1) , 𝑈(𝑛) ) or do it directly yourself: (U ( n ) − U (1)  t ) = (U ( n )  t + U (1) ) = [U ( n )  t + u, U (1)  (u, u + du )] The integrand is the probability that the minimum is about u and the rest of them are ∈ (𝑢, 𝑡 + 𝑢) . There are n possible choices for which one of the variables is the minimum, and all these events have the same probability 𝑑𝑢(min{ 𝑡 + 𝑢, 1} − 𝑢)𝑛−1 . Hence, p (t ) := 1 (U ( n ) − U (1)  t ) = n  (min{t + u,1} − u ) n −1 du . 0 which can be simplified and explictly computed = 𝑛(1 − 𝑡)𝑡 𝑛−1 + 𝑡 𝑛 𝑓𝑜𝑟 𝑡 ∈ [0,1] . Choose 𝑡1 and 𝑡2 such that 𝑝(𝑡2 ) − 𝑝(𝑡1 ) = 0.95 . Then [𝑅/𝑡2 , 𝑅/𝑡1 ] is a 95% CI for 𝛿 . You can try to minimize the length, by minimizing |1/𝑡2 − 1/𝑡1 | subject to the given constraint. Question #7 A clarification on the level of confidence for a Chi-squared distribution 2 Suppose we have 𝜒1−𝛼,29 = 45.22 , where the Chi-squared distribution has 29 degrees of freedom and (1 − 𝛼)% level of confidence. How then do we compute 1 − 𝛼 ? The notation of 𝜒 2 . I've found in books and online sources all come in the form of 𝜒𝑐2 , where c is the degrees of freedom. I know this question seems trivial, but I'm still relatively new to Statistics.. Some clarification will be great! Answer: Recall that for X ~ ChiSquare( ) , that is to say 𝜒𝜈2 , we have 𝑓𝑋 (𝑥) = 𝑥 𝜈/2−1 𝑒 −𝑥/2 , 𝑥 > 0, 2𝜈/2 Γ(𝜈/2) 2 hence, Pr[𝑋 ≤ 𝜒1−𝛼,𝜈 ] 2 In other words, 𝜒1−𝛼,𝜈 = 1 − 𝛼. is the 1 − 𝛼 quantile of the chi-square distribution with ν degrees of freedom. Given this value and ν, the computation for α involves computing an integral; in your case, numeric 45.22 evaluation of ∫𝑥=0 𝑓𝑋 (𝑥) , 𝑑𝑥 ≈ 0.971992 , or 𝑎 ≈ 0.028 . Question #8 A physician who has a group of thirty-eight female patients aged 18 to 24 on a special diet wishes to estimate the effect of the diet on total serum cholesterol. For this group, their average serum cholesterol is 188.4 (measured in mg/100mL). Because of a large-scale government study, the physician is willing to assume that the total serum cholesterol measurements are normally distributed with standard deviation of 𝜎 = 40.7 . Find a 95% confidence interval of the mean serum cholesterol of patients on the special diet. Does the diet seem to have any effect on their serum cholesterol, given that the national average for women aged 18 to 24 is 192.0? Answer: Step 1 Your sample size is 𝑛 = 38 , let X be the serum cholesterol level of your target population. You are assuming that 𝑋 ∼ 𝒩(𝜇, 40.7). So, you hypotheses are 𝐻0 : 𝜇 ≥ 192 vs. 𝐻1 : 𝜇 < 192 . Note that ̅𝑋̅̅𝑛̅ ∼ 𝒩 (𝜇, 𝜎2 𝑛 ) . So the construction of the CI (in confidence level of 1 − 𝑎 ) is given by 𝑃 (𝜇 ∈ [𝑥 ̅̅̅ 𝑛 ± 𝑍1−𝑎/2 𝜎 √𝑛 ]) = 1 − 𝑎, namely, you should check whether you confidence interval [188.4 − 1.96 × 40.7 √38 , 188.4 + 1.96 × contains the 𝜇 under 𝐻0 . Question #9 40.7 √38 ] A random sample of professional wrestlers was obtained, and the annual salary(in dollars) for each was recorded. The summary statistics were 𝑥 = 47, 500 and 𝑛 = 18 . Assume the distribution of annual salary is normal with 𝜎 = 8,500 a. Find a 90% confidence interval for the true mean annual salary for all professional wrestlers. b. How large a sample is necessary in order for the bound on the error of estimation to be 3000? Answer: Step 1 As you know, 𝑎 90% confidence interval is of the form 𝜎 𝑋̅ ± 1.645 . √𝑛 The margin of error is 𝑀 = 1.645 𝜎 √𝑛 . For (b) set 𝑀 = 3000 and solve for n. Question #10 What sample size is necessary for 95% 𝐶𝐼? A state legislator wishes to survey residents of her district to see what proportion of the electorate is aware of her position on using state funds to pay for abortions. a. What sample size is necessary if the 95% 𝐶𝐼 for p is to have a width of at most 0.10 irrespective of p? b. If the legislator has strong reason to believe that at least 2 3 of the electorate know of her position, how large a sample size would you recommend? Answer: If you want a 95% 𝐶𝐼, then it seems to me you need to use 𝑘 = 𝑧𝛼/2 = 1.96 , where k is my abbreviation. Also, in your first computation, I think you mean to set 𝑤 = 0.1 (not 𝑝 = 0.1 , because you assume 𝑝 = 1/2 .) And I am puzzled why you set 𝑛 = 10 , when n is what you're trying to find. Finally, I think 0.25 in your equation is 𝑝(1 − 𝑝) = 𝑝𝑞 = (1/2)(1/2) = 0.25 . With all of that, I do get the same result that you report from the following computation in R. 𝑘 = 1.96; 𝑝 = 0.5; 𝑞 = 0.5; 𝑤 = 0.1 𝑘 2 /2−𝑘 2 ⋅𝑤2 +√𝑘 4 ⋅(𝑝⋅𝑞−𝑤 2 )+𝑤 2 ⋅𝑘 4 𝑤2 The rationale for taking 𝑝 = 𝑞 = 1/2 when p is unknown is that this maximizes the required n and thus insures it is large enough for other possible values of p. 2 1 3 3 However, I do not get 𝑛 = 339 when I change the problem so that 𝑝 = and 𝑞 = . (I'm not sure whether this is the change you made, but it seems reasonable to me.) This should take care of situations where the true p is 2/3 or greater. Here is what I did get: 𝑘 = 1.96; 𝑝 = 2/3; 𝑞 = 1/3; 𝑤 = 0.1 𝑘 2 /2−𝑘 2 ⋅𝑤2 +√𝑘 4 ∗(𝑝∗𝑞−𝑤2 )+𝑤 2 ∗𝑘 4 𝑤2 Because you have what I take to be several typos in your Question, I don't know where to go from here to help you resolve the discrepancy. Please proofread everything (your Question and my Answer), and make any edits, suggestions, or explanations that might help make the next attempt to help more successful. Question #11 What is the radius of a "Gaussian" sphere such that approx. all the population lie within? Let X  be a random vector distributed normally, i.e. 𝐗 ∼ 𝒩(𝟎, 𝜎 2 𝐼𝑛 ) , where \sigma>0 is the standard deviation and 𝐼𝑛 is the identity matrix of order n. n In the case of the univariate Gaussian distribution, i.e., 𝑋 ∼ 𝑁(0, 𝜎 2 ) , we say that approximately 99.7% of the population lie within three standard deviations. What would be an appropriate constant in the case of the multivariate "isotropic" Gaussian distribution, i.e., 𝐗 ∼ 𝒩(𝟎, 𝜎 2 𝐼𝑛 )? That is, what would be the radius of the n-sphere in terms of the standard deviation \sigma , such that approximately all the population lie within? P.s.: Please feel free to edit the title (and/or the body of this question). I was not sure how to express what I wanted in the title. Answer: The distribution of |𝑋|2 /𝜎 2 is 𝜒𝑘2 . Since for 𝑘 = 1 . 𝑃(|𝑥| ≤ 3𝜎) = 0.9973002 I interpret this as the question for which r in the case of 𝑘 = 3 𝑃(|𝑋| ≤ 𝑟𝜎) = 0.9973002. This is 𝑟 = √𝐹 −1 (. 997302) = 3.76205 where F is the cumulative distribution function of 𝑎𝜒32 random variable. Question #12 A t - distribution's convergence to a Normal distribution I'm currently messing around with confidence intervals and I can't really understand how a t distribution converges to a normal distribution for large n. For example, suppose we want to construct a 95% confidence interval when we have a sample mean 𝑋̅ = 74.8 and sample variance 𝑆 = 1.23 with 𝑛 = 143 . I would construct the confidence interval using, (X − z 1−  2 S S ,X +z  ) 1− n n 2 since 𝑛 > 30 . If 𝑛 < 30 I would have used a t-distribution. My question is why does the t - distribution approach a normal distribution for relatively large n? Answer: The t distribution arises because you estimate the population standard deviation 𝜎 by the sample standard deviation S. For smaller n, there is some significant chance that S is quite a bit smaller than 𝜎 ; thus for fixed c and n, there is significant probability that 𝑋 is within 𝑐 𝜎 of 𝜇 and not within S of 𝜇 . (The reverse is possible too, but less likely.) But for large n, S is essentially guaranteed to be very close to 𝜎 , because it is an asymptotically consistent estimator for 𝜎 . And of course, if S is close to 𝜎, 𝑋 being within cσ and being within S of 𝜇 are nearly equivalent. Note that strictly speaking you should always use the t distribution for confidence intervals from a normally distributed population with unknown standard deviation. It is just negligibly different from the same interval constructed with the normal distribution if n is large enough. How large n needs to be really depends on how small a difference can be treated as negligible. Question #13 Accuracy of confidence intervals Suppose I have iid observations 𝑋𝑖 (empirical mean ̅𝑋̅̅𝑛̅ ), drawn from a distribution with unknown mean 𝜇 and known variance 𝜎 2 . To build a confidence interval for 𝜇 I can use the central limit theorem that states: n( X n − )   (0,1) and get the following approximation (if I am not mistaken), with 𝜙 being the quantile function of the standard normal distribution: ( [ X n −  n  1−  ; Xn +  n 2  1−  ])  1 −  2 I've always been told to just provide this as an answer for an interval with 1 − 𝛼 confidence level. But what about the real confidence level? It must be something like 1 − 𝛼 − 𝜖𝑛 , right? What about 𝜖𝑛 ? Answer: However, you should use this method only when you're sure that the CLT applies. It is usually OK if the population distribution is nearly symmetrical and the sample size is fairly large. I am writing this because I have misgivings about the unrestricted language in your Question, which may lead some to believe that this is a "general-purpose CI", which it is not. For example, if the population distribution is nearly exponential, then it is best to observe that the sample mean has nearly a gamma distribution (chi-squared with a little re-scaling) and to use quantiles of that distribution to make a CI. (Also, in the case of an exponential population, if you know 𝜎 then you know μ exactly, and there is no need for a CI for 𝜇 .) In the real world, even for nearly normal data, there are few cases in which 𝜎 is known and \mu is not. If 𝜎 needs to be estimated by the sample SD S, then ̅̅̅̅ √𝑛(𝑋 𝑛 −𝜇) 𝑆 has approximately Student's t distribution with 𝜈 = 𝑛 − 1 degrees of freedom. Then you use Student's t distribution to make the CI. For a 95% CI, results will be about the same as with normal if the sample size is above 30. (But there are different boundaries on the sample size than 30 for confidence levels other than 95%.) When the population distribution is clearly not normal and of an unknown shape, it might be better to use a bootstrap CI than to try to use your normal approximation, even if the sample size is moderately large. Question #14 An approximate 95% confidence interval for the true unemployment rate p is (𝑝̂ − 𝑧 2.5√ ̂ (1−𝑝 ̂) 𝑝 𝑛 , 𝑝̂ − 𝑧 97.5√ ̂ (1−𝑝 ̂) 𝑝 𝑛 ) 0.079(1−0.079) = (0.079 − 1.96√ 4148 , 0.079 + 1.96√ 0.079(1−0.079) ) 4148 = (0.071,0.087) What I'm mainly confused about is: how did they jump immediately to their first line? Answer: Step 1 It's a standard result that an approximate 100(1 − 𝛼)% CI for a proportion p, obtained by observing x successes in a sequence of n independent Bernoulli trials each with success probability p is (𝑝̂ − 𝑧√ 𝑝̂(1−𝑝̂) 𝑛 where 𝑝̂ = 𝑥 𝑝̂(1−𝑝̂) , 𝑝̂ + 𝑧√ 𝑛 ) 𝛼 is the estimate of p and z is the (1 − ) quantile of the standard normal distribution. 𝑛 2 So yes, assumptions have been made about independent Bernoulli trials. Question #15 Are they both used to calculate the confidence interval at 95% for a population proportion? Here are two questions from my book. In a survey in a large city, 170 households out of 250 owned a pet. Find the 95% confidence interval for the proportion of households in the city who own a pet. From a random sample, 136 out of 400 people experience discomfort after receiving a vaccine. Construct a 95% confidence interval for the population proportion who might experience discomfort. For each question do I use: 𝑝̂ ± 1 𝑝(1−𝑝) or 𝑝̂ ± √ √𝑛 𝑛 ? Answer: Step 1 Multiple errors here: Apparently you want a 95% CI for binomial success probability p, using the number of successes X observed in n independent trials, giving 𝑝̂ as an estimate of p. Provided n is large enough for the normal approximation to apply, and for 𝑝̂ to be a reasonably good estimate of p, an approximate 95% confidence interval for p is 𝑝̂ ± 1.96√ 𝑝̂(1−𝑝̂) 𝑛 . 1 𝑝̂(1−𝑝̂) 2 𝑛 In case, 𝑝̂ ≈ , the factor 1.96√ is very nearly 1 √𝑛 . My guess is that you are expected to use the displayed expression to work your problems. Step 2 Considerable theoretical, computational, and simulation evidence has shown that an interval that works much better for sample sizes such as those in your problem is as follows: Use 𝑝̃ = 𝑋+2 𝑛+4 to estimate p and use \tilde n instead of n in the formula above to obtain the 'Agresti-Coull' (or 'Plus Four') confidence interval: 𝑝̃ ± 1.96√ 𝑝̃(1−𝑝̃) 𝑛̃ . Question #16 Confidence interval and standard deviation I'm currently talking an intermediate course in finance where we want to calculate Value-at-Risk for portfolios and bonds. To use this VaR formula I need to know the standard deviation for different confidence intervals. Now my teacher have put up the following standard deviation for different confidence intervals: 𝐶. 𝐼 90 = +/− 1,64 𝑆. 𝑑 𝐶. 𝐼 95 = +/− 1,96 𝑆. 𝑑 𝐶. 𝐼 98 = +/− 2,33 𝑆. 𝑑 𝐶. 𝐼 99,9 = +/− 3,09 𝑆. 𝑑 When I watched an old exam for calculating VaR, the C.I was 99% and the student wrote that the S.d was equal to 2,33. How is this possible? (P:s the student got an A on this exam). Answer: This is correct. Have a look at this standard table of the normal distribution probability. You find that, for 𝑍 ∼ 𝑁(0,1), 𝑃(−∞ < 𝑍 ≤ 2.33) ≈ 0.9901 this comes from a numerical evaluation of  (2.33) = 1 2  2.33 (2.33) = 1 2  2.33 − − e−t /2 , dt  0.99009692444083574978997 2 e−t 2 /2 dt  0.99009692444083574978997 Question #17 Confidence interval for a density function parameter I am trying to solve this problem: Given 𝑋1 , ⋯ , 𝑋𝑛 a random sample of a population of random variables with 𝑝. 𝑑. 𝑓. 𝑓(𝑥, 𝜃) = 𝑒 −(𝑥−𝜃) 𝐼(𝑥)𝑥≥𝜃 , find a confidence interval of level 1 − 𝛼 for 𝜃. SNK Following a hint, I first looked into the distribution of 𝑌 = min(𝑋𝑖 ) − , which I found to be Exponential(n). Taking into account that the indicator function in the pdf limits to values of x greater than 𝜃 , and using the exponential's cdf. and some bounds, I got that the confidence interval should be 𝐼1−𝛼 (𝑋1 , … , 𝑋𝑛 ) = [0, ln(𝛼) 𝑛 ]. Would that be correct? I am not sure I am using the hint correctly. If it is wrong, what else could I try? Answer: I think you were on the right track initially, but may have confused the arithmetic of the exponential distribution If 𝑌 ∼ Exp(𝑛) then 𝑃(𝑌 ≤ 𝑦) = 1 − 𝑒 −𝑛𝑦 so 1 - there is a probability 𝛼 that 0 ≤ 𝑌 ≤ − log 𝑒 (1 − 𝛼) . 𝑛 1 1+𝛼 𝑛 2 - and also a probability 𝛼 that − log 𝑒 ( 1 1−𝛼 𝑛 2 ) ≤ 𝑌 ≤ − log 𝑒 ( ) This gives confidence intervals for 𝜃 . 1 − [min(𝑋𝑖 ) + log 𝑒 (1 − 𝛼) , min(𝑋𝑖 )] 𝑛 1 1−𝛼 𝑛 2 - or [min(𝑋𝑖 ) + log 𝑒 ( 1 1+𝛼 𝑛 2 ) , min(𝑋𝑖 ) + log 𝑒 ( )] The first of these alternatives might be seen as "one-sided" but is narrower than the second which might be seen as "two-sided". As an example, suppose 𝛼 = 95%, 𝑛 = 2, 𝑋1 = 5 and 𝑋2 = 6.2 then this would give confidence intervals for 𝜃 of about [3.502,5] or [3.156,4.987]. Question #18 Confidence interval for parameter 𝑋1 , 𝑋2 , ⋯ , 𝑋𝑛 − 𝑖. 𝑖. 𝑑 observations 𝑋1 = 𝜉 + where 𝜉 ∼ 𝑁(𝜃 2 , 𝜃 2 + 1), 0, 1/ 2 ,  and 𝜂 are independent. 4 , 1/ 2  = Find 𝛼 -confidence interval. The first that I need to do is to find some estimate for 𝜃 . The only one that I find is (5𝑋 − 𝑆 2 + 1)/10 but it is difficult to find distribution. Is it possible to do something else? Answer: A natural choice arises from the method of moments estimator: X = E[ +  ] =  2 + 2 , thus if 𝜃 > 0 , 𝜃̂ = √1 + 𝑋̅ − 1. Then, what is the variance of 𝜃̂? You could use the delta method: Var[ g ( X )]  ( g (E[ X ])) 2 Var[ X ]. This will give you a Wald-type asymptotic confidence interval. Question #19 Confidence Interval for Pareto Distribution A random variable is said to have probability density function 𝑓𝑋 (𝑥) = 𝛼𝑘 𝛼 , 𝑥 𝛼+1 𝛼, 𝑘 > 0 and 𝑥 > 𝑘. Answer: Step 1 𝑘̂ > 𝑥 if and only if [𝑋1 > 𝑥& ⋯ &𝑋𝑛 > 𝑥] and the probability of that is ∞ 𝛼𝑘 𝛼 Pr(𝑋1 > 𝑥) = ∫𝑥 𝑢𝛼+1 (Pr( X 1  x))n 𝑘 𝛼 , 𝑑𝑢 = ( ) , 𝑥 so 𝑛𝛼 𝑘 Pr(𝑘̂ > 𝑥) = ( ) . 𝑥 Thus, 𝑘 𝑛𝛼 Pr(𝑥1 < 𝑘̂ < 𝑥2 ) = ( ) 𝑥1 𝑘 𝑛𝛼 −( ) . 𝑥2 Pr(𝐴𝑘 < 𝑘̂ < 𝐵𝑘) = 𝐴−𝑛𝛼 − 𝐵 −𝑛𝛼 . Pr (𝐴 < 𝑘̂ < 𝐵) = ⋯ 𝑘 1 𝑘 1 Pr ( < < ) = ⋯ 𝐵 𝑘̂ 𝐴 ̂ 𝑘 ̂ 𝑘 𝐵 𝐴 Pr ( < 𝑘 < ) = ⋯ This gives you a confidence interval for k if 𝛼 is known. Since 𝛼 is not known, there is more work to do. (You have to choose A and B to get you the probablities that you want.) Question #20 Confidence interval for population proportion You have a population of 10M people. Suppose you want to determine the share of people with a certain disease. To do so you randomly choose X people and check if they have the disease. You determine that a share P of those people have the disease. How do I compute the 75% / 90% / 95% confidence intervals for the disease prevalence for the whole population? e.g. find y so that there is a 95% chance the disease rate in the whole population is in the range [𝑋 − 𝑦, 𝑋 + 𝑦]. Answer: Suppose n people are chosen at random from a population with N members, where 𝑛/𝑁 < 0.1 . If X of the n have a certain disease, then 𝑝̂ = 𝑋/𝑛 is an estimate of the proportion p of people in the population who have the disease. Wald. Then a traditional Wald 95% confidence interval (CI) for p is 𝑝̂ ± 1.96√𝑝̂ (1 − 𝑝̂ )/𝑛. The 'probability factor' is changed from 1.960 (to 1.645 or 1.150) to get confidence levels 90% or 75%, respectively. Agresti. When the confidence level is 95% and n is small (some say less than 100), then it is better to use the Agresti (CI): Let 𝑛̃ = 𝑛 + 4 and 𝑝̃ = (𝑋 + 2)/𝑛̃ to get the CI 𝑝̃ ± 1.96√𝑝̃(1 − 𝑝̃)/𝑛̃ . This form of the Agrest CI is intended for use only at the 95% confidence level. Wilson. In general, a slightly more accurate form of CI than either of these is the 'Wilson interval for a binomial proportion', which has a somewhat more complicated formula, I refer you to the Wikipedia article for that.