Some Assorted Formulae. Some confidence intervals: σ n. x ± z α/2. x ± t n 1;α/2 n. ˆp(1 ˆp) ˆp ± z α/2 n. χ 2 n 1;1 α/2. n 1;α/2

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1 STA 248 H1S MIDTERM TEST February 26, 2008 SURNAME: SOLUTIONS GIVEN NAME: STUDENT NUMBER: INSTRUCTIONS: Time: 1 hour and 50 minutes Aids allowed: calculator Tables of the standard normal, t and chi-square distributions are on the last 3 pages (pages 10, 11, 12). Total points: 50 Some Assorted Formulae If X Bin(n, p), E(X) = np and Var(X) = np(1 p). Some confidence intervals: x ± z α/2 σ n s x ± t n 1;α/2 n ˆp(1 ˆp) ˆp ± z α/2 n ( (n 1)s 2 (n 1)s 2 ) χ 2, n 1;α/2 χ 2 n 1;1 α/2 1 2abcd 2ef 2g

2 1. A performance analyst measures the sizes of messages sent on two different computer networks (network A and network B). We are interested in comparing the distribution of message size between the two networks. Below are boxplots and empirical distribution functions for the message sizes for each network. The corresponding network is labelled for the empirical distribution functions but not for the boxplots. ECDF for network A ECDF for network B Fn(x) Fn(x) x x 1 2 (a) (1 mark) Indicate on the empirical cumulative distribution function for network A how to estimate the first quartile of message size for this network. First quartile is the value on the horizontal axis that corresponds to 0.25 on the vertical axis. (b) (2 marks) Which boxplot corresponds to which network? How do you know? The first boxplot is for network A. The quartiles from the e.c.d.f. are further apart in the plot for network B. (c) (2 marks) For each of the following indicate whether the statistic will be approximately equal or not for the two networks. If it is not approximately equal, indicate for which network it will be larger: i. the mean approximately equal ii. the standard deviation greater for network B 2

3 2. This question involves the Poisson distribution. Here are some facts about it. Let X be a random variable with a Poisson distribution with parameter µ. X has probability mass function p(x) = e µ µ x for x = 0, 1, 2,... x! and E(X) = µ and Var(X) = µ. Suppose X 1,..., X n are independent and identically distributed random variables Poisson random variables with parameter µ. Then Y = n i=1 X i has a Poisson distribution with parameter nµ. Let X = 1 n ni=1 X i. (a) (1 mark) What is E(X)? µ (b) (1 mark) What is Var(X)? µ n (c) (4 marks) Show that the maximum likelihood estimator of µ is X. For observed data x 1,..., x n the likelihood function is e nµ µ xi xi! The log-likelihood function is Differentiate with respect to µ Setting this equal to 0 gives So the maximum likelihood estimator is X. L(µ) = nµ + x i log(µ) + log(x i ) L µ = n + xi µ ˆµ = x (d) (1 mark) Is your estimator in part (c) unbiased? Why or why not? Yes from the answer to part (a). 3

4 n i=1 (X i X) 2 (Question 2 continued.) (e) (2 marks) Let S 2 = n 1. Is S 2 an unbiased estimator of µ? Justify your answer. You may assume that any results derived in lecture are known. Yes. µ. From lecture, S 2 is an unbiased estimator of the variance and the variance is (f) (4 marks) For S 2 as defined in part (e), Var(S 2 ) = µ (2µn + n 1) n(n 1) Do you prefer the estimator of µ from part (c) or part (e)? Consider at least 3 criteria. Both are unbiased from parts (d) and (e). Both are consistent by the Law of Large Numbers. Consider variance: Var(S 2 ) = µ n So X has smaller variance. So prefer X. ( 2µn n ) > µ since the quantity in brackets is greater than 1 n = Var(X) 4

5 (Question 2 continued.) (g) Below are 2 histograms for randomly generated Poisson random variables. For the first µ = 2 and for the second µ = 200. Poisson observations with mu=2 Poisson observations with mu=200 Frequency Frequency rpois(1000, 2) rpois(1000, 200) i. (2 marks) Describe the shapes of the histograms relative to each other. For µ = 2, the distribution is right-skewed. For µ = 200, the distribution is much more symmetric. ii. (2 marks) What principle is illustrated by the 2 histograms above for the 2 different values of µ? Explain. The Central Limit Theorem. Each observation from the Poisson distribution with µ = 200 can be considered as the sum of 100 observations from the Poisson distribution with µ = 2 and sums of random variables are approximately normally distributed by the CLT. 5

6 3. Consider the probability distribution with density function f(x) = θx θ 1 for 0 < x < 1 and 0 otherwise. θ The mean for a random variable with this distribution is θ+1 and variance is θ (θ+1) 2 (θ+2). Suppose we have a random sample of size n from this distribution. (a) (2 marks) Find the method of moments estimator of θ. Equate moments: gives X = ˆθ = ˆθ ˆθ + 1 X 1 X (b) (4 marks) Describe how you would use the bootstrap to estimate the standard error of the estimator in part (a). For observed data x 1,..., x n draw B samples of size n either by x - estimating θ as 1 x and sample from the above density using this estimate for θ (parametric bootstrap) or - sample with replacement from x 1,..., x n (non-parametric bootstrap). For each sample, calculate ˆθ i (i = 1,..., B) using the formula in (a). The bootstrap estimate of the standard error is the standard deviation of these B estimates of θ. 6

7 4. The summary statistics in the R output below are calculated from salaries obtained from a random sample of 65 computer programmers. > length(salaries) [1] 65 > summary(salaries) Min. 1st Qu. Median Mean 3rd Qu. Max > sd(salaries) [1] (a) (3 marks) Assuming that the salaries are normally distributed, find a 90% confidence interval for the mean salary. t.05,64 = ( ) Confidence interval: ± (b) (3 marks) Assuming that the salaries are normally distributed, find a 95% confidence interval for the variance. χ ,64 = χ 2.975,30 = 16.8,. χ2.025,64 = χ 2 (.025,30 ) = 47.0 Confidence interval: 64(25382) , 64(25382) (c) (3 marks) Based on the summary statistics, what can you say about the shape of the distribution of salaries? How does this affect the accuracy of your answers to parts (a) and (b)? Explain fully. The distribution is right-skewed rather than symmetric like a normal distribution. We can tell this because the mean is greater than the median (and also (maximum - median) > (median - minimum) and (3rd quartile - median) > (median - 1st quartile)). The confidence interval in part (a) is still approximately accurate since confidence intervals for the mean are robust against non-normality but the confidence interval in (b) is not accurate as it relies on having normally distributed data. 7

8 5. Suppose that a factory produces bottles of apple juice. Assume that the volumes of juice in the bottles can be considered independent and identically distributed random observations from a Normal distribution. The label on each bottle states that it contains 500 ml of juice. Suppose that a random sample of 50 bottles of juice is selected from the production and the volume of juice in each is measured. Based on these 50 observations, the resulting 95% confidence interval for the mean volume is (491, 497) ml. (a) (2 marks) Would a 90% confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain. Narrower. It is less likely to capture the true mean. (b) (2 marks) Consider the following statement: We can be highly confident that 95% of all bottles of this type of apple juice have a volume of juice that is between 491 and 497 ml. Is this statement correct? Why or why not? No. The confidence interval is for the mean not for measurements for individual bottles. (c) (2 marks) Suppose that the label on the bottles purports to reflect the mean of the volume of juice in the bottles produced in this factory. Based on the observed confidence interval, is the label correct in stating that the mean is 500 ml? Explain. It might be correct. 5% of all samples result in 95% confidence intervals that miss the mean and this might be one of those. (d) (3 marks) Suppose that legally the bottling machines must be adjusted so that the standard deviation of the amount of juice in the bottles is at most 2 ml. Does it appear that this criterion is being met? Why or why not? t.025,49 = ( ) The half-width of the confidence interval is 3 = s 50 which gives s = Since this is so much larger than 2 it appears that the criterion is not being met. 8

9 6. Let X 1, X 2,..., X n be a random sample from a uniform distribution on the interval [0, θ] so that { 1 f(x) = θ 0 x θ 0 otherwise Then if Y = max(x i ), U = Y/θ has density function We are interested in estimating θ. f U (u) = { nu n 1 0 u 1 0 otherwise (a) (2 marks) Explain why Y θ is a good pivot. It is a function of the data (the maximum observation) and the unknown parameter (θ) and its probability distribution does not depend on any unknown parameters. (b) (2 marks) Using f U (u) it can be shown that P [(α/2) 1/n < Y ] θ (1 α/2)1/n = 1 α Use this to derive a 99% confidence interval for θ. For a 99% confidence interval, α =.01 so we have P ((.005) 1/n Y ) θ (.995)1/n = 1 α Rearranging what s inside the brackets for θ gives Y (.995) 1/n θ Y (.005) 1/n so the confidence interval for data x 1,..., x n is ( max(x1,..., x n ) (.995) 1/n, max(x ) 1,..., x n ) (.005) 1/n 9