UNIVERSITY OF TORONTO SCARBOROUGH Department of Computer and Mathematical Sciences FINAL EXAMINATION, APRIL 2013

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1 UNIVERSITY OF TORONTO SCARBOROUGH Department of Computer and Mathematical Sciences FINAL EXAMINATION, APRIL 2013 STAB57H3 Introduction to Statistics Duration: 3 hours Last Name: First Name: Student number: Aids allowed: - The textbook (Probability and Statistics, Evans, M.J and Rosenthal, J. S) - Class notes and any notes from tutorials - A calculator (No phone calculators are allowed) No other aids are allowed. For example you are not allowed to have any other textbook or past exams. Standard Normal, t, and chi-squared distribution tables are attached at the end. All your work must be presented clearly in order to get credit. Answer alone (even though correct) will only qualify for ZERO credit. Please show your work in the space provided; you may use the back of the pages, if necessary but you MUST remain organized. Show your work and answer in the space provided. There are 14 pages including this page and statistical tables. Please check to see you have all the pages. Good luck!! Question: Total Points: Score:

2 Page 2 of Let X 1, X 2,..., X n be a random sample (i.e. i.i.d.) form the Uniform[ θ, θ] model where θ > 0 is an unknown parameter and T n = 3 n n i=1 X2 i. (a) (6 points) Is T n an unbiased estimator of θ 2? Prove your answer. Solution: E(X 1 ) = 0 and V (X 1 ) = (2θ)2 = θ2 and so 12 3 E(X2 1) = V (X 1 ) = θ2 3 E(T n ) = 3 n n i=1 E(X2 i ) = 3 n θ2 = θ 2. Since E(T n 3 n ) = θ 2, T n is an unbiased estimator of θ 2 (b) (6 points) Is T n consistent in probability for θ 2? Prove your answer. Solution: V (X 2 1) = E(X 4 1) (E(X 2 1)) 2. E(X 4 1) = θ θ x4 1 2θ dx = 1 2θ 1 2θ 5 = θ4. V 2θ 5 5 (X2 1) = E(X1) 4 (E(X1)) 2 2 = θ4 ( θ2 5 3 )2 = 4θ4 V (T n ) = ( 3 n )2 n 4θ4 0 as n. = 4θ2 45 5n 45. x 5 θ = 5 θ Since E(T n ) = θ 2 and V (T n ) 0 as n, T n is consistent in probability for θ 2

3 Page 3 of In a study a researcher took a random sample of 41 butterflies capable of flight early in the morning and measured their resting inactivity period (RIP). It has been theorized that butterflies that fly early in the morning would have a shorter mean RIP. The mean RIP for all butterflies is believed to be 133 seconds. Assume that RIP has a N(µ, σ 2 ) distribution, where both µ and σ > 0 are unknown parameters. The R code below is intended to assess the null hypothesis H 0 : µ 133 using this data set. Assume that the data satisfies the assumptions necessary for the test involved. > x=scan("c:/users/mahinda/desktop/rip.txt") Read 41 items > t.test(x, alternative="less", mu=133) One Sample t-test data: x t = , df = 40, p-value = alternative hypothesis: true mean is less than 133 sample estimates: mean of x (a) (3 points) What do you conclude based on the information in this output? Solution: p-value= < 0.05 and so we reject the null hypothesis. i.e At the 5 percent level of significance, the data provide sufficient evidence of a shorter mean RIP. (b) (7 points) Calculate the value of the t-test statistic for assessing the null hypothesis H 0 : µ 130. Note 1: This hypothesis is not exactly the same as the one tested in the R output above, but the information on this output is sufficient to calculate the value of the t-test statistic for assessing the null hypothesis H 0 : µ 130. Note 2: In this part you don t have to calculate the p-value. Solution: t = s = and so s n n = The required t-value = s n = 1.19

4 Page 4 of (5 points) Let X 1, X 2,..., X n be a random sample (i.e. i.i.d.) form a distribution with p.d.f: { θx 2 if 0 < θ x <, f θ (x) = 0 otherwise. Find the MLE of θ. Solution: L(θ) = n θ i=1 x i I(θ x i ) = θn I(θ x n (1)). This is an increasing i=1 x i function of θ and has the maximum value at θ = x (1), i.e. x (1) is the MLE of θ. 4. Let X 1, X 2,..., X n be a random sample (i.e. i.i.d.) form a distribution with p.d.f: { kθ 5 x 4 e θx if x > 0, f θ (x) = 0 otherwise. where k > 0 is a constant. (a) (5 points) Find the MLE of θ. Solution: L(θ) = n i=1 f θ(x i ) = k 5 θ 5 ( x 4 i ) e θn x ln L = const + 5n ln θ n xθ d ln L dθ = 5n θ set n x = 0 = ˆθ = 5 x (b) (5 points) Determine a minimal sufficient statistic for the model. Explain clearly why your sufficient statistic is minimal sufficient. Solution: L(θ) = n i=1 f θ(x i ) = θ x i I(θ x i ) = θ 5 ( x 4 i ) e θn x = g θ (T (S))h(S) where T (S) = X and so X is a sufficient statistics. This is a function of the MLE and so minimal sufficient. (c) (6 points) Find the constant c (in terms of n) such that c X 2 is an unbiased estimator of ψ(θ) = 1 θ 2 Solution: Note this distribution is Gamma(5, θ) and so E( X) = 5 and V ( X) = θ 5 and E( X 2 ) = (E( X)) 2 + V ( X) = = 5(5n+1) and so c = n nθ 2 θ 2 nθ 2 nθ 2 5(5n+1)

5 Page 5 of (10 points) The following table summarizes the waiting times, in minutes, of a random sample of 200 customers at a bank service counter. Waiting time (x) (0, 0.5] (0.5, 1.0] (1.0, 1.5] (1.5, 2.5] Number of customers Test whether the waiting time (X) can be modeled by a distribution with p.d.f. { x if 0 x < 2.5, f(x) = 0 otherwise. Chapter 5 Goodness of Fit Tests Solution a PX<a ( )= ( x) dx = 0.16a( 5 a), 0 so that PX<2.5 ( )=1. 00 P ( 0.0 X < 0.5 )=0.36 PX<1. ( 5)=0.84 PX<1. ( 0)=0.64 PX<0.5 ( )=0.36 P ( 0.5 X < 1. 0 )=0.28 P ( 1. 0 X < 1. 5 )=0.20 P ( 1. 5 X < 2.5 )=0.16 Using E i = (200 probability) gives the following table of calculations. x O i = f i E i ( O i E i ) ( O i E i ) ( 2 O i E i ) 2 E i H 0 : suggested model is appropriate H 1 : suggested model is not appropriate Significance level, α = 0.10 (say) Degrees of freedom, v = 4 1 = 3 (4 classes; 1 constraint: E i = ) Critical region is χ 2 > O i 0 Accept H 0 Critical region reject H % χ 2 ( ) 2 O Test statistic is χ 2 = i E i =1.758 E i This value does not lie in the critical region. There is no evidence, at the 10% significance level, to suggest that waiting times cannot be modelled by the suggested probability density function. Solution: 104

6 Page 6 of Suppose that s = {1, 1, 4, 1, 2} is an observed sample from a Poisson(λ) distribution, where the prior distribution for λ has p.d.f. { k(1 + λ)e λ if λ > 0 π(λ) = 0 otherwise. where k > 0 is a constant. (a) (7 points) Find the posterior distribution of λ. (b) (3 points) Find the posterior mean of λ. Solution: f λ (s) = n i=1 f(x i) e nλ λ n x π(λ s) = k 1 π(λ)f λ (s) = k 2 (1 + λ)e λ e nλ λ n x k 2 (1 + 0 λ)e λ e nλ λ n x dλ = 1 k 2 ( λ n x e (n+1)λ dλ + λ n x+1 e (n+1)λ dλ) = i.e. k 2 [ Γ(n x+1) (n+1) (n x+1) + i.e. k 2 [ Γ(10) Γ(11) 6 11 ] = 1 Γ(n x+2) (n+1) (n x+2) ] = 1 i.e. k 2 = [ ] 9! + 10! i.e. π(λ s) = [ ] 9! + 10! (1 + λ)e (n+1)λ λ n x 11 E(λ s) = 0 λπ(λ s)dλ = 0 λ [ 9! ! = [ 9! ! 6 11 ] 1 [ 10! ! ] 10 6 = ] 1 6 (1 + λ)e (n+1)λ λ n x dλ 11

7 7. Let S = {0.1, 0.5, 0.2} be an observed sample from a distribution with p.d.f: f(x) = { 2x θ 2 The prior distribution of θ is Uniform[0, 1]. if 0 x θ 0 otherwise. (a) (10 points) Determine a 0.95 HPD interval for θ. Page 7 of 14 Solution: π(θ S) π(θ)f(x) = 1 I θ 2n [x(n),1](θ). This is a decreasing function of θ and so the 0.95 HPD interval is of the form (x (n), u) where. u x (n) 1 x (n) 1 dθ θ 2n 1 dθ = 0.95 θ 2n x 2n+1 (n) u 2n+1 = 0.95 (x 2n+1 (n) 1) For the observed sample, n = 3, and x (n) = 0.5 and u = (b) (5 points) Assess the hypothesis H 0 : θ 0.7. Solution: p-value = Π(θ 0.1 S) = x (n) 1 x (n) dθ θ 2n 1 dθ. θ 2n Calculate this with n = 3 and x (n) = 0.5 and reject H 0 if the p-value is small (smaller than 0.05)

8 Page 8 of (10 points) Let s = {3.690, , 1.989, 0.047, 8.114, 4.996, , 6.975} be an observed sample of size n = 8 from an exponential(λ) distribution. Use this observed sample to calculate a 90% confidence interval for λ. Here are some helpful hints: 1) The mean of this sample is ) Exponential(λ) distribution has p.d.f. f(x) = λe λx for x > 0 and zero otherwise. 3) If (X 1, X 2,..., X n ) i.i.d. Exponential(λ), then T = n i=1 X i Gamma(n, λ). What is the distribution of 2λT? 4) Extended Chi-Squared tables are attached. Solution: Question 8 continues on the next page...

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10 Page 10 of 14 This part consists of multiple choice questions. Just circle your answer. You don t have to show work in this part. 9. (3 points) The test of the hull hypothesis H 0 : µ = 30 against the alternative H a : µ 30 has p-value of What can you say about the confidence intervals for µ calculated using the same sample? A) 30 is inside the 95% confidence interval. B) 30 is inside the 99% confidence interval. C) 30 is inside the 90% confidence interval. D) 30 is outside the 98% confidence interval. E) There is no connection between the test and the confidence interval. 10. (3 points) It is widely thought that there is a high incidence of disability among the homeless population. A random sample of 110 homeless people contained 84 who were disabled on one or more categories (such as psychiatric disability, medical disability etc.). Let p denote the proportion of all homeless people having one or more types of disability. Which of the following numbers is the closest to the value of the test Z-statistic for testing the null hypothesis H 0 : p = 0.75 against H a : p 0.75? A) 0.0 B) 0.3 C) 0.6 D) 0.9 E) 1.2 Solution: p = 84 = p 0.75 = , = (p.75) = (( )/110) 11. (3 points) On the basis of a sample of size 11 from a Normal population, one finds that a 95% confidence interval (using the t-distribution) for the population mean is (78, 112). The width of this interval is = 34. Using the same sample, what is the width of the 99% confidence interval for? A) 51.9 B) 48.4 C) 44.7 D) 55.6 E) 39.3

11 Page 11 of 14 Solution: 34/2.228 = ANS*3.169 = t value for 95% CI = t-value for 99 CI = (3 points) The scores in an examination has a Normal distribution with standard deviation σ = 100. A report says that based on a simple random sample of of size 100, the confidence interval for the population mean score is (486.24, ). What was the confidence level used to calculate this confidence interval? A) 80% B) 90% C) 95% D) 99% E) 99.9% Solution: Margin of error = = SE( X) = σ n = = 10. z = ME = = and z = corresponds to significance level 99%. SE( X) 10 END OF EXAM

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14 Page 14 of 14 Chi-Squared Distribution Quantiles df Chi-Squared Distribution Quantiles df