stab57 midterm

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Department
Statistics
Course
STAB57H3
Professor
Mahinda Samarakoon
Semester
Winter

Description
UNIVERSITY OF TORONTO SCARBOROUGH Department of Computer and Mathematical Sciences Midterm Test, March 2012 STAB57H3 Introduction to Statistics Duration: One hour and ▯fty minutes 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. 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, in ink. Pencil may be used, but then any re-grading will NOT be allowed. There are 11 pages including this page. Please check to see you have all the pages. Circle your tutorial: TUT001(Mon 11:00-12:00) TUT002 (Mon 10:00-11:00) TUT003 (Thu 9:00-10:00) Good Luck! Question: 1 2 3 4 5 6 7 8 9 10 Total Points: 10 5 10 6 11 6 8 11 4 9 80 Score: Page 2 of 11 1. Suppose that a reponse X follows the following probability distribution. x 1 2 3 4 P(X = x) 0.2 0.4 0.3 0.1 (a) (3 points) If we use the expected value of the response as a predictor, then what is the prediction of a future response X? Solution: predicted value =1 ▯ 0:2 + 2 ▯ 0:4 + 3 ▯ 0:3 + 4 ▯ 0:1 = 2:3 (b) (4 points) Calculate the mean-squared error of the prediction in part (a) above. 2 2 2 Solution: MSE = E(X ▯ 2:3) = (1 ▯ 2:3) ▯ 0:2 + (2 ▯ 2:3) ▯ 0:4 + (2 ▯ 2:3) ▯ 0:3 + (4 ▯ 2:3) ▯ 0:1 = 0:81 (c) (3 points) The mode of the distribution of the response X is 2. If we use the mode of the response X as a predictor of X, then what is the mean-squared error of this prediction? Solution: MSE = E(X ▯ 2) = (1 ▯ 2) ▯ 0:2 + (2 ▯ 2) ▯ 0:4 + (2 ▯ 2) ▯ 2 2 0:3 + (4 ▯ 2) ▯ 0:1 = 0:9 Page 3 of 11 2. (5 points) Suppose that the life length (X, in years) of a machine is known to have an Exponential(2) distribution. Determine the shortest interval that contains the value of a future X with probability 0.99. [Recall that the pdf of exponential(▯ ) distribution is: f(x) = ▯e ▯▯x for x ▯ 0 and zero otherwise.] Solution: This is very similar to example 5.2.1 p 244 text. Ru ▯2x 0 2e dx = 0:99 i.e. 1 ▯ e2u = 0:99 ln0:01 u = ▯2 = 2:3 The shortest interval is (0;2:3) Page 4 of 11 3. Suppose that we have a population ▯ = f▯ ;▯ ;▯ ;▯ ;▯ g and a quantitative measure- 1 2 3 4 5 ment X given by: i 1 2 3 4 5 X(▯ )i 1.5 1.0 2.0 1.5 3.5 (a) (4 points) Calculate F X1:7) and f (X:5). 3 2 Solution: F (X:7) = 5 and f X1:5) = .5 (b) (3 points) Suppose you are going to to select a simple random sample of size 2 (i.e. n = 2) from this population. Find the probability that f▯ ;▯ 2 w5ll become your simple random sample. Solution: 5 = 1▯2 = 2 = 0:1 (2) 5▯4 20 (c) (3 points) Suppose you are going to to select an i.i.d. sample (i.e. sampling with replacement) of size 2 (i.e. n = 2) from this population. Find the probability that f▯ 2▯ 5 will become your sample. 1 1 1 Solution: 5▯ 5 = 25 Page 5 of 11 4. The R code and the output below are intended to identify outliers in the grades of a statistics class (not STAB57). grades=scan("C:/Users/Admin/Desktop/grades.txt") a=quantile(grades, 0.25) b=quantile(grades, 0.75) a b c = grades < (a - 1.5 * (b -a)) d = grades > (b + 1.5 * (b -a)) e = sum(c) + sum(d) e R output > grades=scan("C:/Users/Admin/Desktop/grades.txt") Read 40 items > a=quantile(grades, 0.25) > b=quantile(grades, 0.75) > a 25% 60 > b 75% 77 > c = grades < (a - 1.5 * (b -a)) > d = grades > (b + 1.5 * (b -a)) > e = sum(c) + sum(d) > e [1] 3 (a) (2 points) Based on 1:5▯IQR rule, how many outliers are there in this data set? Solution: 3 (This is e in the given R output) (b) (4 points) One student in this class had 39 for this course (i.e. his grade in this course is 39). Test whether this grade is an outlier based on 1:5 ▯ IQR rule. Show all related calculations clearly. Solution: Q1 ▯ 1:5 ▯ IQR = a ▯ 1:5 ▯ (b ▯ a) = 34:5 < 39 and so 39 is NOT an outlier. Page 6 of 11 5. Let (X ;1 ) 2e a random sample (i.e i.i.d.) form the distribution with p.d.f. ( 1 ▯x=▯ ▯e if 0 < x < ▯, f▯(x) = 0 otherwise. where ▯ > 0 is unknown. Let T = (X + X )=51 2 2 Note: This distribution has mean ▯ and variance ▯ (a) (5 points) Show that T is a biased estimator of ▯ and compute its bias when used as an estimator of ▯. (Hint: Show that Bias (T) is a▯function of ▯ of the form Bias (T) = k▯ and calculate the value of k.) ▯ 2 Solution: EX = EX1= ▯ =) 2T = ▯ and so Bias (T5 = ET▯▯ = ▯0:6▯▯ T is biased since bias is not equal to 0. (b) (6 points) Calculate the mean squared error of T when it is used an estimator of 2 ▯. (Hint Show that MSE (T) is▯a function of ▯ of the form MSE (T) = c▯ and ▯ calculate the value of c.) 2 2 2 2 2
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