Tiêu đề ST104b - Statistics 2 - 2011 Examiners Commentaries - Zone-B.pdf ✅

04b Statistics 2 (b) The random variables X1 and X2 are each normally distributed with mean 1 and variance 1. Their correlation coefficient is 0.25. Find P(2X1 > X2). The probability can be rewritten as P (2X1 − X2 > 0). 2X1 − X2 is normally distributed with mean 2 × 1 − 1 = 1 and variance 2 2 × 1 + 12 × 1 − 2 × 2 × 1 × 0.25 = 4. So P (2X1 − 2X2 > 0) = 1 − Φ 0 − 1 √ 4 = Φ (0.5) = 0.6915. Comment: To find the mean of 2X1 − X2 we use formulae (4.12) and (4.14) on p.69. Its variance we deduce from formula (5.15) for Var(aX + bY ) on p.94. The result then follows by standardising, similarly to Activity 3.13 on p.56. See Sample examination question 1 on p.98. It is important to know how to use statistical tables. (c) The random variables εi, i = 1, 2, 3, are independent and normally distributed with mean 0 and variance 1 and α is an unknown parameter. Suppose that you are given observations y1 and y2 such that y1 = α + 1, y2 = −2α + 2. Find the least squares estimator αˆ, verify it is unbiased and calculate its variance. We need to find ˆα that minimises (y1 − α) 2 + (y2 + 2α) 2 . Differentiating with respect to α, we have −2 (y1 − αˆ) + 2 × 2 (y2 + 2ˆα) = 0 and therefore αˆ = y1 − 2y2 5 . To verify it is unbiased, E (ˆα) = E (y1) − 2E (y2) 5 = α − 2 (−2α) 5 = α. Also Var (ˆα) = 1 5 2 + 2 5 2 = 1 5 . Comment: This question requires the application of the ‘least squares’ technique (Chapter 11). This is an optimisation problem, hence we need to differentiate with respect to the unknown parameter (α) and equate the derivative to zero to obtain the minimum of the sum of the ε 2 i s. Note that the parameter α is unknown and hence this quantity may not appear in the expression for the estimator ˆα. Recall that ˆθ is an unbiased estimator of θ if E(ˆθ) = θ, see p.116. The variance of ˆα is calculated using the formula for Var(aX + bY ) on p.94, taking into account that ε1 and ε2 are independent and standard normally distributed. Candidates were given partial marks if they indicated how to find ˆα (but, for example, a calculation error was made; a common mistake was for candidates to omit the factor 2 in the derivative of (y2 + 2α) 2 ) or by giving the definition of an unbiased estimator. 2

04b Statistics 2 Here is the ANOVA table: Source Degrees of Freedom Sum of Squares Mean Square F-statistic Between samples 2 82.5 41.25 4.452 Within Samples 17 157.5 9.265 Total 19 240 As 4.452 > 3.59 which is the 5% point of the F2,17 distribution, we see that there is evidence that the means are not equal. Comment: This is straightforward one-way ANOVA. A worked-out example is Activity 10.1 on pp.171–3. Partial credit was given for intermediate results. For example, if the correct conclusion was made based on an incorrect F-statistic, candidates still received some marks. It is important to use the right distribution with the correct degrees of freedom. SECTION B Answer all three questions in this section (60 marks in total). Question 2 Let X be a random variable with a binomial distribution with parameters 5 and θ; let also Y be a random variable with a binomial distribution with parameters 4 and θ. X and Y are independent. (a) Consider the estimators X 5 , Y 4 and X+Y 9 . Show that they are all unbiased estimators for θ. It holds that E X 5 = 1 5 5θ = θ, E Y 4 = 1 4 4θ = θ, E X + Y 9 = 5θ + 4θ 9 = θ. They are all unbiased. Comment: Binomial random variables are introduced on p.36. They have mean nθ as shown on pp.73–4. (b) Which of the three estimators would you choose and why? We will calculate the variances of the three estimators. Var X 5 = 1 5 2 5θ (1 − θ) = θ (1 − θ) 5 . Var Y 4 = 1 4 2 4θ (1 − θ) = θ (1 − θ) 4 . Var X + Y 9 = 5θ (1 − θ) + 4θ (1 − θ) 9 2 = θ (1 − θ) 9 . The third one is the best, since it has the smallest variance. Comment: In Chapter 7.7 the mean squared error (MSE) is discussed. For an estimator we prefer a small MSE. According to formula (7.4) on p.119 this means that for unbiased estimators we prefer the one with the smallest variance. Candidates who wrote the latter statement received partial credit. The variance of a Bin(n, θ) distributed random variable is given by nθ(1 − θ). This fact is included in the list of formulae given at the end of the examination paper. Note that to find the variance of X + Y we use that they are assumed to be independent. 4