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

04b Statistics 2 iv. False. For example if it is a two-sided interval, it has a finite length and if it is a one-sided interval it has an infinite length. Comment: See the discussion concerning confidence intervals on p.134: a 95% confidence interval (Tl , Tu) covers the unknown parameter with a probability of 95% over repeated samples. Such an interval need not be symmetric about the unknown parameter and hence we can have intervals of different lengths. (b) The random variables X1 and X2 are each normally distributed with mean 4 and variance 2. Their correlation coefficient is 5 12 . Find P(3X1 > 2X2). The probability can be rewritten as P (3X1 − 2X2 > 0). 3X1 − 2X2 is normally distributed with mean 3 × 4 − 2 × 4 = 4 and variance 3 2 × 2 + 22 × 2 − 2 × 3 × 2 × 5 12 × 2 = 16. So P (3X1 − 2X2 > 0) = 1 − Φ 0 − 4 √ 16 = Φ (1) = 0.8413. Comment: To find the mean of 3X1 − 2X2 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 = 3α + 2. Find the least squares estimator αˆ, verify it is unbiased and calculate its variance. We need to find ˆα that minimises ε 2 1 + ε 2 2 = (y1 − α) 2 + (y2 − 3α) 2 . Differentiating with respect to α, we have −2 (y1 − αˆ) − 2 × 3 (y2 − 3ˆα) = 0 and therefore αˆ = y1 + 3y2 10 . To verify it is unbiased, E (ˆα) = E (y1) + 3E (y2) 10 = α + 3 × 3α 10 = α. Also Var (ˆα) = 1 102 + 3 102 = 1 10 . 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 2
Examiners’ commentaries 2011 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 3 in the derivative of (y2 − 3α) 2 ) or by giving the definition of an unbiased estimator. (d) Continuing from (c) suppose you observed y1 = 0.6 and y2 = 1.8. Suppose now y3 = α + ε3. Provide a 95% prediction interval for y3. The prediction interval would be 0.6 + 3 × 1.8 10 ± 1 √ 10 + 1 × 1.96 = 0.6 ± 2.58, which is (−1.98, 3.18). Comment: Many candidates did not answer this question correctly. In fact, it is a direct application of formula (12.9) on p.210. Note that 1 is added to the variance of ˆα as the confidence interval is longer since we are dealing with prediction here. We use the value z = 1.96 since P(Z < 1.96) = 0.975 (and thus P(−1.96 < Z < 1.96) = 0.95) where Z is a standard normally distributed random variable. (e) Arthur and Bernice throw one fair die each. Given that the score of Arthur is larger than that of Bernice, what is the probability he scored a 5 or a 6? The probability that they score the same is 6 36 = 1 6 . So the probability that the 1st player will score more is 1 − 1 6 2 = 5 12 . The probability that the first player scores a 5 or 6 and the other one scores lower (i.e. not a 12) is 1 6 × 5 6 + 1 6 × 4 6 = 1 4 . Hence, the required probability is 1 4 5 12 = 3 5 = 0.6. Comment: Full marks were awarded as well if the correct answer was found based on a table of all possible outcomes. (f) Three independent samples were taken. Sample A consists of 4 observations taken from a normal distribution with mean μA and variance σ 2 ; sample B consists of 6 observations taken from a normal distribution with mean μB and variance σ 2 ; sample C consists of 5 observations taken from a normal distribution with mean μC and variance σ 2 . The average value of the first sample was 24, the average value of the second sample was 20 and the average value of the third sample was 18. The sum of the squared observations (all of them) was 6722.4. Test the hypothesis H0 : μA = μB = μC against the alternative that this is not so. We will perform a one-way ANOVA. First we calculate the overall mean. This is 4 × 24 + 6 × 20 + 5 × 18 15 = 20.4. 3
04b Statistics 2 We can now calculate the sum of squares between procedures. This is 4 × (24 − 20.4)2 + 6 × (20 − 20.4)2 + 5 × (18 − 20.4)2 = 81.6. The total sum of squares is 6722.4 − 15 × 20.4 2 = 480. Here is the ANOVA table: Source Degrees of Freedom Sum of Squares Mean Square F-statistic Between samples 2 81.6 40.8 1.229 Within Samples 12 398.4 33.2 Total 14 480 As 1.229 < 3.89 which is the 5% point of the F2,12 distribution, we see that there is no 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. Note that no evidence that the means are not equal does not mean that there is evidence that the means are equal. 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 Poisson distribution with parameter 2λ; let also Y be a random variable with a Poisson distribution with parameter 4λ. X and Y are independent. (a) Consider the estimators X 2 , Y 4 and X+Y 6 . Show that they are all unbiased estimators for λ. It holds that E X 2 = 1 2 2λ = λ, E Y 4 = 1 4 4λ = λ, E X + Y 6 = 2λ + 4λ 6 = λ. They are all unbiased. Comment: Poisson random variables are introduced on p.44. They have mean λ as shown on p.78. (b) Which of the three estimators would you choose and why? We will calculate the variances of the three estimators. Var X 2 = 1 2 2 2λ = λ 2 . Var Y 4 = 1 4 2 4λ = λ 4 . Var X + Y 6 = 1 6 2 2λ + 1 6 2 4λ = λ 6 . 4