Title ST104b - Statistics 2 - 2010 Examiners Commentaries - Zone-A.pdf ✅

04b Statistics 2 Comment: It is very important that you know the definition of conditional probability; you might want to try learning activity 2.8. iv. False. Suppose that Cov(X, Y ) = −2 and Var(X) = Var(Y ) = 1. Then we find ρ = Corr (X, Y ) = Cov(X, Y ) √ Var(X) Var(Y ) = −2, which is impossible as ∣ Corr (X, Y )∣ ≤ 1. Comment: To get full marks for this question the candidate needed to mention that the correlation coefficient always lies between −1 and 1. See the discussion of the correlation coefficient at the top of p.51. (b) Let μij be the expected value of cell (i, j). The difference μij1−μij2 is the same for all i. Also the difference μi1j−μi2j is the same for all j. Comment: Many candidates struggled with this question. The model for two-way ANOVA is μij = μ = αi + βj as given in Section 9.5 on p.105. This means that if you look at the difference between cell means in a certain row between two columns, than this does not depend on the row you look at. Similarly, if you look at the difference between cell means in a certain column between two rows, this does not depend on the column you pick. Example 63 gives a worked out application of two way analysis of variance of three varieties of potatoes on four locations. (c) Since for random variables X and Y and constants a and b it holds that Var(aX + bY ) = a 2 Var(X) + 2ab Cov(X, Y ) + b 2 Var(Y ) we find that (due to independence) the variance of kX1 + X2 is equal to 4 ( k 2 + 1) and we find that kX1 + X2 is normally distributed with mean zero and variance 4 ( k 2 + 1) . It follows that we should find k such that P(kX1 + X2 ≤ 5) = Φ ( 5 √ 4 (k 2 + 1)) = 0.8413, where Φ denotes the cumulative distribution function of a standard normal random variable. From the statistical tables we read Φ(1) = 0.8413. This leads to the equation 5 2 √ k 2 + 1 = 1 which is solved by k = ±2.29. Comment: Equation (5.4) on p.59 tells us that the sum of two independent normally distributed random variables is again normally distributed with new mean given by the sum of the two means and the new variance given by the sum of the two variances. It is important to know the formula for variance of a sum of random variables and that independence of X and Y implies that Cov(X, Y ) = 0. It is also important to know how to use statistical tables. (d) Since U = 1 can occur only when either X or Y is equal to 1 but neither is equal to 2, the relevant cells are the ones with values (0, 1), (1, 0) and (1, 1) that have probabilities 0.05, 0.1 and 0.1, respectively So P (U = 1) = 0.05 + 0.1 + 0.1 = 0.25 and P (X = 0, Y = 1∣U = 1) = 0.05 0.25 = 1 5 , P (X = 1, Y = 0∣U = 1) = 0.10 0.25 = 2 5 , P (X = 1, Y = 1∣U = 1) = 0.10 0.25 = 2 5 . We deduce that P (X = 1∣U = 1) = 4 5 and P (X = 0∣U = 1) = 1 5 2

04b Statistics 2 The probability that the dish is available and does not order it is 1 2 × 1 2 = 1 4 . So the probability that he does not order it is 1 2 + 1 4 = 3 4 and hence the probability that the dish was not available given he did not order it is. 1 2 3 4 = 2 3 . Comment: Candidates should know the definition of conditional probability. Note that if you are asked to calculate a probability and you find a number greater than 1, you have made a mistake! See remark 4 on p.10. The answer can be given numerically or as a fraction. (b) Since it is given that availability of goat stew is independent of days, we find that the probability that goat stew was unavailable on the last three occasions given that the customer did not order this dish on the last three occasions is equal to (2/3)3 = 8/27. Alternatively, we can also work through the following more lengthy calculations to arrive at the same result: The probability that the goat was not available on all occasions and he did not order it is 1 2 × 1 × 1 2 × 1 × 1 2 × 1 = 1 8 . The probability that goat was available exactly once and he did not order it is 3 × 1 2 × 1 2 × 1 2 × 1 × 1 2 × 1 = 3 16 . Note the factor 3 here is due to the fact that that the availability of the goat dish could have been on either the first occasion, the second occasion or the final one. The probability that the goat was available exactly twice and he did not order it is 3 × 1 2 × 1 2 × 1 2 × 1 2 × 1 2 × 1 = 3 32 The probability that the goat was available on all three occasions and he did not order it is 1 2 × 1 2 × 1 2 × 1 2 × 1 2 × 1 2 = 1 64 Hence, the probability he did not order it is 1 8 + 3 16 + 3 32 + 1 64 = 27 64 and so the probability it was never available given he did not order it is 1 8 27 64 = 8 27 . Comment: Even though the calculations can be a bit lengthy, the method for finding the answer to such questions is always the same and candidates are recommended to practice similar exercises such as example 9 on p.17 and learning activity 2.9 on p.15. (c) From above we have P (Y = 3∣X = 3) = 1 8 27 64 = 8 27 P (Y = 2∣X = 3) = 3 16 27 64 = 4 9 4