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

Examiners’ commentaries 2010 Examiners’ commentaries 2010 04b Statistics 2 The comments refer to the Statistics 2 subject guide. Important note This commentary reflects the examination and assessment arrangements for this unit in the academic year 2009–10. The format and structure of the examination may change in future years, and any such changes will be publicised on the virtual learning environment (VLE). Specific comments on questions – Zone B Question 1 (a) i. False. For example, toss three fair coins and let A be the event “1st coin is heads”, B be the event “2nd coin is heads” and C be the event “3rd coin is heads”. They are independent but their intersection is not empty (it is equal to throwing heads three times). Using the next question is also acceptable. Comment: A number of candidates thought “disjoint” and “independent” are the same concepts which they are not! Consider for example tossing a fair coin twice. Then the probability that both tosses are head is given by 1/4 > 0 (hence not disjoint), which is the probability that the first toss is head times the probability that the second toss is head (hence independent). See also example learning activity 2.7] ii. True. Otherwise P (A ∩ B ∩ C) = 0 ∕= P (A) P (B) P (C) since P (A) > 0, P (B) > 0 and P (C) > 0. Comment: It is very important to have a good understanding of the notion when events are independent, see the definition on p.12. iii. False. For example: throw two fair dice. Let A be the event “The first die is even”, B be the event “The second die is even” and C be the event “The total score is even”. A and B are independent since P(A ∩ B) = 1/4 = P(A)P(B), B and C are independent since P(B ∩ C) = 1/4 = P(B)P(C), and A and C are independent since P(A ∩ C) = 1/4 = P(A)P(C). Their intersection is not empty, since A ∩ B ∩ C = A ∩ B the event that both dice are even. Comment: Since this statement is false, any counterexample will do, and many other solutions could be given here. iv. False. Throw two fair dice. Let A be the event “The first die is even”, B be the event “The first die is even” and C be the event “The total score is odd”. A and B are independent, B and C are independent, and A and C are independent (see part iii.). However, the intersection of A, B and C is empty since throwing two even dice leads to an even total score. Comment: Again, there are many other possible counterexamples that can be given here. See Example 6 concerning hat, scarf and gloves. (b) Let Xij be the random variable representing the (i, j) th cell. The model is Xij = μ + αi + βj + εij 1
04b Statistics 2 where μ, αi , βj are parameters and εij are independent and identically normally distributed random variables. Comment: Many candidates struggled with this question. Two-way ANOVA means that if you look at the difference between cell means in a certain row between two columns, then 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. See Section 9.5. All definitions are important. (c) From the independence of {X1, X2, Y } it follows that the distribution of X2 1 + X2 2 + Y is chi-square with 1 + 1 + 5 = 7 degrees of freedom. From the table we read that the value at 5% level is given by k = 14.067. Comment: The first line in Section 7.3 states that the sum of the squares of ν independent standard normal random variables has a χ 2 ν distribution (ν degrees of freedom). Hence X2 1 + X2 2 ∼ χ 2 2 and thus X2 1 + X2 2 + Y has 7 degrees of freedom. (d) Since the sum of X and Y should be equal to one, the relevant cells are those with values (0, 2), (1, 1) and (2, 0), which have probabilities 0.1, 0.2 and 0.2, respectively. So P (X + Y = 2) = 0.1 + 0.2 + 0.2 = 0.5 and thus P (X = 0, Y = 2∣X + Y = 2) = 0.1 0.5 = 1 5 , P (X = 1, Y = 1∣X + Y = 2) = 0.2 0.5 = 2 5 , P (X = 2, Y = 0∣X + Y = 2) = 0.20 0.5 = 2 5 . We then have E (X∣X + Y = 2) = 1 × 2 5 + 2 × 2 5 = 6 5 , E (Y ∣X + Y = 2) = 2 ⋊ 1 5 + 1 ⋊ 2 5 = 4 5 and E (XY ∣X + Y = 2) = 1 × 1 × 2 5 = 2 5 , hence Cov (X, Y ∣X + Y = 2) = 2 5 − 6 5 4 5 = − 14 25 . Comment: You should think of objects of the form X∣X + Y = 2, Y ∣X + Y = 2 and XY ∣X + Y = 2 as new random variables (see for example learning activity 4.2). To find their expected value it is best to first write down their distribution function (see above). (e) We find E (X) = ∫ ∞ 1 x 4 x 5 dx = 4 ∫ ∞ 1 1 x 4 dx = 4 3 , E ( X2 ) = ∫ ∞ 1 x 2 4 x 5 dx = 4 ∫ ∞ 1 1 x 3 dx = 4 2 = 2 and E ( X2X ) = E ( X3 ) = ∫ ∞ 1 x 3 4 x 5 dx = 4 ∫ ∞ 1 1 x 2 dx = 4. Therefore, Cov ( X, X2 ) = 4 − 2 × 4 3 = 4 3 . Comment: Recall that the expected value of h(X) when X has density fX is given by ∫ h(x)fX(x) dx, see equation (3.8) on p.36. A common mistake in this questions was that some candidates were sloppy with the region of integration. You should be able to calculate integrals of standard functions (such as polynomials). 2
Examiners’ commentaries 2010 (f) We cannot calculate this covariance because it would require us to find Var ( X2 ) and therefore need to calculate E ( X4 ) , but E ( X4 ) = ∫ ∞ 1 x 4 4 x 5 dx = 4 ∫ ∞ 1 1 x dx which is a divergent integral. Comment: The definition of the correlation coefficient is given on p.51. Question 2 (a) The probability that the ball is black and he says it is black is 4 10 × 1 = 4 10 . The probability that it is white and he says it is black is 6 10 × 1 2 = 3 10 . and so the probability he says it is black is 4 10 + 3 10 = 7 10 and the probability it is black given he says it is black is 4 10 7 10 = 4 7 . 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) The probability that both balls were black and he said they were black is 4 10 × 1 × 3 9 × 1 = 2 15 . The probability that one was black and one was white and he said they were both black is 4 10 × 1 × 6 9 × 1 2 + 6 10 × 1 2 × 4 9 × 1 = 4 15 . The probability that both balls were white and he said they were black is 6 10 × 1 2 × 5 9 × 1 2 = 1 12 . The probability he said they were both black is therefore 2 15 + 4 15 + 1 12 = 29 60 and the probability they were black given he said they were black is 2 15 29 60 = 8 29 . 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 practise similar exercises such as example 9 on p.17 learning activity 2.9 on p.15. 3
04b Statistics 2 (c) The probability that all three balls were black and he said they were black is 4 10 × 1 × 4 10 × 1 × 4 10 × 1 = 0.064. The probability that exactly two balls were black and he said they were all black is 3 × 4 10 × 1 × 4 10 × 1 × 6 10 × 1 2 = 0.144. The probability that exactly one ball was black and he said they were all black is 3 × 4 10 × 1 × 6 10 × 1 2 × 6 10 × 1 2 = 0.108. The probability that no ball was black and he said they were all black is 6 10 × 1 2 × 6 10 × 1 2 ⋊ 6 10 × 1 2 = 0.027. So the probability he said they were all black is 0.064 + 0.144 + 0.108 + 0.027 = 0.343 and so the probability that at least two were white given he said they were all black is 0.108+0.027 0.342 = 135 343 =0.394 Comment: This question concerns conditional probability and hence we apply Bayes’ Theorem (Section 2.6). Question 3 (a) We have to minimise (y1 − α − β) 2 + (y2 − β − γ) 2 + (y3 − γ − α) 2 + (y4 − α − β − γ) 2 . Differentiating with respect to α and equating to 0 we get −2 ( y1 − αˆ − βˆ ) − 2 (y3 − αˆ − γˆ) − 2 ( y4 − αˆ − βˆ − γˆ ) = 0 which can be rewritten as 3ˆα + 2βˆ + 2ˆγ = y1 + y3 + y4 Differentiating with respect to β and equating to 0 we get −2 ( y1 − αˆ − βˆ ) − 2 ( y2 − βˆ − γˆ ) − 2 ( y4 − αˆ − βˆ − γˆ ) = 0 which can be rewritten as 2ˆα + 3βˆ + 2ˆγ = y1 + y2 + y4 Differentiating with respect to γ and equating to 0 we get −2 ( y2 − γˆ − βˆ ) − 2 (y3 − αˆ − γˆ) − 2 ( y4 − αˆ − βˆ − γˆ ) = 0 which can be rewritten as 2 ˆα + 2βˆ + 3ˆγ = y2 + y3 + y4 Adding them all up 7 ˆα + 7βˆ + 7ˆγ = 2y1 + 2y2 + 2y3 + 3y4 so 2ˆα + 2βˆ + 2ˆγ = 4y1 + 4y2 + 4y3 + 6y4 7 4