S15.3 False. The selection ratio for the pooled data is 68.5%. Women: 28/181 = 15.5 % Men: 581/2571= 22.6 % 15.5/22.6 = 68.6 % More women took the test in 1977, when they had a higher pass rate. s15.4 A is correct. B has too much area. C has incorrect units. s15.6 False, these data are cross-sectional, not longitudinal. (Education was less, on average, for the older men). s15.9 a) The 70th percentile is about .52 standard units, while the 80th percentile is about .84 standard units. These represent original units of about 60 and 67, respectively, for a 7 point difference. b) (Using same reasoning) 9 points. S15.12 a) True. b) True. c) False. d) False. S15.15 RMS error of the regression line of first-year scores on LSAT scores is [1-(.55)^2]^.5 * 12 = about 10. Therefore about 16% of the people (greater than one RMS error) will do more than 10 points better than their prediction. S15.16 Statistics... Regression effect. S15.20 This is the same as getting 7 heads. Using the binomial formula: (# of combinations)*(chance of each) 10!/(7!*3!) * (.5)^10 = 12 %. 16.1 (iii). The number of 1s will probably not equal exactly 6,000, but it should be close, compared to the overall sample size (10,000). 16.2 Without replacement, every ticket will be drawn, so the answer will be (i). 16.3 Both are wrong; the chances are the same in Roulette (with a fair wheel) every time you play. 16.4 a) 60 rolls. Since an ace is only expected 1/6 of the time (16%), you want as few rolls as possible. b) 600 rolls. You now have probability in your favor, so you want as many rolls as possible. c) 600 rolls. (Same reasoning as b). d) 60 rolls. Your chances of getting any one exact value decrease with the number of rolls. 16.7 25 draws from [ 4 -1 -1 -1 -1 ] 16.8 50 draws from box with 4 $8 tickets and 34 -$1. 16.9 (ii). B gives a better chance of winning. Since there are more reds in the box, you want as many draws as possible (see problem 4b). 16.12 There is no problem 16.12. 17.1 a) 100, 1000. b) The average of the box is 7 and the SD is 3. So, the expected value for the sum is 100 * 7 = 700 and the SE is sqrt(100) * 3 = 30. The sum will be around 700, give or take 30 or so. The chance, then, is about 90%. 17.3 a) (iii), see section 4. b) (i) c) (v) d) (iv) e) (ii) 17.4 The number of aces in 180 rolls of a die is like the sum of 180 draws from a box with 1 ticket marked ``1'' and 5 tickets marked ``0''. The number of aces will be around 30, give or take 5 or so. There is about a 99.7% chance that the number of aces will be in the range of 15 to 45. About 99.7% of the people should get a number in that range. 17.6 12 chance error in the sum of the draws 45 observed value for the number of 1's 187 observed value for the sum of the draws 25 expected value for the number of 3's 50 expected value for the number of 1's 175 expected value for the sum of the draws 5 standard error for the number of 1's 32 observed value for the number of 3's 17.8 a) Option (ii) is right. The sum is equally likely to go up or down 1 on each draw, just like the difference. Option (i) is out, you can't add words; with option (iii), the sum can't go up; with option (iv), the sum can't go down; with option (v), the sum has a chance to stand still, but difference has to go up or down. b) Expected value = 0, SE = sqrt(100) = 10 17.9 a) is false b) is true c) is true Reason: if you play (i), the net gain is like the sum of 1000 draws from a box with 12 tickets marked $2 and 26 marked -$1. the average of the box is -$2/38 and the SD is about $1.39. The expected value for the net gain is -$53 and the SE is $44. If you play (ii), there is another box, with a bigger SD. The net gain has the same expected value but the SE is $182. Chance variability helps you overcome the negative expected value, so you are more likely to come out ahead with B. you are more likely to win big - or lose big. The chance of coming out ahead in (i) is 12%. The chance of coming out ahead with (ii) is about 38%. 17.10 Statement (iii) is false. The SE does not go up by the full factor of 2, but only by sqrt(2) ~= 1.4. Unless there is something rather strange about the box, the chance that the sum is between 700 and 900 will be quite a bit more than 75%. 17.11 You have to change to box to: 0 0 0 1 3. The average of the box is 0.8, and the SD is about 1.2. So the sum will be around 80, give or take 12 or so. 18.1 20, 25 18.2 a) The average of the box is 4 and the SD is about 2.24; the expected value for the sum is 1600 and the SE is about sqrt(400)*2.24~=45. The chance is about 99%. b) The number of 3's is like the sum of 400 draws from the box: 0 1 0 0. The expected number is 100, and the SE is 8.66 The chance is about 12%. (NB: The continuity correction could be used for part (b) if more accuracy were needed.) 18.4 If you get 12 heads, you automatically get 13 tails. Forget the tails, they're a distraction. The problem can be done by the method of chapter 15: 25!/(12!*13!) * (1/2)^25 = 5,200,300/33,554,432 ~= 15.50% Or, it can be done by the method of example 1 on page 317. The expected number of heads is 12.5 and the SE is 2.5. The normal curve gives the (very good) approximation of 15.54%. 18.5 (i) is the probability histogram for the sum, it's like the normal curve. (ii) is the probability histogram for the product, it's like figure 10, pg 323. (iii) is the histogram for the numbers drawn, it's like a histogram for the numbers in the box. 18.6 Something is wrong with COIN. With a million tosses, the number of heads should be around 500,000 give or take 500 or so. COIN is 4 SE's too high. 18.8 (a) is true, (b) is false, (c) is false, (d) is true. The expected value is computed without error, as 50. The 5 represents the likely size of the chance error in the number of heads, not in the expected value. 18.9 (a) is true, (b) is false. You don't have enough draws to use the normal approximation. See exercise 6 on page 324. 18.10 Both statements are true. With more tosses, the probability histogram gets closer to the normal curve. You get a feeling for how many draws are enough by looking at the pictures in the chapter. 18.12 (a) Can't be done. For example the box could have 4 tickets marked ``1'' and 6 tickets marked ``-10'', or 4 marked ``3'' and 6 marked ``-2''. With these two boxes, the chances would be quite different. (b) Can be done, using the normal approximation -- that's why the average and SD are so useful.