Solutions to the Exercises Chapter 1 Solution 1.1 (a) Your computer may be programmed to allocate borderline cases to the next group down, or the next group up; and it may or may not manage to follow this rule consistently, depending on its handling of the numbers involved. Following a rule which says 'move borderline cases to the next group up', these are the five classifications. (i) 1.0-1.2 1.2-1.4 1.4-1.6 1.6-1.8 1.8-2.0 2.0-2.2 2.2-2.4 6 6 4 8 4 3 4 2.4-2.6 2.6-2.8 2.8-3.0 3.0-3.2 3.2-3.4 3.4-3.6 3.6-3.8 6 3 2 2 0 1 1 (ii) 1.0-1.3 1.3-1.6 1.6-1.9 1.9-2.2 2.2-2.5 10 6 10 5 6 2.5-2.8 2.8-3.1 3.1-3.4 3.4-3.7 7 3 1 2 (iii) 0.8-1.1 1.1-1.4 1.4-1.7 1.7-2.0 2.0-2.3 2 10 6 10 7 2.3-2.6 2.6-2.9 2.9-3.2 3.2-3.5 3.5-3.8 6 4 3 1 1 (iv) 0.85-1.15 1.15-1.45 1.45-1.75 1.75-2.05 2.05-2.35 4 9 8 9 5 2.35-2.65 2.65-2.95 2.95-3.25 3.25-3.55 3.55-3.85 7 3 3 1 1 (V) 0.9-1.2 1.2-1.5 1.5-1.8 1.8-2.1 2.1-2.4 6 7 11 7 4 2.4-2.7 2.7-3.0 3.0-3.3 3.3-3.6 3.6-3.9 7 4 2 1 1 (b) Computer graphics: the diagrams are shown in Figures 1.9 to 1.11. Solution 1.2 (a) Computer graphics: see Figure 1.12. (b) Computer graphics: see Figure 1.13. If your computer gives graphics that are text-character based (otherwise known as low-resolution graphics) then the scatter plots you obtain will not be as precise as those appearing in the text and the fitted line will not be displayed. However, the main message of the data should still be apparent. Elements of Statistics Solution 1.3 (a) In order of decreasing brain weight to body weight ratio, the species are as follows. Species Body weight Brain weight Ratio Rhesus Monkey Mole Human Mouse Potar Monkey Chimpanzee Hamster Cat Rat Mountain Beaver Guinea Pig Rabbit Goat Grey Wolf Sheep Donkey Gorilla Asian Elephant Kangaroo Jaguar Giraffe Horse Pig Cow African Elephant Triceratops Diplodocus Brachiosaurus (b) (i) Computer graphics: see Figure 1.14. (ii) Computer graphics: see Figure 1.15. Solution 1.4 There were 23 children who survived the condition. Their birth weights are 1.130, 1.410, 1.575, 1.680, 1.715, 1.720, 1.760, 1.930, 2.015, 2.040, 2.090, 2.200, 2.400, 2.550, 2.570, 2.600, 2.700, 2.830, 2.950, 3.005, 3.160, 3.400, 3.640. The median birth weight for these children is 2.200 kg (the 12th value in the sorted list). There were 27 children who died. The sorted birth weights are 1.030, 1.050, 1.100, 1.175, 1.185, 1.225, 1.230, 1.262, 1.295, 1.300, 1.310, 1.500, 1.550, 1.600, 1.720, 1.750, 1.770, 1.820, 1.890, 1.940, 2.200, 2.270, 2.275, 2.440, 2.500, 2.560, 2.730. The middle value is the 14th (thirteen either side) so the median birth weight for these children who died is 1.600 kg. Solutions to Exercises Solution 1.5 The ordered differences are 3.8, 10.3, 11.8, 12.9, 17.5, 20.5, 20.6, 24.4, 25.3, 28.4, 30.6. The median difference is 20.5. Solution 1.6 Once the data are entered, most computers will return the sample median at a single command. It is 79.7 inches. Solution 1.7 (a) The mean birth weight of the 23 infants who survived SIRDS is 1.130 + 1.575 + . + 3.005 53.070 -xs = --- = 2.307 kg; Notice the subscripts S, D and T 23 23 used in this solution to label and the mean birth weight of the 27 infants who died is distinguish the three sample means. It was not strictly - 1.050 + 1.175 + . + 2.730 - 45.680 necessary to do this here, since we XD = - -= 1.692 kg. 27 27 will not be referring to these The mean birth weight of the entire sample is numbers again in this exercise, but it is a convenient labelling system when a statistical analysis becomes more complicated. Solution 1.8 The mean 'After - Before' difference in ~able'l.11is - 25.3 + 20.5 + . + 28.4 206.1 X = - -= 18.74 pmol/l. 11 11 Solution 1.9 The mean snowfall over the 63 years was 80.3 inches. Solution 1.10 (a) The lower quartile birth weight for the 27 children who died is given by 4~ = x(+(n+l))= x(7) = 1.230kg; the upper quartile birth weight is = x(;(,+~))= ~(21)= 2.200kg. (b) For these silica data, the sample size is n = 22. The lower quartile is qL = x(;(n+l))= X(?) = x(5;) which is three-quarters of the way between x(~)= 26.39 and X(G) = 27.08. This is say, 26.9. The sample median is m=x(5(n+l)) 1 = = x(~l$), which is midway between x(ll) = 28.69 and = 29.36. This is 29.025; say, 29.0. Elements of Statistics The upper quartile is Qa = X($(,+,)) = X(?) = x(17a), one-quarter of the way between ~(17)= 33.28 and x(18) = 33.40. This is qv = 33.28 + i(33.40 - 33.28) = i(33.28) + a(33.40) = 33.31; say, 33.3. Solution 1.1 1 For the snowfall data the lower and upper quartiles are q~ = 63.6 inches and qu = 98.3 inches respectively. The interquartile range is qu - q~ = 34.7 inches. Solution 1.12 Answering these questions might involve delving around for the instruction manual that came with your calculator! The important thing is not to use the formula-let your calculator do all the arithmetic. All you should need to do is key in the original data and then press the correct button. (There might be a choice, one of which is when the divisor in the 'standard deviation' formula is n, the other is when the divisor is n - 1. Remember, in this course we use the second formula.) (a) You should have obtained s = 8.33, to two decimal places. (b) The standard deviation for the silica data is s = 4.29. (c) For the collapsed runners' ,L? endorphin concentrations, s = 98.0. Solution 1.13 (a) The standard deviation s is 0.66 kg. (b) The standard deviation s is 23.7ikhes. Solution 1.14 Summary measures for this data set are x(~)= 23, q~ = 34, m = 45, q~ = 62, = 83. The sample median is m = 45; the sample mean is 31 = 48.4; the sample standard deviation is 18.1. The range is 83 - 23 = 60; the interquartile range is 62 - 34 = 28. Solution 1.15 The first group contains 19 completed families. Some summary statistics are m = 10, Z = 8.2, s = 5.2, interquartile range = 10. For the second group of 35 completed families, summary statistics are m = 4, = 4.8, s = 4.0, interquartile range = 4. The differences are very noticeable between the two groups. Mothers educated for the longer time period would appear to have smaller families. In each case the mean and median are of comparable size. For the smaller group, the interquartile range is much greater than the standard deviation. If the three or four very large families are removed from the second data set, the differences become even more pronounced. Solutions to Exercises Solution 1.16 (a) The five-figure summary for the silica data is given by A convenient scale sufficient to cover the extent of the data is from 20 to 40. The i.q.r. is 33.31 - 26.91 = 6.40. Then and this exceeds the sample maximum, so the upper adjacent value is the sample maximum itself, 34.82. Also This value is less than the sample minimurn, so the lower adjacent value is the sample minimum itself. For these data there are no extreme values. The boxplot is shown in Figure S1.l. Percentage silica Figure S1.1 (b) For the snowfall data the lower adjacent value is 39.8; the minimum is 25.0. The upper adjacent value is equal to the maximum, 126.4. The boxplot is shown in Figure S1.2. Annual snowfall (inches) Figure S1.2 Solution 1.1 7 The sample skewness for the first group of mothers is -0.29. Solution 1.18 (a) The five-figure summaries for the three groups are normal: (14, 92, 124.5, 274.75, 655) alloxan-diabetic: (13, 70.25,139.5, 276, 499) insulin-treated: (18, 44, 82, 133, 465). The normal group has one very high recording at 655; the next highest is 455, which is more consistent with the other two groups.