chaps07_08 - 172 Chapter 7 Inference for Distributions...

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172 Chapter 7 Inference for Distributions Chapter 7 Solutions 7.1. (a) A stemplot (right) reveals that the distribution has two peaks and a high value (not quite an outlier). Both the stemplot and quantile plot show that the distribution is not normal. (b) Maybe: We have a large enough sample to overcome the nonnormal distribution, but we are sampling from a small population. (c) The mean is x = 27 . 29 cm, s . = 17 . 7058 cm, and the margin of error is t · s / 40: df t Interval Table D 30 2.042 27 . 29 ± 5 . 7167 = 21.57 to 33.01 cm Software 39 2.0227 27 . 29 ± 5 . 6626 = 21.63 to 32.95 cm (d) One could argue for either answer. We chose a random sample from this tract, so the main question is, can we view trees in this tract as being representative of trees elsewhere? 0 222244 0 579 1 0113 1 678 2 2 2 6679 3 112 3 5789 4 0033444 4 7 5 112 5 6 6 9 0 10 20 30 40 50 60 70 –3 –2 –1 0 1 2 3 Diameter (cm) z score 7.2. (a) The distribution is extremely skewed to the right, with two or three high outliers. (b) Means are typically not the best measure of center for skewed distributions. (c) The mean CRP is x . = 10 . 0323 mg / l, s . = 16 . 5632 mg / l, and the margin of error is t · s / 40: df t Interval Table D 30 2.042 x ± 5 . 3477 = 4.68 to 15.38 mg / l Software 39 2.0227 x ± 5 . 2972 = 4.74 to 15.33 mg / l The skewness of the distribution makes this methodology somewhat suspect. 0 00000000000000003334 0 55555677899 1 2 1 5 2 02 2 6 3 0 3 4 4 6 5 5 9 6 6 7 3
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Solutions 173 7.3. Shown are the results for taking natural logarithms; common (base 10) logarithms would differ by a factor of about 0.434. (a) The distribution is still skewed to the right, but considerably less so than the original data. There do not appear to be any outliers in the transformed data (although the group of 0s stands out). (b) Means are typically not the best measure of center for skewed distributions. (c) The mean log-CRP is x . = 1 . 4952, s . = 1 . 3912, and the margin of error is t · s / 40: df t Interval Table D 30 2.042 1 . 4952 ± 0 . 4492 = 1.0460 to 1.9443 Software 39 2.0227 1 . 4952 ± 0 . 4449 = 1.0502 to 1.9401 After undoing the transformation (that is, exponentiating these limits and then subtracting 1), this is about 1.85 to 5.99 mg/l. 0 0000000000000000 0 1 1 555788888 2 001233 2 58 3 0134 3 8 4 13 7.4. The distribution is skewed to the right, with two peaks—clearly not normal. However, the sample size of 40 should be sufficient to overcome this, so the t methods should be fairly reliable. The mean is x = 0 . 76475 µ mol / l, s . = 0 . 3949 µ mol / l, and the margin of error is t · s / 40: df t Interval Table D 30 2.042 0 . 76475 ± 0 . 1275 = 0.6372 to 0.8923 µ mol / l Software 39 2.0227 0 . 76475 ± 0 . 1263 = 0.6384 to 0.8911 µ mol / l 0 2333333333333 0 455 0 6667 0 88889999 1 00011111 1 23 1 4 1 1 9 7.5. (a) The distribution is not normal—there were lots of 1s and 10s—but the nature of the scale means that there can be no extreme outliers, so with a sample of size 60, the t methods should be acceptable. (b) The mean is x . = 5 . 9, s . = 3 . 7719, and the margin of error is t · s / 60: df t Interval Table D 50 2.009 5 . 9 ± 0 . 9783 = 4.9217 to 6.8783 Software 59 2.0010 5 . 9 ± 0 . 9744 = 4.9256 to 6.8744 (c) Because this is not a random sample, it may not represent other children well.
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