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Solution Manual for Statistical Methods, 3rd Edition

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    CHAPTER 1: SAMPLE PROBLEMS FOR HOMEWORK, CLASS OR EXAMS These problems are designed to be done without access to a computer, but they may require a calculator. 1. The boxplots shown below summarize data for vocabulary scores from two samples of first-graders, some of whom attended Pre-K, and some of who did not (No Pre-K). a. Among Pre-K children, approximately what proportion have vocabulary scores greater than 41? b. Among no Pre-K children, approximately what proportion have vocabulary scores between 24 and 28? c. Which group apparently has the higher typical vocabulary score? d. Which group shows the greater variability in their vocabulary scores? e. Summarize the shape of the distribution for each group. 2. You jotted down some summary values for a data set of 15 observations of wave heights, in feet, at a monitoring site. Your notes state that:   i y 37 and   2 i y 173. a. Give the mean and the standard deviation of the wave heights, in feet. b. Give the mean and the standard deviation of the wave heights, in meters. (Hint: a meter = 3.3 feet). c. You also have written down that the median wave height, in feet, was 2 . Is the distribution more likely negatively skewed, positively skewed, or symmetric?

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    3. Volunteers in a psychology experiment watch a video re-enactment of a crime. After a delay, they answer questions about the scene. The researcher records both the length of the delay (in hours) and the number of questions the volunteer answered correctly. A graphical display of the data is shown below. Write a sentence, in simple language, that summarizes the relationship between the two variables. 4. A school is concerned that the playground area may have contaminated soil left from a time when a creosote plant was nearby. The school can not afford to test all the soil on the playground, so a statistician divides a map of the playground into 1000 equally sized rectangles. A random number generator is used to select 30 rectangles. A soil specimen is taken from each selected rectangle. a. Identify the population, the sampling frame, and the sample. b. Is this study observational or a designed experiment? c. For each variable, say whether its scale is nominal, ordinal, interval or ratio. Identify one type of graph that can be used to summarize the data. Contamination level (none, trace, moderate, high) Creosote concentration, in parts per billion Usage (general play, organized sports, drainage structure, etc.) 5. Barometric pressures in tropical cyclones are very negatively (left) skewed. You wrote down 1005 and 985 as the mean and median, but forgot to label which was which. Label the values properly.

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    6. The two histograms below summarize total cholesterol levels for large samples of elderly men and women. Write a short paragraph comparing the two groups. Be sure to contrast the typical values, variability and shape of the distributions. 0 5 10 15 20 25 30 P e r c e n t F 112. 5 137. 5 162. 5 187. 5 212. 5 237. 5 262. 5 287. 5 312. 5 337. 5 362. 5 387. 5 0 5 10 15 20 25 30 P e r c e n t M chol 7. The frequency table below summarizes the results for mercury concentrations in a sample of Florida lakes. Display the relative frequencies using an appropriate graph. Mercury Concentration Number of cases Trace (very low) 11 Low 27 Borderline 8 Dangerous 3

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    8. The data below shows electricity costs in cents per KWH for a sample of 14 utility companies. The data has already been sorted, reading across each row. 12.4 12.8 13.1 13.6 13.8 13.9 14.0 14.2 14.2 14.4 14.6 14.8 15.0 15.3 a. Calculate the first quartile, median, and third quartile. b. Calculate the range and the interquartile range. c. Identify the outliers, if there are any. Show your computations. d. Construct the boxplot for this data, and comment on the shape of the distribution. 9. The boxplots show NOx emissions from your power plant (expressed as pounds per MWh of electricity generated), using two different settings for the air intakes. Compare the emissions, being sure to address the typical values, variability, shape and/or other special features. If your goal is to lower typical NOx emissions, which setting should you select? If your goal is to avoid any extremely high values, which setting should you select? setting HIGH LOW NOx 8.00 6.00 4.00 2.00 10. a. In a data set with a sample mean of 10 and a standard deviation of 2, you note that 78% of observations fall in the interval (6, 14). Is this distribution approximately bell- shaped? Why or why not? b. In a data set with a sample mean of 10 and a standard deiation of 2, you wrote down the percent of the observations that fall in the interval (6, 14). Later, you find your writing is hard to read, but the number is either 63% or 83%. Which must it be and why? SOLUTIONS 1. a. 25% b. 25% c. The pre-K group apparently has higher typical values. d. The No pre-K group shows slightly higher variability. e. Scores in the pre-K group are nearly symmetrically distributed, but in the No pre-K group they are positively (or right) skewed.

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    2. a. mean = 2.47, SD = 2.42 b. this is a simple change of scale. Mean = 2.47/3.3 = 0.75, SD = 2.42/3.3 = 0.73 c. Since the mean is substantially greater than the median, the distribution is most likely right, or positively, skewed. Alternately, one might see that the size of the SD compared to the mean, combined with the fact that there are no negative wave heights, implies a positive skew. 3. As the time delay increases, the number of questions answered correctly tends to decrease. 4. a. The population is all soil in the playground, or alternatively all 1000 rectangles of soil. The sampling frame is the list of the 1000 rectangles. The sample is the 30 rectangles selected for study. b. Observational c. Contamination level is ordinal, use bar chart or pie chart; Creosote concentration is ratio, use a histogram, boxplot or stem-and-leaf plot. Usage is nominal, use a bar chart or pie chart. 5. Since the data is very left skewed, the mean must be less than the median. Mean = 985, median = 1005. 6. Typical values for the females are slightly higher than those for the men. Females show slightly higher variability, though the difference in variability is not large. Both distributions show some right skew, but the skew is stronger in the males. 7. The bar chart is shown below. Pie charts are possible, but are harder to draw and harder to compare when there is more than one sample. 8. a. median = (14.0+14.2)/2 = 14.1 Using the algorithm described in the text, Q1 = 13.35 and Q3 = 14.5 However, a variety of algorithms exist, on the TI83/84 calculators, Q1 = 13.6 and Q3 = 14.6. b. Range = 2.9. IQR = 1.15 using algorithm for quartiles given in book, but IQR = 1 using TI83/84. c. Using algorithm for quartiles given in book, lower fence = 11.625 and upper fence = 16.225. There are no outliers. d. will depend slightly on algorithm for outliers. Nearly symmetric, but some students will not a slight negative skew. 9. The typical NOx values are lower at the High air intake setting, as shown by the lower median. However, the NOx values are more variable at the high air intake setting, and somewhat right-skewed. At the low intake setting, the values are more symmetric and less variable. Hence, to lower the typical NOx values one should choose the high air intake setting, but to avoid extremely high values, one needs to use the low air intake setting.

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    10. a. This distribution cannot be approximately bell-shaped because if it were the percentage in this interval (mean ± 2 std.dev) would be close to 95%. b. It must be 83%, because this interval (mean ± 2 std. dev) must include at least 75% of the observations, using Tchebysheff’s Theorem.

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    Chapter 1 – Page 1 CHAPTER 1 EXERCISE 1 a) Mean = 17.00, SD = 5.53, median = 17.00, Range = 22, IQR = 20 – 13 = 7. b) See boxplot. c) A typical value for X is 17 which is both the mean and the median. About half the values are between 13 and 20. The boxplot shows that the upper tail is slightly longer, but difference is small and mean and median are equal. This distribution is roughly symmetric.

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    Chapter 1 – Page 2 EXERCISE 2 Variable Mean Median Variance Standard Deviation Shape WATER 7.125 1.500 452.864 21.28 extremely positively skewed VEG 1.120 0.00 4.327 2.080 extremely positively skewed FOWL 75.635 11.5 42197.33 205.420 extremely positively skewed WATER Stem Leaf # Boxplot 14 9 1 * 13 12 11 10 9 8 7 6 5 4 3 13 2 * 2 1 05667 5 0 0 00000000000001111111111112222222222345556779 44 +--+--+ ----+----+----+----+----+----+----+----+---- Multiply Stem.Leaf by 10**+1 VEG Stem Leaf # Boxplot 9 0 1 * 8 8 0 1 * 7 7 0 1 * 6 6 5 5 2 1 0 4 4 0 1 | 3 | 3 0 1 | 2 8 1 | 2 00002 5 | 1 58 2 +-----+ 1 0002 4 | + | 0 58 2 | | 0 00000000000000000000000000000002 32 *-----* ----+----+----+----+----+----+--

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    Chapter 1 – Page 3 FOWL Stem Leaf # Boxplot 14 1 1 * 13 12 11 10 9 8 7 6 5 4 3 6 1 * 2 124 3 0 1 227888 6 0 0 00000000000000000000001111112222223335678 41 +--+--+ ----+----+----+----+----+----+----+----+- Multiply Stem.Leaf by 10**+2 b) Frequency distribution for FOWL Fowl Frequency Midpoint approximate mean = 5500/52 0 ≤ x < 100 41 50.0 = 105.8 100 ≤ x < 200 6 150.0 200 ≤ x < 300 3 250.0 approximate variance = 50746 300 ≤ x < 400 1 350.0 1400 ≤ x < 1500 1 1450.0 c) Scatterplots Aside from the single point with extraordinary number of waterfowl, there is little relationship. fo w l 0 1 0 0 2 0 0 3 0 0 4 0 0 5 0 0 6 0 0 7 0 0 8 0 0 9 0 0 1 0 0 0 1 1 0 0 1 2 0 0 1 3 0 0 1 4 0 0 1 5 0 0 ve g 0 1 2 3 4 5 6 7 8 9

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    Chapter 1 – Page 4 EXERCISE 3 a) INDEX: mean = -0.149 median = -0.155 variance = 1.771 IQR = 1.59 (show stem and leaf, histogram, or boxplot as part of answer) FUTURE: mean = -0.208 median = -0.3 variance = 0.601 IQR = 0.81 (show stem and leaf, histogram, or boxplot as part of answer) b) The Scatter plot is shown below. There is a clear trend for positive changes in the FUTURE contract value to be associated with positive changes in the NYSE INDEX. Yes, FUTURE can be used to help predict changes in the NYSE index. index -4 -3 -2 -1 0 1 2 3 future -3 -2 -1 0 1 2

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    Chapter 1 – Page 5 EXERCISE 4 Y1 Stem Leaf # Boxplot 8 14 2 | 7 3 1 | 6 1237889 7 +-----+ 5 338 3 *-----* 4 025 3 | + | 3 112 3 +-----+ 2 478 3 | 1 668 3 | ----+----+----+----+ Y2 Stem Leaf # Boxplot 6 5679 4 | 6 4 1 | 5 789 3 +-----+ 5 22 2 | | 4 678 3 *--+--* 4 0024 4 | | 3 559 3 +-----+ 3 12333 5 | ----+----+----+----+ Y3 Stem Leaf # Boxplot 9 4 1 | 8 0 1 | 7 | 6 777 3 | 5 444 3 +-----+ 4 000 3 | | 3 | + | 2 7777777 7 *-----* 1 333333 6 +-----+ 0 0 1 | ----+----+----+----+

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    Chapter 1 – Page 6 Y4 Stem Leaf # Boxplot 7 4 1 | 6 | 5 0 1 | 4 578 3 | 3 99 2 +-----+ 2 123357 6 *--+--* 1 01468 5 | | 0 1113699 7 +-----+ ----+----+----+----+ Y1 is slightly negatively (or left) skewed; Y2 is nearly symmetrically distributed, though without a well defined peak; Y3 is positively (or right) skewed; Y4 is extremely positively skewed. comparison to empirical rule, actual percentage of observations in interval mean Std dev Shape ±1 SD (expect 68%) ±2 SD (expect 95%) ±3 SD (expect all) Y1 4.840 2.107 slight left skew 68% 100% 100% Y2 4.748 1.262 near symm. 60% 100% 100% Y3 3.696 2.434 right skew 76% 96% 100% Y4 2.328 1.888 extreme right skew 64% 96% 100% The empirical rule works reasonably well even for the skewed distributions.

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    Chapter 1 – Page 7 EXERCISE 5 a) DAYS: Mean= 15.854 median= 17, variance= 24.324 TEMP: Mean= 39.348 median= 40 variance= 11.152 b) Scatterplot: days 0 10 20 30 temp 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47

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    Chapter 1 – Page 8 From the scatterplot, there appears to be no definitive relationship between the average temperature and the number of rainy January days. EXERCISE 6 a) The distribution of EXPEND is extremely right skewed with a mean ($1816M) that is much greater than the median ($1169M), so the best measure of location is the median and the best measure of dispersion is the IQR ($1938M). There are three outlier states with California being an extreme outlier at $15348M. b) Per Capita expenditures are also right skewed, but slightly less so than expenditures. The median is $0.269k ($2,690) and the IQR is $0.139k. c) For Per Capita expenditures, there is one state that is an extreme outlier (Alaska) and another that is a moderate outlier (Delaware). Both these are states with small populations, which may mean that there are a lot of fixed costs in the systems. EXERCISE 7 Scatter of DFOOT vs. HCRN: Scatter of dfoot 2 3 4 5 6 hcrn 1 2 3 4 5 6 7 8 9

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    Chapter 1 – Page 9 dfoot vs. ht: Scatter of HCRN vs. HT: The strongest relationship exists between DFOOT, the diameter of the tree at one foot above ground level and HT, the total height of the tree. (notice from the scatter plot that it is closest to a linear relationship) One would expect that as the base of the tree increases in diameter the tree would increase in height as well. dfoot 2 3 4 5 6 ht 20 21 22 23 24 25 26 27 28 29 30 31 32 hcrn 1 2 3 4 5 6 7 8 9 ht 20 21 22 23 24 25 26 27 28 29 30 31 32

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