fundamentals of biostatistics 8th edition rosner solutions manual
fundamentals of biostatistics 8th edition rosner
fundamentals of biostatistics
solutions manual
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Fundamentals of Biostatistics 8th Edition Rosner Solutions Manual Contents Chapter 2 Descriptive Statistics ................................ ................................ ................................ ........ 2 Chapter 3 Probability ................................ ................................ ................................ ...................... 21 Chapter 4 Discrete Probability Distributions ................................ ................................ .................. 43 Chapter 5 Continuous Probability Distributions ................................ ................................ ............ 65 Chapter 6 Estimation ................................ ................................ ................................ ...................... 93 Chapter 7 Hypothesis Testing: One -Sample Inference ................................ ............................... 119 Chapter 8 Hypothesis Testing: Two -Sample Inference ................................ ............................... 146 Chapter 9 Nonparametric Method s ................................ ................................ .............................. 192 Chapter 10 Hypothesis Testing: Categorical Data ................................ ................................ ....... 216 Chapter 11 Regression and Correlation Methods ................................ ................................ ......... 267 Chapter 12 Multisample Inference ................................ ................................ ............................... 322 Chapter 13 Design and Analysis Techniques for Epidemiologic Studies ................................ .... 358 Chapter 14 Hypothesis Testing: Person -Time Data ................................ ................................ .... 413 (x − x ) 20.1 We have x = xi = 215 = 8.6 days DESCRIPTIVE STATISTICS n 25 median = (n + 1) th largest observation = 13th largest observation = 8 days 2 20.2 We have that 25 2 i s2 = i=1 = 24 (5 − 8.6)2 +
+ (4 − 8.6)2 24 = 784 24 = 32.67 s = standard deviation = variance = 5.72 days range = largest − smallest observation = 30 − 3 = 27 days 20.3 Suppose we divide the patients according to whether or not they received antibiotics, and calculate the mean and standard deviation for each of the two subsamples: x s n Antibiotics 11.57 8.81 7 No antibiotics 7.44 3.70 18 Antibiotics - x7 8.50 3.73 6 It appears that antibiotic users stay longer in the hospital. Note that when we remove observation 7, the two standard deviations are in substantial agreement, and the difference in the means is not that impressive anymore. This example shows that x and s2 are not robust; that is, their values are easily affected by outliers, particularly in small samples. Therefore, we would not conclude that hospital stay is different for antibiotic users vs. non -antibio tic users. 2 CHAPTER 2/DESCRIPTIVE STATISTICS 3 2.4-2.7 Changing the scale by a factor c will multiply each data value xi by c, changing it to cxi . Again the same individual’s value will be at the median and the same individual’s value will be at the mode, but these value s will be multiplied by c. The geometric mean will be multiplied by c also, as can easily be shown: Geometric mean = [(cx1)(cx2 )
(cxn )]1/n = (cn x1 x2 xn )1/n = c( x1 x2 xn )1/n = c old geometric mean The range will also be multiplied by c. For example, if c = 2 we have: xi –3 –2 –1 0 1 2 3 xi Original Scale –6 –4 –2 0 2 Scale 2 4 6 2.8 We first read the data file “running time” in R > require(xlsx) > running< -na.omit(read.xlsx("C:/Data_sets/running_time.xlsx",1, header=TRUE)) Let us print the first observations > head(running) week time 1 1 12.80 2 2 12.20 3 3 12.25 4 4 12.18 5 5 11.53 6 6 12.47 The mean 1-mile running time over 18 weeks is equal to 12.09 minutes: > mean(running$time) [1] 12.08889 2.9 The standard deviation is given by > sd(running$time) [1] 0.3874181 2.10 Let us first create the variable “time_100” and then calculate its mean and standard deviation > running$time_100=100*running$time > mean(running$time_100) [1] 1208.889 > sd(running$time_100) [1] 38.74181 2.11 Let us to construct the stem -and-leaf plot in R using the stem.leaf command from the package “aplpack” > require(aplpack) CHAPTER 2/DESCRIPTIVE STATISTICS 4 23 282.43 (x − x ) > stem.leaf(running$time_100, unit=1, trim.outliers=FALSE) 1 | 2: represents 12 leaf unit: 1 n: 18 2 115 | 37 3 116 | 7 5 117 | 23 7 118 | 03 8 119 | 2 (1) 120 | 8 9 121 | 8 8 122 | 05 6 123 | 03 4 124 | 7 3 125 | 5 2 126 | 7 127 | 1 128 | 0 Note: one can also use the standard command stem (which does require the “aplpack” package) to get a similar plot > stem(running$time_100, scale = 4) 2.12 The quantiles of the running times are > quantile(running$time) 0% 25% 50% 75% 100% 11.5300 11.7475 12.1300 12.3225 12.8000 An outlying value is identify has any value x such that x upper quartile+1.5 (upper quartile -lower quartile) = 12.32 +1.5 (12.32 −11.75) = 12.32 + 0.85 = 13.17 Since 12.97 minutes is smaller than the largest nonoutlying value (13.17 minutes), this running time recorded in his first week of running in the spring is not an outlying value relative to the distribution of running times recorded the previous year. 2.13 The mean is x = xi = 469 = 19.54 mg dL 12.8 12.6 12.4 12.2 12.0 11.8 11.6 Box plot of running times 24 2.14 We have that 24 2 i s2 = i=1 = 24 (49 −19.54)2 +
+ (12 −19.54) 2 23 6495.96 23 = 282.43 s = = 16.81 mg/dL 2.15 We provide two rows for each stem corresponding to leaves 5-9 and 0-4 respectively. We have Time =
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