Assignment 1 - Grace Hughes

Application 1 Porcellini et. al. (A-8) studied 13 HIV positive patients who were treated with highly active antiretroviral therapy (HAART) for at least 6 months. The CD4 T cell counts ( × 10 6 / L ¿ at baseline for the 13 subjects are listed below. 230 205 313 207 227 245 173 58 103 181 105 301 169 Compute (a) the mean, (b) the median, (c) the mode, (d) the range, (e) the variance, (f) the standard deviation, (g) the coefficient of variation, (h) the interquartile range. Select the measure of central tendency that you think would be most appropriate for describing the data. 58, 103, 105, 169, 173, 181, 205, 207, 227, 230, 245, 301, 313 a) mean = 2517/13 = 193.62 b) median = 13+1/2 = 14/2 = 7; therefore, the median is 205 c) there is no mode d) range = 313 - 58 = 255 e) variance = [(58−193.62) 2 +(103−193.62) 2 +(105−193.62) 2 +(169−193.62) 2 +(173−193.62) 2 +(181−193.62) 2 +(205−193.62) 2 +(2 07−193.62) 2 +(227−193.62) 2 +(230−193.62) 2 +(245−193.62) 2 +(301−193.62) 2 + (313−193.62) 2 ] / (13-1) = 5568.09 f) standard deviation = √ (66817.08)/12= 74.62 g) coefficient of variation = 74.62/193.62 x 100% = 38.54% h) interquartile range = 245 - 169 = 76 i) The measure of central tendency that would be most appropriate for describing the data is the median. Application 2 Cardosi et. al. (A-12) performed a 4-year retrospective review of 102 women undergoing radical hysterectomy for cervical or endometrial cancer. Catheter associated urinary tract infection was observed in 12 of the subjects. Below is the number of postoperative days until diagnosis of the infection for each subject experiencing an infection. 16 10 49 15 6 15 8 19 11 22 13 17 Compute (a) the mean, (b) the median, (c) the mode, (d) the range, (e) the variance, (f) the standard deviation, (g) the coefficient of variation, (h) the interquartile range. Select the measure of central tendency that you think would be most appropriate for describing the data. 6, 8, 10, 11, 13, 15, 15, 16, 17, 19, 22, 49 a) mean = 201/12 = 16.75 b) median = 12+1/2 = 6.5; therefore, the median is 15 c) the mode is 15 d) range = 49 - 6 = 43 e) variance = (6-16.75) 2 + (8-16.75) 2 + (10-16.75) 2 + (11-16.75) 2 + (13-16.75) 2 + (15-16.75) 2 + (15-16.75) 2 + (16-16.75) 2 + (17-16.75) 2 + (19-16.75) 2 + (22-16.75) 2 + (49-16.75) 2 / (12-1) = 124.02

f) standard deviation = √ 1364.22/11 = √124.02 = 11.14 g) coefficient of variation = 11.14/16.75 x 100 = 66.5% h) Q1 = 10, Q3 = 19, IQR = 19-10 = 9 i) The measure of central tendency that would be most appropriate for describing the data would be median since the set of data has such a large outlier, 49. Application 3 Krieser et al. (A-17) examined glomerular filtration rate (GFR) in pediatric renal transplant recipients. GFR is an important parameter of renal function assessed in renal transplant recipients. The following are measurements from 19 subjects of GFR measured with diethylenetriamine pentacetic acid. (Note: some subjects were measured more than once.) 18 42 21 43 21 43 23 48 27 48 27 51 30 55 32 58 32 60 32 62 36 67 37 68 41 88 42 63 Compute (a) the mean, (b) the median, (c) the variance, (d) the standard deviation, (e) the coefficient of variation. (f) Select the measure of central tendency that you think would be most appropriate for describing the data. (g) What percentage of the measurements is within one standard deviation of the mean? Two standard deviations of the mean? Three standard deviations of the mean? 18, 21, 21, 23, 27, 27, 30, 32, 32, 32, 36, 37, 41, 42, 42, 43, 43, 48, 48, 51, 55, 58, 60, 62, 63, 67, 68, 88 a) mean = 1215/28 = 43.39 b) the median is 42 c) variance = (18 - 43.39) 2 + (21 - 43.39) 2 + (21 - 43.39) 2 + ... + (67 - 43.39) 2 + (68 - 43.39) 2 + (88 - 43.39) 2 = 291.95 28 - 1 d) standard deviation = √ 291.95 = 17.09 e) coefficient of variation = 17.09/43.39 x 100% = 39.4% f) The measure of central tendency that would be most appropriate for describing the data would be the median. g) percent of measurements within one standard deviation of the mean: 68% mean ± 1 standarddeviation

43.93 + 17.09 = 62.01 43.94 - 17.09 = 26.84 between (26.84, 62.01) percent of measurements within two standard deviations of the mean: 96.4% mean ± 2 standard deviation 43.93 + (17.09 + 17.09 )= 78.11 43.94 - (17.09 - 17.09) = 9.76 between (9.76, 78.11) percent of measurements withing three standard deviations of the mean: 99.6% 43.93 + 17.09 + 17.09 + 17.09 = 95.2 43.94 - 17.09 - 17.09 - 17.09= -7.33 (28/28) · 100 = 100% between (-7.33, 95.2) Application 4 Spivack (A-18) investigated the severity of disease associated with C. difficile in pediatric inpatients. One of the variables they examined was number of days patients experienced diarrhea. The data for the 22 subjects in the study appear below. 3 11 3 4 14 2 4 5 3 11 2 2 3 2 1 1 7 2 1 1 3 2 Compute (a) the mean, (b) the median, (c) the variance, (d) the standard deviation, (e) the coefficient of variation. (f) Select the measure of central tendency that you think would be most appropriate for describing the data. (g) What percentage of the measurements is within one standard deviation of the mean? Two standard deviations of the mean? Three standard deviations of the mean? 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 5, 7, 11, 11, 14 a) mean = 87/22 = 3.95 b) median = 3 c) variance = ( ( 3 − 3.95 ) 2 + ( 11 − 3.95 ) 2 + ( 3 − 3.95 ) 2 + ( 4 − 3.95 ) 2 + ( 14 − 3.95 ) 2 + ( 2 − 3.95 ) 2 + ( 4 − 3.95 ) 2 + ( 5 − 3.95 ) 2 + ( 3 − 3.95 ) 2 + ( 11 −