Essentials of Biostatistics for Public Health - Lecture Notes

Answer Key Essentials of Biostatistics for Public Health Solutions to Problems 1. A study is run to estimate the mean total cholesterol level in children 2-6 years of age. A sample of 9 participants is selected and their total cholesterol levels are measured as follows. 185 225 240 196 175 180 194 147 223 Generate a 95% confidence interval for the true mean total cholesterol levels in adults with a history of hypertension. X 185 225 240 196 175 180 194 147 223 1,765 X X2 34,225 50,625 57,600 38,416 30,625 32,400 37,636 21,609 49,729 352,865 ΣX 1,765   196.1 n 9 ΣX2  (ΣΣ 2 )/n 352,865  (1,765)2 /9  n 1 8 352,865  346,136.1 6728.89   841.1  29.0  8 8 s , df = n – 1 = 9 – 1 = 8 Xt n s For 95% confidence, t = 2.306 s Xt n 29.0 196.1 (2.306) 9 196.1 + 22.29 (173.8, 218.4) 2. A clinical trial is planned to compare an experimental medication designed to lower blood pressure to a placebo. Before starting the trial, a pilot study is conducted involving 10 participants. The objective of the study is to assess how systolic blood pressure changes over time untreated. Systolic blood pressures are measured at baseline and again 4 weeks later. Compute a 95% confidence interval for the difference in blood pressures over 4 weeks. Baseline 120 4 Weeks 122 Baseline 120 145 130 160 152 143 126 121 115 135 145 130 142 135 160 158 4 Weeks 122 142 135 158 155 140 130 120 124 130 152 155 143 140 126 130 121 120 Difference = 4 Weeks-Baseline 2 -3 5 -2 3 -3 4 -1 9 -5 9 135 130 Difference2 4 9 25 4 9 9 16 1 81 25 183 ΣDifference 9   0.9 n 10 2 2 ΣDifferences  (ΣΣDiffereces) /n 183  (9) 2 /10   19.43  4.4 n 1 9 s Xd  t d , df = n – 1 = 10 – 1 = 9 n Xd  s 115 124 For 95% confidence, t = 2.262 s Xd  t d n 4.4 0.9  (2.262) 10 0.9 + 3.1 (-2.2, 4.0) 3. The main trial is conducted and involves a total of 200 patients. Patients are enrolled and randomized to receive either the experimental medication or the placebo. The data shown below are data collected at the end of the study after 6 weeks on the assigned treatment. Mean (Std Dev) Systolic Blood Pressure % Hypertensive % With Side Effects Experimental (n=100) 120.2 (15.4) 14% 6% Placebo (n=100) 131.4 (18.9) 22% 8% a. Generate a 95% confidence interval for the difference in mean systolic blood pressures between groups. 1 1 (X1 - X 2 )  ZSp  n1 n 2 Check s12/s22 = 15.42/18.92 =0.66 Sp  Sp  (n1  1)s12  (n2  1)s22 n1  n 2  2 (100 1)15.42  (100 - 1)18.92 100  100  2 Sp  297.19 =17.2 (120.2 - 131.4)  1.960(17.2) 1 1 .  100 100 -11.2 + 4.77 (-16.0, -6.4) b. Generate a 95% confidence interval for the difference in proportions of patients with hypertension between groups. (p̂1 - p̂ 2 )  Z p̂ 1(1 - p̂ 1) p̂2 (1  p̂ 2 )  n1 n2 Sample size is adequate with at least 5 successes and failures in each group. (0.14  0.22)  1.96 0.14(1 0.14) 0.22(1  0.22)  100 100 -0.08 + 1.96 (0.054) -0.08 + 0.106 (-0.19, 0.03) c. Generate a point estimate for the relative risk of side effects in patients assigned to the experimental group as compared to placebo. Generate a 95% confidence interval for the relative risk. R̂R = p̂ 1/ p̂ 2 = 0.06/0.08 = 0.75 Ln(R̂R)  Z (n1 - x 1 )/x1 (n2 - x 2 )/x2  n1 n2 Sample size is adequate with at least 5 successes and failures in each group. 94/6 92/8 Ln(0.75)  1.96  100 100 -0.29 + 1.96 (0.521) -0.29 + 1.02 (-1.31, 0.73) The 95% confidence interval for Ln(RR) is (-1.31, 0.73). To generate the confidence interval for the relative risk, we take the antilog (exp) of the lower and upper limits: (exp(-1.31), exp(0.73)) (0.27, 2.08) 4. The following data are collected as part of a study of coffee consumption among undergraduate students. The following reflect cups per day consumed: 3 4 6 8 2 1 0 2 X2 9 16 36 64 4 1 0 4 134 X 3 4 6 8 2 1 0 2 26 a. Compute the sample mean X ΣX 26   3.3 n 8 b. Compute the sample standard deviation s ΣX 2  (ΣΣ 2 )/n 134  (26)2 /8 134  84.5    7.07  2.7 n 1 7 7 c. Construct a 95% confidence interval for the mean number of cups of coffee consumed among all undergraduates. s , df = n – 1 = 8 – 1 = 7 Xt n For 95% confidence, t = 2.365 s Xt n 2.7 3.3  (2.365) 8 3.3 + 2.26 (1.0, 5.6) 5. A clinical trial is conducted comparing a new pain medication for arthritis. Participants are randomly assigned to receive the new medication or a placebo. The outcome is pain relief within 30 minutes. The data are shown below. New Medication Placebo Pain Relief 44 21 No Pain Relief 76 99 p̂ 1 = 44/120 = 0.367, p̂ 2 = 21/120 = 0.175 a. Estimate the risk difference in pain relief between treatments R̂D = p̂ 1- p̂ 2 = 0.367 – 0.175 = 0.192 b. Estimate the relative risk in pain relief between treatments R̂R = p̂ 1/ p̂ 2 = 0.367/0.175 = 2.097 c. Estimate the odds ratio in pain relief between treatments ÔR = p̂ 1/(1 p̂1 ) 0.367 / 0.633 0.580   = 2.74 p̂ 2 /(1 p̂ 2 ) 0.175 / 0.825 0.212 d. Construct a 95% confidence interval for the odds ratio 1 1 1 1    Ln(ÔR)  Z x 1 (n1  x 1 ) x 2 (n2  x 2 ) Sample size is adequate with at least 5 successes and failures in each group. Ln(2.74)  1.96 1 1 1 1    44 76 21 99 1.01 + 1.96 (0.306) 1.01 + 0.60 (0.41, 1.61) The 95% confidence interval for Ln(OR) is (0.41, 1.61). To generate the confidence interval for the odds ratio, we take the antilog (exp) of the lower and upper limits: (exp(0.41), exp(1.61)) (1.51, 5.00) 6. The following data are collected in a clinical trial evaluating a new compound designed to improve wound healing in trauma patients. The new compound is compared against a placebo. After treatment for 5 days with the new compound or placebo the extent of wound healing is measured and the data are shown below. Treatment New Compound (n=125) Placebo (n=125) Wound Healing: % Reduction in Size of Wound None 1-25% 26-50% 51-75% 76-100% 4 11 37 32 41 12 24 45 34 10 Suppose that clinicians feel that if the percent reduction in the size of the wound is greater than 50% then the treatment is a success. a. Generate a 95% confidence interval for the percent success in patients receiving the new compound. p̂ = 73/125 = 0.584 p̂(1 - p̂) n Sample size is adequate with at least 5 successes and failures in the sample. p̂  Z 0.584(1 - 0.584) 125 0.584 + 0.086 0.584  1.960 (0.498, 0.670) b. Generate a 95% confidence interval for the difference in the percent success between the new compound and placebo p̂ 1 = 73/125 = 0.584, p̂ 2 = 44/125 = 0.352 (p̂1 - p̂2 )  Z p̂ 1(1 - p̂ 1 ) p̂ 2 (1  p̂ 2 )  n1 n2 Sample size is adequate with at least 5 successes and failures in each group. (0.584  0.352)  1.96 0.584(1  0.584) 0.352(1  0.352)  125 125 0.232 + 1.96 (0.061) 0.232 + 0.120 (0.112, 0.352) c. Generate a 95% confidence interval for the relative risk of treatment success between treatments R̂R = p̂ 1/ p̂ 2 = 0.584/0.352 = 1.66 Ln(R̂R)  Z (n1 - x 1 )/x1 (n2 - x 2 )/x2  n1 n2 Sample size is adequate with at least 5 successes and failures in each group. 52/73 81/44 Ln(1.66)  1.96  125 125 0.507 + 1.96 (0.143) 0.507 + 0.28 (0.227, 0.787) The 95% confidence interval for Ln(RR) is (0.227, 0.787). To generate the confidence interval for the relative risk, we take the antilog (exp) of the lower and upper limits: (exp(0.227), exp(0.787)) (1.25, 2.20) d. Generate a 95% confidence interval for the odds ratio of treatment success between treatments. ÔR = p̂ 1/(1 p̂1 ) 0.584 / 0.416 1.404   = 2.59 p̂ 2 /(1 p̂ 2 ) 0.352 / 0.648 0.543 Ln(ÔR)  Z 1 1 1 1    x 1 (n1  x 1 ) x 2 (n2  x 2 ) Sample size is adequate with at least 5 successes and failures in each group. Ln(2.59)  1.96 1 1 1 1    73 52 44 81 0.952 + 1.96 (0.261) 0.952 + 0.51 (0.44, 1.46) The 95% confidence interval for Ln(OR) is (0.44, 1.46). To generate the confidence interval for the odds ratio, we take the antilog (exp) of the lower and upper limits: (exp(0.44), exp(1.46)) (1.55, 4.31) 7. A clinical trial is conducted to compare an experimental medication to placebo to reduce the symptoms of asthma. Two hundred participants are enrolled in the study and randomized to receive either the experimental medication or placebo. The primary outcome is self-reported reduction of symptoms. Among 100 participants who receive the experimental medication, 38 report a reduction of symptoms as compared to 21 participants of 100 assigned to placebo. a. Generate a 95% confidence interval for the difference in proportions of participants reporting a reduction of symptoms between the experimental and placebo groups. p̂ 1 = 38/100 = 0.38, p̂ 2 = 21/100 = 0.21 (p̂1 - p̂2 )  Z p̂ 1 (1 - p̂ 1 ) p̂ 2 (1  p̂ 2 )  n1 n2 Sample size is adequate with at least 5 successes and failures in each group . 0.38(1  0.38) 0.21(1 0.21)  (0.38  0.21) 1.96 100 100 0.17 + 1.96 (0.063) 0.17 + 0.12 (0.05, 0.29) b. Estimate the relative risk for reduction in symptoms between groups R̂R = p̂ 1/ p̂ 2 = 0.38/0.21 = 1.81 c. Estimate the odds ratio for reduction in symptoms between groups ÔR = p̂1 /(1 p̂1 ) 0.38 / 0.62 0.61   = 2.26. p̂2 /(1 p̂ 2 ) 0.21 / 0.79 0.27 d. Generate a 95% confidence interval for the relative risk Ln(R̂R)  Z (n1 - x 1 )/x1 (n2 - x 2 )/x2  n1 n2 Sample size is adequate with at least 5 successes and failures in each group. Ln(1.81)  1.96 62/38 79/21  100 100 0.59 + 1.96 (0.232) 0.59 + 0.455 (0.135, 1.045) The 95% confidence interval for Ln(RR) is (0.135, 1.045). To generate the confidence interval for the relative risk, we take the antilog (exp) of the lower and upper limits: (exp(0.135), exp(1.045)) (1.14, 2.84) 8. The following table displays descriptive statistics on the participants involved in the study described in Problem 7. Characteristic Mean (SD) Age, years % Males Mean (SD) Educational Level, years Mean (SD) Annual Income, $000s Mean Body Mass Index Experimental Medication (n=100) 47.2 (4.3) 46% 13.1 (2.9) $36.560 ($1,054) 24.7 (2.7) Placebo (n=100) 46.1 (5.1) 58% 14.2 (3.1) $37.470 ($998) 25.1 (2.4) a. Generate a 95% confidence interval for the mean age among participants assigned to the placebo. s XZ n 5.1 46.1 (1.960) 100 46.1 + 0.999 (45.1, 47.1) b. Generate a 95% confidence interval for the difference in mean ages in participants assigned to the experimental versus the placebo groups. 1 1 (X1 - X 2 )  ZSp  n1 n 2 Check s12/s22 = 4.32/5.12 =0.71 Sp  Sp  (n1  1)s12  (n2  1)s22 n1  n 2  2 (100  1)4.32  (100 - 1)5.12 100  100  2 (47.2 - 46.1)  1.960(4.72)  22.25  4.72 1 1 .  100 100 1.1 + 1.31 (-0.21, 2.41) c. Generate a 95% confidence interval for the difference in mean body mass index in participants assigned to the experimental versus the placebo groups. (X1 - X 2 )  ZSp 1 1  n1 n 2 Check s12/s22 = 2.72/2.42 =1.26 Sp  (n1  1)s12  (n2  1)s22 n1  n 2  2 Sp  (100 1)2.7 2  (100 - 1)2.4 2  6.52  2.55 100  100  2 1 1  . 100 100 (24.7 - 25.1)  1.960(2.55) -0.4 + 0.71 (-1.11, 0.31) 9. A crossover trial is conducted to compare an experimental medication for migraine headaches to a currently available medication. A total of 50 patients are enrolled in the study and each patient receives both treatments. The outcome is the time, in minutes, until the headache pain resolves. Following each treatment, patients record the time it takes until pain is resolved. Treatments are assigned in random order (i.e., some patients receive the currently available medication first and then the experimental medication and others receive the experimental medication first and then the currently available medication). The mean difference in times between the experimental and currently available medication is -9.4 minutes, with a standard deviation of 2.8 minutes. Construct a 95% confidence interval for the mean difference in times between the experimental and currently available medication. Xd Z sd n - 9.4  (1.960) 2.8 50 -9.4 + 0.78 (-10.2, -8.6) 10. Suppose in the study described in Problem 9, each participant is also asked if the assigned medication causes any stomach upset. Among the 50 participants, 12 report stomach upset with the experimental medication. Construct a 90% confidence interval for the proportion of participants who might experience stomach upset with the experimental medication. p̂ = 12/50 = 0.24 p̂(1 - p̂) n Sample size is adequate with at least 5 successes and failures in the sample. p̂  Z 0.24(1- 0.24) 50 0.24 + 0.099 0.24  1.645 (0.14, 0.34) 11. A 95% confidence interval for the mean diastolic blood pressure in women is 62 to 110. What is the margin of error? What is the standard error? CI has form: 86 + 24 Margin of error is 24 (1.96*Standard Error = 24), thus Standard Error = 12.24. 12. The following data are collected from 10 randomly selected patients undergoing physical therapy following knee surgery. The data represent the percent gain in range of motion after 3 weeks of therapy. Generate a 95% confidence interval for the mean percent gain in range of motion. 24% 32% 50% 62% 21% X s 45% 80% 24% 30% 10% 378 = 37.8 10 ΣX2  ( X) 2 /n 18,386  (378)2 /10 18,386  14,288.4    455.29  21.3 n 1 10 1 9 Xt s , df = n – 1 = 10 – 1 = 9 n For 95% confidence, t = 2.262 s Xt n 21.3 37.8  (2.262) 10 37.8 + 15.24 (22.6, 53.0) 13. The following data are collected in a randomized trial to test the efficacy of a new drug for migraine headaches. The following are characteristics of study participants overall and then organized by the treatment to which they are assigned. Mean (SD) Age, years % Male % High School Graduate Severity of Migraine Headaches % Mild % Moderate % Severe Median (Q1-Q3) Number of Days Missed Work in Past Year Due to Migraine Min-Max Number of Days Missed Work in Past Year Due to Migraine a. New Drug (n=100) 32.8 (4.7) 54% 76% Placebo (n=100) 31.9 (5.1) 48% 80% All (n=200) 32.0 (4.9) 51% 78% 22% 38% 40% 20% 42% 38% 21% 39% 39% 5 (3-12) 6 (2-18) 6 (3-17) 0-35 0-48 0-48 Generate a 95% confidence interval for the difference in mean ages between groups. (X1 - X 2 )  ZSp 1 1  n1 n 2 Check s12/s22 = 4.72/5.12 =0.85 Sp  Sp  (n1  1)s12  (n2  1)s22 n1  n 2  2 (100 1)4.7 2  (100 - 1)5.12 100  100  2 (32.8 - 31.9)  1.960(4.90) 0.9 + 1.36 (-0.46, 2.26)  24.05  4.90 1 1 .  100 100 b. Generate a 95% confidence interval for the difference in proportions of males between groups. (p̂1 - p̂2 )  Z p̂ 1 (1 - p̂ 1 ) p̂ 2 (1  p̂ 2 )  n1 n2 Sample size is adequate with at least 5 successes and failures in each group. 0.54(1  0.54) 0.48(1  0.48)  100 100 (0.54 0.48) 1.96 0.06 + 1.96 (0.071) 0.06 + 0.14 (-0.08, 0.20) c. Generate a 95% confidence interval for the difference in proportions of patients with severe migraine headaches between groups. (p̂1 - p̂ 2 )  Z p̂ 1 (1 - p̂ 1 ) p̂ 2 (1  p̂ 2 )  n1 n2 Sample size is adequate with at least 5 successes and failures in each group. (0.40 0.38) 1.96 0.40(1  0.40) 0.38(1  0.38)  100 100 0.02 + 1.96 (0.069) 0.02 + 0.13 (-0.11, 0.15) 14. A clinical trial is run to assess the efficacy of a new pacemaker device in patients with atrial fibrillation (AF). Two hundred participants are randomized to receive the new pacemaker or a currently available pacemaker. There are two primary outcomes of interest – the number of days in a three month period with an atrial fibrillation event and hospitalization for atrial fibrillation over the three month follow-up period. Data on baseline characteristics and the outcomes are shown below. Baseline Characteristics Mean (SD) Age, years % Male New Pacemaker (n=100) 67.3 (5.9) 48% Available Pacemaker (n=100) 66.9 (5.6) 52% 8.4 (3.2) 14.9 (3.9) 4% 9% Outcomes Mean (SD) Number of days with AF event % Hospitalized for AF a. Compute a 95% confidence interval for the difference in mean number of days with an AF event between participants receiving the new pacemaker as compared to the available pacemaker. (X1 - X 2 )  ZSp 1 1  n1 n 2 Check s12/s22 = 3.22/3.92 = 0.67 Sp  Sp  (n1  1)s12  (n2  1)s22 n1  n 2  2 (100 1)3.22  (100 - 1)3.92 100  100  2  12.73  3.56 1 1  . 100 100 (8.4 - 14.9)  1.960(3.56) -6.5 + 0.99 (-7.49, -5.51) b. Compute a 95% confidence interval for the mean number of days with an AF event among participants receiving the new pacemaker. XZ s 8.4  (1.960) n 3.2 100 8.4 + 0.63 (7.77, 9.03) 15. A pilot study is run to investigate the feasibility of recruiting pregnant women into a study of risk factors for preterm delivery. Women are invited to participate at their first clinical visit for prenatal care. The following represent the gestational ages in weeks of women who consented to participate in the study. Compute a 95% confidence interval for the mean gestational age of women enrolling in the study. 11 14 21 22 X s 9 10 13 18 118 8 = 14.8 ΣX2  ( X) 2 /n 1916  (118)2 /8 1916 - 1740.5    25.07  5.0 n 1 8 1 7 Xt s n , df = n – 1 = 8 – 1 = 7 For 95% confidence, t = 2.365 s Xt n 5.0 14.8  (2.365) 8 14.8 + 4.18 (10.6, 19.0) 16. A clinical trial is run comparing a new drug for high cholesterol to a placebo. A total of 40 participants are randomized (with equal assignment to treatments) to receive either the new drug or placebo. Their total serum cholesterol levels are measured after eight weeks on the assigned treatment. Participants receiving the new drug reported a mean total serum cholesterol level of 209.5 (std dev = 21.6) and participants receiving the placebo reported a mean total serum cholesterol level of 228.1 (std dev = 19.7). a. Construct a 95% confidence interval for the difference in mean total serum cholesterol levels between participants receiving the new drug versus placebo. 1 1  n1 n 2 df = n1 + n2 -2 = 20 + 20 -2 = 38 (X1 - X 2 )  tSp Check s12/s22 =21.62/19.72 = 1.20 Sp  Sp  (n1  1)s12  (n2  1)s22 n1  n 2  2 (20  1)21.62  (20 - 1)19.72 20  20  2 (209.5 - 228.1)  2.024(20.7)  427.33  20.7 1 1  . 20 20 -18.6 + 13.25 (-31.85, -5.35) b. Is the new drug effective? Justify briefly. Yes. The 95% confidence interval for the difference in means does not include the null value (0) and thus there is a statistically significant difference in mean total serum cholesterol levels between participants receiving the new drug versus placebo.