Sample Standard Deviation
 The following data are collected as part of study of coffee consumption among undergraduate students. The following reflect cups per day consumed:
 4 6 8 2 1 0 2
 Compute the sample mean
 Compute the sample standard deviation
 Construct a 95% confidence interval for the mean number of cups of coffee consumed among all undergraduates.
 A clinical trial is conducted to evaluate 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, and the data are shown below
Pain Relief  No Pain Relief  
New Medication  44  76 
Placebo  21  99 
 Estimate the RD in pain relief between treatments
 Estimate the RR in pain relief between treatments
 Estimate the OR in pain relief between treatments
 Construct 95% CI for OR
 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 a selfreported 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 the placebo.
 Generate 95% CI for the difference in proportions of participants reporting a reduction of symptoms between the experimental and placebo groups.
 Estimate the RR for reduction in symptoms between groups
 Estimate OR for reduction in symptoms between groups
 Generate 95% CI for the RR
 A crossover trial is conducted to compare an experimental medication for migraine headaches to a currently available medication. A total 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 the pain is resolved. Treatments are assigned in random order. The mean different in times between the experimental and currently available medication is 9.4 minutes with a standard deviation of 2.8 minutes. Construct a 95% CI for the mean difference in times between the experimental and currently available medication.
 A 95% CI for the mean diastolic blood pressure in women is 62 to 110. What is the margin of error? What is the standard error?
 The data in Table 627 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.
New Drug
(n = 100) 
Placebo
(n = 100) 
All
(n = 200) 

Mean (SD) age, years  32.8 (4.7)  31.9 (5.1)  32.0 (4.9) 
% Male  54%  48%  51% 
% High school graduate  76%  80%  78% 
Severity of migraine headaches  
% Mild  22%  20%  21% 
% Moderate  38%  42%  39% 
% Severe  40%  38%  39% 
Median (Q1Q3)
Number of days missed work in past year due to migraine 
5 (3 – 12)  6 (2 – 18)  6 (3 – 17) 
Min – Max number of days missed work in past year due to migraine  0 – 35  0 – 48  0 – 48 
 Generate a 95% CI for the difference in mean ages between groups.
 Generate a 95% CI for the difference in proportion of male between groups.
 Generate a 95% CI for the difference in proportion of patients with severe migraine headaches between groups.
Solution
 The following data are collected as part of study of coffee consumption among undergraduate students. The following reflect cups per day consumed:
 4 6 8 2 1 0 2
 Compute the sample mean
Sample Mean = ∑x_{i} / n
= 3+4+6+8+2+1+0+2/8
= 3.25
 Compute the sample standard deviation
Sample Standard Deviation = ( ∑(x_{i} – µ)^{2}/(n1))^{1/2}
= ((33.25)^{2} + (43.25)^{2} + (63.25)^{2} + (83.25)^{2} + (23.25)^{2} + (13.25)^{2} + (03.25)^{2} + (23.25)^{2}/7)^{1/2}
= (0.0625 + 0.5625 + 7.5625 + 22.5625 + 1.5625 + 5.0625 + 10.5625 + 1.5625/7)^{1/2}
= 2.659
 Construct a 95% confidence interval for the mean number of cups of coffee consumed among all undergraduates.
95% CI = µ ± z x s/n^{1/2}
= 3.25 ± 1.96 x 2.659/8^{1/2}
= 3.25 ± 1.843
= (1.407 ,5.093)
 A clinical trial is conducted to evaluate 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, and the data are shown below
Pain Relief  No Pain Relief  
New Medication  44  76 
Placebo  21  99 
 Estimate the RD in pain relief between treatments
Risk Difference = 44/(44+76) – 21/(21+99)
= 0.192
 Estimate the RR in pain relief between treatments
Relative Risk = 44*(21+99) / 21*(44+76)
= 2.095
 Estimate the OR in pain relief between treatments
Odds Ratio = 44*99 / 21*76
= 2.729
 Construct 95% CI for OR
ln(OR) = ln(2.729) = 1.004
SE{ln(OR)} = (1/44 + 1/76 + 1/21 + 1/99)^{1/2}
= 0.306
95% CI = exp(ln(OR) ± 1.96*SE{ln(OR)})
= exp(1.004 ± 1.96*0.306)
= exp(1.004±0.5998)
= (exp(0.404), exp(1.604))
= (1.498, 4.973)
 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 a selfreported 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 the placebo.
Symptom Reduction  No Symptom Reduction  
Medication  38  62 
Placebo  21  79 
 Generate 95% CI for the difference in proportions of participants reporting a reduction of symptoms between the experimental and placebo groups.
RD = 38/(38+62) – 21(21+79) = 0.17
SE{RD} = (38×62/100^{3} + 21×79/100^{3})^{1/2} = 0.063
95% CI = RD ± 1.96 x SE{RD}
= 0.17 ± 1.96×0.063
= 0.17 ± 0.124
= (0.046, 0.294)
 Estimate the RR for reduction in symptoms between groups
RR = 38 x (21+79) / 21 x (38+62)
RR = 1.809
 Estimate OR for reduction in symptoms between groups
OR = 38×79/21×62 = 2.306
 Generate 95% CI for the RR
RR = 1.809
ln(RR) = ln(1.809) = 0.593
SE{ln(RR)} =(1/38 + 1/21 – 1/100 – 1/100)^{1/2} = 0.232
95% CI = exp(ln(RR) ± 1.96 x SE{ln(RR)} )
= exp( 0.593 ± 1.96×0.232)
= exp(0.593 ± 0.455)
= (exp(0.138), exp(1.048))
= (1.148, 2.852)
 A crossover trial is conducted to compare an experimental medication for migraine headaches to a currently available medication. A total 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 the pain is resolved. Treatments are assigned in random order. The mean different in times between the experimental and currently available medication is 9.4 minutes with a standard deviation of 2.8 minutes. Construct a 95% CI for the mean difference in times between the experimental and currently available medication.
95% CI = µ ± z x s/n^{1/2}
= 9.4 ± 1.96x(2.8/50^{1/2})
= 9.4 ± 0.776
= (10.176, 8.624)
 A 95% CI for the mean diastolic blood pressure in women is 62 to 110. What is the margin of error? What is the standard error?
Margin of error = (11062)/2 = 24
Standard Error = Margin of error/ 1.96 = 12.245
 The data in Table 627 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.
New Drug
(n = 100) 
Placebo
(n = 100) 
All
(n = 200) 

Mean (SD) age, years  32.8 (4.7)  31.9 (5.1)  32.0 (4.9) 
% Male  54%  48%  51% 
% High school graduate  76%  80%  78% 
Severity of migraine headaches  
% Mild  22%  20%  21% 
% Moderate  38%  42%  39% 
% Severe  40%  38%  39% 
Median (Q1Q3)
Number of days missed work in past year due to migraine 
5 (3 – 12)  6 (2 – 18)  6 (3 – 17) 
Min – Max number of days missed work in past year due to migraine  0 – 35  0 – 48  0 – 48 
 Generate a 95% CI for the difference in mean ages between groups.
M1M2 = 32.8 – 31.9 = 0.9
MSE = s1^{2} + s2^{2} /2 = 4.7^{2} + 5.1^{2} /2 = 24.05
S = (2MSE/n)^{1/2} = ((2*24.05)/100)^{1/2} = 0.693
95% CI = 0.9 ± 1.96×0.693
= 0.9 ± 1.358
= (0.458, 2.258)
 Generate a 95% CI for the difference in proportion of male between groups.
p1 = 0.54 p2 = 0.48
SE1 = (0.54*(10.54)/100)^{1/2} = 0.0498
SE2 = (0.48*(10.48)/100)^{1/2} = 0.04996
SE = (SE1^{2} + SE2^{2} )^{1/2} = 0.0706
95% CI = p1p2 ± 1.96 x SE
= 0.06 ± 1.96×0.0706
= (0.078, 0.198)
 Generate a 95% CI for the difference in proportion of patients with severe migraine headaches between groups.
p1 = 0.4 p2 = 0.38
SE1 = (0.4*(10.4)/100)^{1/2} = 0.04899
SE2 = (0.38*(10.38)/100)^{1/2} = 0.04854
SE = ( SE1^{2} + SE2^{2} ) ^{½} = 0.06896
95% CI = p1p2 ± 1.96 x SE
= 0.02 ± 1.96×0.06896
= 0.02 ± 0.135
= (0.115, 0.155)