A clinical trial is being conducted in order to determine the efficacy of a new drug that will be used to treat Crohn’s disease. The efficacy of the medication will not only be determined by the physical improvement of symptoms but will also be determined by using a blood test to examine the concentration of C-reactive protein (an inflammatory marker) in an individual’s blood. If the researcher wants the margin of error for the level of C-reactive protein to be less than or equal to 2.9 mg/dL and the standard deviation for C-reactive protein concentrations among Crohn’s disease patients was previously documented at 10.1 mg/dL, how many patients should be recruited for each group of individuals in the study, assuming a 95% confidence interval will be used to quantify the mean differences between the control group and the treatment group? n for the treatment group = 12 and n for the control group = 12 n for the treatment group = 14 and n for the control group = 14 n for the treatment group = 47 and n for the control group = 47 n for the treatment group = 94 and n for the control group = 94 --------I'm confused which formula to use-------- For 95% confidence, z = 1.96, Standard deviation = 10.1, E = 2.9 Sample size: n = ((1.96 x 10.1) / 2.9)² = 47 ------OR------- For 95% CI crtiical Z = 1.96 first sample standard deviation σ1 = 10.1 second sample standard deviation σ2 = 10.1 margin of error E = 2.9 required sample size n = (z/E)²(σ1²+σ2²) = 94
A clinical trial is being conducted in order to determine the efficacy of a new drug that will be used to treat Crohn’s disease. The efficacy of the medication will not only be determined by the physical improvement of symptoms but will also be determined by using a blood test to examine the concentration of C-reactive protein (an inflammatory marker) in an individual’s blood. If the researcher wants the margin of error for the level of C-reactive protein to be less than or equal to 2.9 mg/dL and the standard deviation for C-reactive protein concentrations among Crohn’s disease patients was previously documented at 10.1 mg/dL, how many patients should be recruited for each group of individuals in the study, assuming a 95% confidence interval will be used to quantify the
n for the treatment group = 12 and n for the control group = 12
n for the treatment group = 14 and n for the control group = 14
n for the treatment group = 47 and n for the control group = 47
n for the treatment group = 94 and n for the control group = 94
--------I'm confused which formula to use--------
For 95% confidence, z = 1.96, Standard deviation = 10.1, E = 2.9
n = ((1.96 x 10.1) / 2.9)² = 47
------OR-------
For 95% CI crtiical Z = 1.96
first sample standard deviation σ1 = 10.1
second sample standard deviation σ2 = 10.1
margin of error E = 2.9
required sample size n = (z/E)²(σ1²+σ2²) = 94
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