The effect of a drug to change in heart beat by 5 beats per minute over 48 hours in either direction. A two sided test conducted at the 0,05 level of significance will be used. What sample size is needed for this study.
The mean for the population u = 75 beats per minute STD = 10 beats per minute .05 level of significance Power = 80%
What is the sample size needed?
Z = 1.645 because of the level of significance is 0.05 for H Z = beta 1-beta = .80 1-.80 = .20 = Z = .84
X = 75
n =[ (1.645 + .84)(10)}/70-75}]2 {(2.485 * 10)/-5}2= 24.5/-5 = 4.972 = 25 = n
To calculate the sample size needed for this study, we can use the formula:
n = [((Z + Zβ) * STD) / E]^2
Where:
- n is the sample size needed
- Z is the critical value (also known as the z-score) corresponding to the desired level of significance (α). In this case, α = 0.05, so Z = 1.645 for a two-sided test.
- Zβ is the critical value (z-score) corresponding to the desired power (1 - β). In this case, the power is 80%, so 1 - β = 0.80, and Zβ = 0.84.
- STD is the standard deviation of the population, which is 10 beats per minute.
- E is the desired margin of error, which is the change in heart rate of 5 beats per minute.
Putting in the values:
n = [((1.645 + 0.84) * 10) / 5]^2
= (2.485 * 10 / 5)^2
= (24.85 / 5)^2
= 4.97^2
= 24.7029
Since we need a whole number for the sample size, we can round up to the nearest whole number.
Therefore, the sample size needed for this study is approximately 25.