obesity is defined as a body mass index of 30kg/m2 or more. A 95% confidence interval for the percentage of U.S. adults aged 20 years and over who were obesity was found to be 22.4 to 23.5%. What was the sample size????
Can you please show your work?? Thanks
I think you need to provide the correct formulae for this one. I would link to some useful info, but this site doesn't allow linking.
To determine the sample size, we need to use the formula for confidence intervals given by:
n = (Z * σ / E)^2
Where:
- n is the sample size
- Z is the Z-value corresponding to the desired confidence level (in this case, 95% confidence corresponds to a Z-value of approximately 1.96)
- σ is the standard deviation (or the population standard deviation if known)
- E is the margin of error, which is half of the width of the confidence interval
In this case, we have the confidence interval (22.4% to 23.5%), which represents the margin of error. We can calculate the margin of error as half of the width:
E = (23.5% - 22.4%) / 2 = 0.55%
Now, we substitute the values into the formula:
n = (1.96 * σ / 0.55%)^2
Unfortunately, we don't have information about the standard deviation (σ), so we cannot calculate the exact sample size without that information. However, if we assume that the standard deviation is 1 (a standard value often used when the population standard deviation is unknown), we can proceed with the calculation:
n = (1.96 * 1 / 0.55%)^2
n = (1.96 * 100 / 0.55)^2
n ≈ 7565
Therefore, based on the assumption of a standard deviation of 1, the estimated sample size is approximately 7565.