Obesity is defined as a body mass index of 3 kg/m2 or more. A 95% confidence interval for the percentage of U.S. adults aged 29 years and over who were obese was found to be 22.4% to 22.5%. What was the sample size?
To find the sample size, we need the margin of error and the formula for calculating it. The margin of error (ME) can be calculated using the formula:
ME = (Z * σ) / √n
Where:
- Z is the Z-value corresponding to the desired confidence level (in this case, it is the Z-value for a 95% confidence interval, which is usually 1.96).
- σ is the standard deviation of the population (which we don't have in this case).
Since we don't have the standard deviation (σ) of the population, we can use the conservative estimate of p(1-p) as an approximation. Here, p is the proportion of the population that is obese, which lies within the confidence interval.
Given that the confidence interval is 22.4% to 22.5%, we can calculate the sample size as follows:
ME = (Z * √[p(1-p)]) / √n
Since the margin of error for a 95% confidence interval is usually 0.5%, we can write:
0.005 = (1.96 * √[0.224 * (1 - 0.224)]) / √n
Simplifying the equation, we get:
0.005 = (1.96 * √[0.224 * 0.776]) / √n
Now, we can solve for n:
√n = (1.96 * √[0.224 * 0.776]) / 0.005
n = [(1.96 * √[0.224 * 0.776]) / 0.005]^2
Calculating this formula gives us the sample size.
To determine the sample size, we need to understand the formula for calculating the margin of error and the confidence interval. The formula is as follows:
n = (Z * σ / E)^2
Where:
n = sample size
Z = Z-score (critical value) corresponding to the desired level of confidence (in this case, 95%)
σ = standard deviation (which acts as a measure of variability in the population)
E = margin of error
Now, let's examine the given information:
The 95% confidence interval is given as 22.4% to 22.5%. We can use the midpoint of this interval as an estimate for the population parameter. Thus, the point estimate for the percentage of U.S. adults aged 29 years and over who were obese is:
Point estimate = (22.4% + 22.5%) / 2 = 22.45%
The margin of error (E) is the difference between the upper and lower bounds of the confidence interval divided by 2:
E = (22.5% - 22.4%) / 2 = 0.05% / 2 = 0.025%
Now, we need to find the Z-score for a 95% confidence level. The Z-score corresponds to the area under the standard normal distribution curve. For a 95% confidence level, the Z-score is approximately 1.96.
Substituting the values into the sample size formula, we have:
n = (1.96 * σ / E)^2
Since we don't know the standard deviation (σ) of the population, we need to use an approximation based on the assumed worst-case scenario where p = q = 0.5 (maximum variance). This is often used when the true standard deviation is unknown.
Plugging in the values and solving for n:
n = (1.96 * 0.5 / 0.025)^2
n = 3841.69
Rounding up to the nearest whole number, the sample size for the study is approximately 3842.
Therefore, the sample size was approximately 3842.