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Statistics question with confidence intervals?
Obesity is defined as a body mass index (BMI) of 3 kg/m2 or more. A 95% confidence
interval for the percentage of California adults aged 20 years and over who were obese was
found to be 21% to 24.9%. What was the sample size?
If I'm right, then we are talking about a proportion here. I'm having trouble figuring out what p (or q) is so that I can work backwards and find n.
1 Answer
- Anonymous10 years agoFavorite Answer
We are given the interval [21%, 24.9%]. Since the confidence interval is defined as CI = p ± z*SE we can conclude that p itself must be halfway between the two ends of the interval as z*SE fixed. Therefore:
p = (21% + 24.9%) / 2 = 22.95% = 0.2295
q = 1 - p = 1 - 0.2295 = 0.7705
Furthermore, by comparing p to either of the two limits of the interval we find the value of z*SE:
z*SE = 0.249 - 0.2295 = 0.0195
Now, using your calculator or a table of values you find that z* ≈ 1.96 for 95%, hence we have got the following:
0.0195 = 1.96 * sqrt(pq / n) = 1.96 * sqrt(0.2295 * 0.7705 / n)
Rearranging this to solve for n:
0.0195 / 1.96 = sqrt(0.2295 * 0.7705 / n)
(0.0195 / 1.96)^2 = 0.2295 * 0.7705 / n
n = (0.2295 * 0.7705) / (0.0195 / 1.96)^2 ≈ 1786