Hi evry1, im really stuck with this question for confidence intervals as i dont have a sample stand. deviation?
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A study showed that out of 500 people that took a weight loss drug, 70 developed side effects.Compute a 95% confidence interval for the proportion of people that developed side effects after taking the drug.
How do i do this without a standard deviation of any kind?
Thanks so much!
ANSWER: Manufacturer can claim to a 95% confidence interval only (11.4% , 16.6%) of people will develop side effects after taking the drug.
Why???
POPULATION PROPORTION, CONFIDENCE INTERVAL, NORMAL DISTRIBUTION
95% CONFIDENCE INTERVAL p +/- (z critical value) * SQRT[p * (1 - p)/n]
p = POPULATION PROPORTION [0.14] (14%) 70/500 = 0.14
z critical value [1.645] from Table Look-up Normal Distribution
n = SAMPLE SIZE [500]
95% CONFIDENCE INTERVAL = 0.14+/- 1.645 * SQRT [ 0.14 * (1 - 0.14)/500] = [0.114,0.166] (11.4% , 16.6%)
CONCLUSION: 95% CONFIDENCE INTERVAL of true POPULATION PROPORTION = [0.114, 0.166] (11.4% , 16.6%)
Manufacturer can claim to a 95% confidence interval only (11.4% , 16.6%) of people will develop side effects after taking the drug.