The upper limit of a confidence interval at the 99% level of confidence for the population proportion if a sample of size 100 had 40 successes is about:Group of answer choices0.696.0.496.0.726.0.526. PreviousNext
Question
The upper limit of a confidence interval at the 99% level of confidence for the population proportion if a sample of size 100 had 40 successes is about:Group of answer choices0.696.0.496.0.726.0.526. PreviousNext
Solution
To find the upper limit of a confidence interval for a population proportion, we can use the formula:
p̂ ± Z * sqrt[(p̂(1 - p̂) / n]
where:
- p̂ is the sample proportion (number of successes divided by the sample size)
- Z is the Z-score for the desired confidence level
- n is the sample size
Here, p̂ = 40/100 = 0.4, n = 100, and for a 99% confidence level, Z is approximately 2.576 (you can find this value in a standard normal distribution table).
Substituting these values into the formula, we get:
0.4 ± 2.576 * sqrt[(0.4 * 0.6) / 100]
To find the upper limit, we add the second part to p̂:
0.4 + 2.576 * sqrt[(0.4 * 0.6) / 100] = 0.4 + 2.576 * sqrt[0.24 / 100] = 0.4 + 2.576 * sqrt[0.0024] = 0.4 + 2.576 * 0.049 = 0.4 + 0.12624 = 0.52624
So, the upper limit of the confidence interval at the 99% level of confidence for the population proportion is approximately 0.526.
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