The margin of error for the 92.5% confidence interval to measure the population estimate is 0.0349 .
We have the following confidence interval of proportions in a sample of n people who were polled, with a success probability of and a confidence level of.
In which n = 515
z is the z-score that has a p-value of 1 - 92.5 = 7.5
The margin of error is of:
80 of the 515 participants in a clinical trial exhibited improvement after the treatment.
This means that
92.5% confidence level
So , z is the value of Z that has a p-value of , so .
Margin of error:
[tex]Z_{\frac{\alpha}{2}}\cdot \sqrt{\frac{p(1-p)}{n} }[/tex] = 0.0349
The margin of error for the 92.5% confidence interval used to estimate the population proportion is of 0.0349 .
In statistics, a confidence interval describes the likelihood that a population parameter would fall between a set of values for a given percentage of the time. Confidence ranges that include 95% or 99% of anticipated observations are frequently used by analysts.
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