12.1.31 sample size without educated guess
For the specified margin of error and confidence level, obtain a sample size that will ensure a margin of error of at most the one specified.
margin of error = 0.05; confidence level = 95%Since confidence level is .95, \(\alpha = 1 - .95 = .05\). So \(\alpha/2=.05/2\)
We can find \(z_{\alpha/2}\) by using qnorm()
alpha = .05
zalpha2= abs(qnorm(alpha/2))
E = .05From question 12.1.25 we can find the margin of error for a 95% confidence interval by using the formular \(E=z_{\alpha/2}.\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)
This question is reverse, instead of finding E, the question gives us margin of error E and confidence level, ask us to find sample size n to meet the requirement.
As professor Covert introduces in chapter 12 worksheet, we can derive the formula \(E=z_{\alpha/2}.\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)
to \(n=\hat{p}.(1-\hat{p}).(\frac{z_{\alpha/2}}{E})^2\)
Since the proportion \(\hat{p}\) runs from 0 to 1 so \(\hat{p}.(1-\hat{p})\) will have the maximum value at \(\hat{p}=.5\) or \(\hat{p}.(1-\hat{p}) =.25\)
For example, if we let \(\hat{p} = x\) and \(y= \hat{p}.(1-\hat{p})\) we have \(y = -x^2 + x\). This parabola has vertex at \(x=-\frac{b}{2a}=-\frac{1}{2.(-1)}=\frac{1}{2}\) and \(y = -(\frac{1}{2})^2+\frac{1}{2}=\frac{1}{4}\)
Since we do not know the sample proportion \(\hat{p}\), we know that \(n \le \frac{1}{4}.(\frac{z_{\alpha/2}}{E})^2\)
The question asks obtain a sample size that will ensure a margin of error of at most the one specified.
We find the sample size by using the formula \(n = \frac{1}{4}.(\frac{z_{\alpha/2}}{E})^2\)
n = 0.25 * (zalpha2 / E)^2
n## [1] 384.1459Round up to nearest integer
ceiling(n)## [1] 385
Hope that helps!