12.1.37 sample size with educated guess

For the specified margin of error, confidence level, and educated guess for the observed value, obtain a sample size that will ensure a margin of error of at most the one specified (provided, of course, that that observed value of the sample proportion is further from 0.5 than the educated guess).

margin of error = 0.03; confidence level = 90%;educated guess = 0.56

Since confidence level is .90, \(\alpha = 1 - .9 = .1\). So \(\alpha/2=.1/2\)

We can find \(z_{\alpha/2}\) by using qnorm()

alpha = .1
zalpha2= abs(qnorm(alpha/2))
p = .56
E = .03

This question is similar to 12.1.31, in this case we have proportion \(\hat{p} = .56\)

From question 12.1.25 we can find the margin of error for a 90% 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 \(n=\hat{p}.(1-\hat{p}).(\frac{z_{\alpha/2}}{E})^2\)

We find the sample size by using the formula \(n=\hat{p}.(1-\hat{p}).(\frac{z_{\alpha/2}}{E})^2\)

n = p*(1-p) * (zalpha2 / E)^2
n

## [1] 740.7177

Round up to nearest integer

ceiling(n)

## [1] 741

Hope that helps!