27. making an argument your friend says that a margin of error can be any value between 0 and 100 percent…

Explanation:

Step1: Recall margin - of - error concept

The margin of error (E = z\frac{\sigma}{\sqrt{n}}) (for a confidence interval of the mean, where (z) is the z - score, (\sigma) is the standard deviation and (n) is the sample size). The margin of error can be (0) (when the sample is the entire population, (n = N)) and can approach (100%) (when the sample is very small and provides almost no information). So the statement that the margin of error can be any value between 0 and 100 percent, not including 0 or 100 percent is incorrect.

Step2: Analyze sample - size and margin - of error relationship

We know that (E = z\frac{\sigma}{\sqrt{n}}). If we want to change the margin of error from (E) to (E_1=\frac{1}{2}E), let the original sample size be (n) and the new sample size be (n_1). Then (E = z\frac{\sigma}{\sqrt{n}}) and (\frac{1}{2}E=z\frac{\sigma}{\sqrt{n_1}}). Substitute (E = z\frac{\sigma}{\sqrt{n}}) into (\frac{1}{2}E=z\frac{\sigma}{\sqrt{n_1}}), we get (\frac{1}{2}(z\frac{\sigma}{\sqrt{n}})=z\frac{\sigma}{\sqrt{n_1}}). Cross - multiply and simplify: (\sqrt{n_1}=2\sqrt{n}), then (n_1 = 4n).

Answer:

1. No
2. The expression is (4n)
3. The sample size must be increased 4 times to cut the margin of error in half.