Question

1. making an argument your friend says that a margin of error can be any value between 0 and 100 percent, not including 0 or 100 percent. is your friend correct? yes no explain your reasoning 0 / 10000 word limit 28. abstract reasoning suppose a random sample of size n is required to produce a margin of error of ±e. write an expression in terms of n for the sample size needed to produce a margin of error to ±1/2e. how many times must the sample size be increased to cut the margin of error in half? the expression is 1 the sample size must be increased 2 times to cut the margin of error in half. correct answers

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.