Question

1. several factors are involved in the creation of a confidence interval. among them are the sample size, the level of confidence, and the margin of error. which statements are true?

a) for a given sample size, higher confidence means a smaller margin of error.
b) for a specified confidence level, larger samples provide smaller margins of error.
c) for a fixed margin of error, larger samples provide a greater confidence.
d) for a given confidence level, halving the margin of error requires a sample twice as large.
e) for a given sample size reducing the margin of error will mean lower confidence.
f) for a certain confidence level, you can get a smaller margin of error by selecting a bigger sample.
g) for a fixed margin of error, smaller samples will mean lower confidence.
h) for a given confidence level, a sample 9 times as large will make a margin of error one third as big.

Explanation:

Step1: Recall the formula for margin of error

The formula for margin of error \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\), where \(z_{\alpha/2}\) is related to the confidence level, \(\sigma\) is the standard deviation (assumed constant here), and \(n\) is the sample size.

Step2: Analyze statement (a)

For a given \(n\), higher confidence means a larger \(z_{\alpha/2}\). Since \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\), a larger \(z_{\alpha/2}\) gives a larger \(E\). So statement (a) is false.

Step3: Analyze statement (b)

For a specified confidence level (fixed \(z_{\alpha/2}\)), as \(n\) increases, \(\frac{\sigma}{\sqrt{n}}\) decreases (because \(\sqrt{n}\) increases). Since \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\), \(E\) decreases. So statement (b) is true.

Step4: Analyze statement (c)

For a fixed \(E\), if \(n\) increases, \(\frac{\sigma}{\sqrt{n}}\) decreases. To keep \(E\) fixed (\(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\)), \(z_{\alpha/2}\) must increase. A larger \(z_{\alpha/2}\) corresponds to a greater confidence level. So statement (c) is true.

Step5: Analyze statement (d)

If \(E_1=z_{\alpha/2}\frac{\sigma}{\sqrt{n_1}}\) and \(E_2 = \frac{E_1}{2}=z_{\alpha/2}\frac{\sigma}{\sqrt{n_2}}\). Then \(\frac{E_1}{2}=z_{\alpha/2}\frac{\sigma}{\sqrt{n_2}}\) and \(E_1=z_{\alpha/2}\frac{\sigma}{\sqrt{n_1}}\). Substituting \(E_1\) gives \(\frac{z_{\alpha/2}\frac{\sigma}{\sqrt{n_1}}}{2}=z_{\alpha/2}\frac{\sigma}{\sqrt{n_2}}\), simplifying to \(n_2 = 4n_1\). So statement (d) is false.

Step6: Analyze statement (e)

For a given \(n\), if \(E\) is reduced ( \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\)), then \(z_{\alpha/2}\) must be reduced. A smaller \(z_{\alpha/2}\) corresponds to a lower confidence level. So statement (e) is true.

Step7: Analyze statement (f)

For a certain confidence level (fixed \(z_{\alpha/2}\)), as \(n\) increases (bigger sample), \(\frac{\sigma}{\sqrt{n}}\) decreases. Since \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\), \(E\) decreases. So statement (f) is true.

Step8: Analyze statement (g)

For a fixed \(E\), if \(n\) decreases (smaller sample), \(\frac{\sigma}{\sqrt{n}}\) increases. To keep \(E\) fixed (\(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\)), \(z_{\alpha/2}\) must decrease. A smaller \(z_{\alpha/2}\) corresponds to a lower confidence level. So statement (g) is true.

Step9: Analyze statement (h)

Let \(E_1=z_{\alpha/2}\frac{\sigma}{\sqrt{n_1}}\) and \(n_2 = 9n_1\). Then \(E_2=z_{\alpha/2}\frac{\sigma}{\sqrt{n_2}}=z_{\alpha/2}\frac{\sigma}{\sqrt{9n_1}}=\frac{1}{3}z_{\alpha/2}\frac{\sigma}{\sqrt{n_1}}=\frac{E_1}{3}\). So statement (h) is true.

Answer:

b) True, c) True, e) True, f) True, g) True, h) True