Question

1. which of the following would result in the widest confidence interval?

a. a sample size of 30 with 99% confidence.
b. a sample size of 100 with 95% confidence
c. a sample size of 100 with 99% confidence.
d. a sample size of 30 with 95% confidence.

Explanation:

Step1: Recall the formula for confidence interval width

The width of a confidence interval (for a mean, assuming normal distribution and known or large - sample variance) is \(2\times z_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}\). Here, \(z_{\alpha/2}\) is the critical value, \(\sigma\) is the standard deviation (assumed constant), and \(n\) is the sample size.

Step2: Analyze the effect of confidence level and sample size on \(z_{\alpha/2}\) and \(\frac{1}{\sqrt{n}}\)

• For confidence level:
• The higher the confidence level, the larger the \(z_{\alpha/2}\) value. For a \(95\%\) confidence level, \(z_{\alpha/2}=1.96\). For a \(99\%\) confidence level, \(z_{\alpha/2} = 2.576\).
• For sample size:
• The function \(y=\frac{1}{\sqrt{n}}\) is a decreasing function of \(n\). As \(n\) increases, \(\frac{1}{\sqrt{n}}\) decreases.

Step3: Compare the options

• Option a: \(n = 30\), \(z_{\alpha/2}=2.576\), and \(\frac{1}{\sqrt{n}}=\frac{1}{\sqrt{30}}\approx0.183\)
• Option b: \(n = 100\), \(z_{\alpha/2}=1.96\), and \(\frac{1}{\sqrt{n}}=\frac{1}{10}=0.1\)
• Option c: \(n = 100\), \(z_{\alpha/2}=2.576\), and \(\frac{1}{\sqrt{n}}=\frac{1}{10}=0.1\)
• Option d: \(n = 30\), \(z_{\alpha/2}=1.96\), and \(\frac{1}{\sqrt{n}}=\frac{1}{\sqrt{30}}\approx0.183\)

• Calculate the product \(z_{\alpha/2}\times\frac{1}{\sqrt{n}}\) for each option:
• Option a: \(2.576\times0.183\approx0.471\)
• Option b: \(1.96\times0.1 = 0.196\)
• Option c: \(2.576\times0.1=0.258\)
• Option d: \(1.96\times0.183\approx0.359\)

Answer:

A. A sample size of 30 with 99% confidence.