Question

which of the following would result in the widest confidence interval?
a sample size of 100 with 95% confidence
a sample size of 100 with 99% confidence.
a sample size of 30 with 95% confidence.
a sample size of 30 with 99% confidence.
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Explanation:

Step1: Recall the formula for confidence interval

The formula for a confidence interval for a population mean (when population standard deviation \(\sigma\) is known) is \(\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\), and the width of the confidence interval is \(2z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\). Here, \(z_{\alpha/2}\) depends on the confidence level, and \(n\) is the sample size.

Step2: Analyze the effect of confidence level

As the confidence level increases, the value of \(z_{\alpha/2}\) increases. For a \(95\%\) confidence level, \(z_{\alpha/2}=1.96\), and for a \(99\%\) confidence level, \(z_{\alpha/2} = 2.576\). So, higher confidence level leads to a larger \(z_{\alpha/2}\) value.

Step3: Analyze the effect of sample size

As the sample size \(n\) decreases, the value of \(\frac{\sigma}{\sqrt{n}}\) increases (since \(\sigma\) is a constant). For example, if \(n_1 = 30\) and \(n_2=100\), \(\frac{1}{\sqrt{30}}\approx0.1826\) and \(\frac{1}{\sqrt{100}} = 0.1\)

Step4: Compare the options

• Option 1: \(n = 100\), \(z_{0.025}=1.96\), width \(W_1=2\times1.96\times\frac{\sigma}{\sqrt{100}}= 0.392\sigma\)
• Option 2: \(n = 100\), \(z_{0.005}=2.576\), width \(W_2=2\times2.576\times\frac{\sigma}{\sqrt{100}}=0.5152\sigma\)
• Option 3: \(n = 30\), \(z_{0.025}=1.96\), width \(W_3=2\times1.96\times\frac{\sigma}{\sqrt{30}}\approx2\times1.96\times0.1826\sigma\approx 0.716\sigma\)
• Option 4: \(n = 30\), \(z_{0.005}=2.576\), width \(W_4=2\times2.576\times\frac{\sigma}{\sqrt{30}}\approx2\times2.576\times0.1826\sigma\approx0.94\sigma\)

Answer:

A sample size of 30 with 99% confidence.