Question

question
at a sample size of 30, you have a 95% confidence level that the sample mean has a confidence interval of 0. if you increase the sample size to 60, what do you expect to happen?
your response
type answer here...

Explanation:

Step1: Recall the formula for the confidence interval

The formula for the confidence interval for the population mean (when population standard deviation \(\sigma\) is known) is \(CI=\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\), where \(z_{\alpha/2}\) is the critical value, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. The margin of error \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\).

Step2: Analyze the effect of sample size on margin of error

We know that \(E\propto\frac{1}{\sqrt{n}}\). When \(n_1 = 30\), \(E_1=z_{\alpha/2}\frac{\sigma}{\sqrt{30}}\). When \(n_2 = 60\), \(E_2=z_{\alpha/2}\frac{\sigma}{\sqrt{60}}\).

We can find the ratio \(\frac{E_2}{E_1}=\frac{z_{\alpha/2}\frac{\sigma}{\sqrt{60}}}{z_{\alpha/2}\frac{\sigma}{\sqrt{30}}}=\sqrt{\frac{30}{60}}=\frac{1}{\sqrt{2}}\approx0.707\)

Since the confidence interval is \(2E\) (because \(CI=\bar{x}-E,\bar{x} + E\)), when the sample size \(n\) increases, the margin of error \(E\) decreases.

Answer:

The confidence interval will decrease.