Question

a study of 50 people living in crescent beach, bc, showed the average age as 42 years with a standard deviation of 12 years. (note: round invnorm answer to 2 decimal places for your calculations)
a) find the 95% confidence interval of the mean age for all the people living in crescent beach.
b) if the 95% confidence interval of the study stays the same, but we have 100 people instead of 50, what happens to the confidence interval? why?

Explanation:

Step1: Find the 95% confidence interval for part a)

The formula for the confidence interval for the population mean when the population standard deviation \(\sigma\) is known is \(\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\). For a 95% confidence interval, \(\alpha = 1 - 0.95=0.05\), and \(z_{\alpha/2}=z_{0.025}\). From the standard normal table, \(z_{0.025} = 1.96\). Given \(\bar{x}=42\), \(\sigma = 12\), and \(n = 50\).

The margin of error \(E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=1.96\times\frac{12}{\sqrt{50}}\)

First, calculate \(\frac{12}{\sqrt{50}}\approx\frac{12}{7.07}\approx1.697\)

Then \(E = 1.96\times1.697\approx3.33\)

The confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\), so \(42- 3.33<\mu<42 + 3.33\), which is \(38.67<\mu<45.33\)

Step2: Analyze the effect of sample size change in part b)

The formula for the margin of error \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\). We can see that \(E\) is inversely proportional to \(\sqrt{n}\) (since \(z_{\alpha/2}\) and \(\sigma\) are constant in this case). When \(n\) increases from \(n_1 = 50\) to \(n_2=100\)

Let \(E_1\) be the margin of error for \(n = 50\) and \(E_2\) be the margin of error for \(n = 100\)

\(E_1=z_{\alpha/2}\frac{\sigma}{\sqrt{50}}\) and \(E_2=z_{\alpha/2}\frac{\sigma}{\sqrt{100}}\)

\(\frac{E_2}{E_1}=\frac{\sqrt{50}}{\sqrt{100}}=\frac{1}{\sqrt{2}}\approx0.707\)

The confidence interval becomes narrower. As the sample size \(n\) increases, the standard error \(\frac{\sigma}{\sqrt{n}}\) decreases. Since the confidence interval is \(\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\), a smaller standard error leads to a narrower interval.

Answer:

a) The 95% confidence interval is \(38.67 <\mu<45.33\)

b) The confidence interval becomes narrower. Because the margin of error \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\) and as \(n\) increases (from \(50\) to \(100\)), the standard error \(\frac{\sigma}{\sqrt{n}}\) decreases.