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 critical value

For a 95% confidence interval, the critical value \(z\) (assuming normal distribution) is \(z = 1.96\) (from standard normal distribution tables).

Step2: Calculate margin of error for \(n = 50\)

The formula for margin of error \(E=z\times\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 12\), \(n = 50\).

$$ E_1=1.96\times\frac{12}{\sqrt{50}}\approx1.96\times\frac{12}{7.07}\approx1.96\times1.697\approx3.33 $$

The confidence interval is \(\bar{x}\pm E_1\), where \(\bar{x} = 42\). So the interval is \((42 - 3.33,42 + 3.33)=(38.67,45.33)\)

Step3: Calculate margin of error for \(n = 100\)

Using the same formula \(E=z\times\frac{\sigma}{\sqrt{n}}\), with \(n = 100\)

$$ E_2=1.96\times\frac{12}{\sqrt{100}}=1.96\times1.2 = 2.35 $$

The confidence interval is \(\bar{x}\pm E_2\). So the interval is \((42- 2.35,42 + 2.35)=(39.65,44.35)\)

Answer:

a) The 95% confidence interval is \((38.67,45.33)\)
b) When \(n\) increases from \(50\) to \(100\), the confidence interval narrows. This is because as the sample size \(n\) increases, the margin of error \(E = z\times\frac{\sigma}{\sqrt{n}}\) decreases (since \(\sqrt{n}\) increases). A larger sample size provides more information and reduces the uncertainty, resulting in a more precise estimate of the population mean.