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 critical value

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

Step2: Calculate the standard error for \( n = 50\)

The formula for the standard error \( SE=\frac{\sigma}{\sqrt{n}}\), where \(\sigma = 12\) and \(n = 50\).

$$SE=\frac{12}{\sqrt{50}}\approx\frac{12}{7.07}\approx1.70$$

Step3: Calculate the confidence interval for \( n = 50\)

The formula for the confidence interval is \(\bar{x}\pm z\times SE\). Assuming \(\bar{x} = 42\), the confidence interval is \(42\pm1.96\times1.70\).

$$42 - 1.96\times1.70=42- 3.33 = 38.67$$

$$42 + 1.96\times1.70=42 + 3.33=45.33$$

The 95% confidence interval is \((38.67,45.33)\)

Step4: Calculate the standard error for \( n = 100\)

Using the formula \(SE=\frac{\sigma}{\sqrt{n}}\), with \(\sigma = 12\) and \(n = 100\)

$$SE=\frac{12}{\sqrt{100}}=\frac{12}{10} = 1.2$$

Step5: Calculate the confidence interval for \( n = 100\)

Using \(\bar{x}\pm z\times SE\) with \(z = 1.96\), \(\bar{x}=42\)

$$42-1.96\times1.2=42 - 2.35=39.65$$

$$42 + 1.96\times1.2=42+2.35 = 44.35$$

The 95% confidence interval is \((39.65,44.35)\)

Answer:

a) The 95% confidence interval for \(n = 50\) is \((38.67,45.33)\)
b) When \(n\) increases from \(50\) to \(100\) (while keeping the confidence level and \(\bar{x},\sigma\) the same), the confidence interval becomes narrower. This is because the standard error \(SE=\frac{\sigma}{\sqrt{n}}\) decreases as \(n\) increases. A smaller standard error leads to a smaller margin of error (\(z\times SE\)), so the interval is more precise.