The owner of Limp Pines Resort wanted to know the average age of its clients. A random sample of 25 tourists is taken. It shows a mean age of 46 years with a standard deviation of 5 years. The width of a 98 percent CI for the true mean client age is approximately:
A. ± 1.711 years.
B. ± 2.326 years.
C. ± 2.492 years.
D. ± 2.797 years.
The correct answer and explanation is :
To determine the width of the 98% confidence interval (CI) for the true mean age of clients at Limp Pines Resort, we can follow these steps:
Step 1: Formula for Confidence Interval (CI)
The formula for a confidence interval for the population mean is given by:
[
\text{CI} = \bar{x} \pm (z_{\alpha/2} \times \frac{s}{\sqrt{n}})
]
Where:
- (\bar{x}) = sample mean
- (s) = sample standard deviation
- (n) = sample size
- (z_{\alpha/2}) = critical value from the standard normal distribution corresponding to the confidence level
The width of the confidence interval is given by:
[
\text{Width} = 2 \times (z_{\alpha/2} \times \frac{s}{\sqrt{n}})
]
Step 2: Identify Known Values
From the question, we know:
- Sample mean (\bar{x} = 46)
- Sample standard deviation (s = 5)
- Sample size (n = 25)
- Confidence level = 98%, so (\alpha = 0.02)
Step 3: Find the Critical Value (z_{\alpha/2})
The confidence level is 98%, which means (\alpha = 1 – 0.98 = 0.02). Since we are working with a two-tailed test, the critical value (z_{\alpha/2}) corresponds to the value that leaves 1% in each tail (0.02/2 = 0.01).
From standard normal tables or a z-table, the z-value for 98% confidence (i.e., the value that leaves 1% in each tail) is approximately 2.326.
Step 4: Calculate the Standard Error of the Mean
The standard error of the mean (SE) is given by:
[
SE = \frac{s}{\sqrt{n}} = \frac{5}{\sqrt{25}} = \frac{5}{5} = 1
]
Step 5: Calculate the Width of the Confidence Interval
Now, we can calculate the width of the confidence interval:
[
\text{Width} = 2 \times (z_{\alpha/2} \times SE) = 2 \times (2.326 \times 1) = 2 \times 2.326 = 4.652
]
Conclusion:
The correct width of the confidence interval is approximately 4.652, and since the options are in terms of “±,” the correct answer corresponds to the closest value for the margin of error (half the width of the confidence interval).
So, the answer is B. ± 2.326 years.
This means that the margin of error for the true mean age of clients at Limp Pines Resort is about 2.326 years in either direction from the sample mean of 46 years. This is the width for the 98% confidence interval.