The owner of Torpid Oaks B&B wanted to know the average distance its guests had traveled. A random sample of 16 guests showed a mean distance of 85 miles with a standard deviation of 32 miles. The 90 percent confidence interval (in miles) for the mean is approximately:
A. (71.0, 99.0)
B. (71.8, 98.2)
C. (74.3, 95.7)
D. (68.7, 103.2)
The correct answer and explanation is :
To determine the 90% confidence interval for the mean distance guests traveled to reach Torpid Oaks B&B, we use the t-distribution because the population standard deviation is unknown and the sample size is small (n = 16).
Given:
- Sample size (n) = 16
- Sample mean (x̄) = 85 miles
- Sample standard deviation (s) = 32 miles
- Confidence level = 90%
- Degrees of freedom (df) = n − 1 = 15
Step 1: Find the t-critical value for 90% confidence and 15 degrees of freedom.
From a t-distribution table (or calculator), the t₀.05 (two-tailed) for df = 15 is approximately:
t* ≈ 1.753
Step 2: Use the formula for the confidence interval:
[
\text{Confidence Interval} = \bar{x} \pm t^* \cdot \frac{s}{\sqrt{n}}
]
Substitute values:
[
\text{Margin of Error} = 1.753 \cdot \frac{32}{\sqrt{16}} = 1.753 \cdot \frac{32}{4} = 1.753 \cdot 8 = 14.024
]
[
\text{Lower Limit} = 85 – 14.024 = 70.976 \approx 71.0
]
[
\text{Upper Limit} = 85 + 14.024 = 99.024 \approx 99.0
]
Final Answer:
[
\boxed{(71.0,\ 99.0)}
]
Correct Option: A. (71.0, 99.0)
Explanation (300+ words):
Confidence intervals estimate the range in which the true population parameter (in this case, the mean distance traveled) lies, with a specified level of certainty. Here, the owner of the B&B wants to estimate the average distance their guests travel, based on a sample of 16 guests.
Because the sample size is small and the population standard deviation is unknown, we apply the t-distribution rather than the normal distribution. The t-distribution accounts for added variability and is broader than the normal curve, especially for small samples.
We used a confidence level of 90%, meaning we are 90% confident that the actual population mean falls within the calculated interval. This level of confidence implies that in 100 similar samples, 90 of the calculated intervals would contain the true population mean.
To find the confidence interval, we calculated the margin of error, which depends on the t-critical value and the standard error of the mean. The standard error is the standard deviation divided by the square root of the sample size, which quantifies how much the sample mean is expected to vary from the true mean.
Using a t-critical value of 1.753 (based on 15 degrees of freedom), we computed the margin of error to be approximately 14.024 miles. When added and subtracted from the sample mean of 85 miles, it yields a range from 71.0 to 99.0 miles.
Thus, the correct 90% confidence interval is (71.0, 99.0), and the answer is A.