We are tasked with finding a 98% confidence interval for the population mean amount of ground toxins based on the provided sample data. The steps to calculate this confidence interval are as follows:

1. Determine the sample mean ($\bar{x}$): First, we calculate the mean of the given data points.

2. Determine the sample standard deviation ($s$): We need the standard deviation of the sample to estimate the population standard deviation.

3. Find the critical value for 98% confidence level: Since the population standard deviation is not given and the sample size is less than 30, we'll use the t-distribution. The degrees of freedom for the t-distribution will be $df = n - 1 = 17$.

4. Calculate the margin of error (ME): The margin of error is calculated as: $\text{ME} = t_{\alpha/2} \times \frac{s}{\sqrt{n}}$ where $t_{\alpha/2}$ is the t-critical value for 98% confidence, $s$ is the sample standard deviation, and $n$ is the sample size.

5. Construct the confidence interval: The confidence interval is given by: $\left( \bar{x} - \text{ME}, \bar{x} + \text{ME} \right)$

Let me now calculate the sample mean, sample standard deviation, and the confidence interval.The calculations yield the following results:

• Sample Mean ($\bar{x}$): 20.1 ppm
• Sample Standard Deviation (s): 11.2 ppm
• T-critical value (for 98% confidence with 17 degrees of freedom): 2.567
• Margin of Error (ME): 6.8 ppm

Thus, the 98% confidence interval for the population mean amount of ground toxins is:

$13.3 \, \text{ppm} < \mu < 26.9 \, \text{ppm}$

Would you like more details on any step or have questions?

Here are 5 related questions you could explore:

1. How does increasing the sample size affect the confidence interval?
2. What if the confidence level was 95% instead of 98%? How would the interval change?
3. How does the t-distribution differ from the normal distribution?
4. Why do we use the t-distribution instead of the z-distribution in this case?
5. How is the margin of error related to the sample standard deviation?

Tip: The margin of error decreases as the sample size increases, providing a more precise estimate of the population mean.