To find a 99% confidence interval for the mean pH in rainfall based on the sample data, we can use the following formula for the confidence interval:

$\text{CI} = \bar{x} \pm t \cdot \frac{s}{\sqrt{n}}$

where:

• $\bar{x}$ is the sample mean,
• $s$ is the sample standard deviation,
• $n$ is the sample size,
• $t$ is the critical value from the $t$-distribution with $n - 1$ degrees of freedom.

Step 1: Identify Given Information

• $\bar{x} = 3.1$
• $s = 0.6$
• $n = 40$
• Confidence level = 99%

Step 2: Find the Critical Value $t$

Since we need a 99% confidence interval, the significance level is $\alpha = 1 - 0.99 = 0.01$. With $n - 1 = 39$ degrees of freedom, we look up the critical $t$-value for a two-tailed test with $\alpha/2 = 0.005$.

Using a $t$-table or calculator:

• $t_{0.005, 39} \approx 2.708$.

Step 3: Calculate the Margin of Error

The margin of error is calculated as: $\text{Margin of Error} = t \cdot \frac{s}{\sqrt{n}}$

Substituting in our values: $\text{Margin of Error} = 2.708 \cdot \frac{0.6}{\sqrt{40}} \approx 2.708 \cdot 0.0949 \approx 0.257$

Step 4: Calculate the Confidence Interval

Now, we can find the confidence interval: $\text{CI} = \bar{x} \pm \text{Margin of Error} = 3.1 \pm 0.257$

So, the 99% confidence interval for the mean pH is approximately: $(3.100 - 0.257, 3.100 + 0.257) = (2.843, 3.357)$

Answer

The 99% confidence interval for the mean pH in rainfall is: $(2.843, 3.357)$

Would you like further explanation on any part?

Here are five related questions you might consider:

1. How does changing the confidence level affect the width of the confidence interval?
2. Why is the $t$-distribution used instead of the normal distribution in this case?
3. What would happen to the confidence interval if we had a larger sample size?
4. How do different levels of pH in rain impact various environments?
5. What factors contribute most to acid rain formation?

Tip: For greater accuracy in estimating population parameters, increase the sample size whenever possible.