The following data represent the pH of rain for a random sample of 12 rain dates in a particular region. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. The sample standard deviation is sequals0.332. Construct and interpret a 99​% confidence interval for the standard deviation pH of rainwater in this region. 4.66
5.22
5.10
4.71
4.77
4.86
5.82
4.85
5.10
4.84
4.76
4.64



To construct a 99% confidence interval for the standard deviation of the pH of rainwater, we will use the chi-square distribution because the data is assumed to be normally distributed, and we are working with a sample standard deviation.

### Given:
- Sample size \( n = 12 \)
- Sample standard deviation \( s = 0.332 \)
- pH values:  
  4.66, 5.22, 5.10, 4.71, 4.77, 4.86, 5.82, 4.85, 5.10, 4.84, 4.76, 4.64

### Steps to Calculate the Confidence Interval for the Standard Deviation:

1. **Calculate the sample variance**:
   \[
   s^2 = (0.332)^2 = 0.110224
   \]

2. **Degrees of freedom**:
   \[
   df = n - 1 = 12 - 1 = 11
   \]

3. **Determine the chi-square critical values** for a 99% confidence interval. Using the chi-square distribution table (or a calculator), we find the chi-square critical values for 11 degrees of freedom at the 0.5% and 99.5% percentiles (since we want 99% confidence, we look at the outer 0.5% tails):
   - \(\chi^2_{0.005,11} \approx 3.053\)
   - \(\chi^2_{0.995,11} \approx 26.757\)

4. **Compute the confidence interval for the population variance** \(\sigma^2\):
   \[
   \frac{(n - 1) s^2}{\chi^2_{0.995}} \leq \sigma^2 \leq \frac{(n - 1) s^2}{\chi^2_{0.005}}
   \]
   Substituting the values:
   \[
   \frac{11 \times 0.110224}{26.757} \leq \sigma^2 \leq \frac{11 \times 0.110224}{3.053}
   \]
   \[
   0.0453 \leq \sigma^2 \leq 0.3976
   \]

5. **Take the square root** to find the confidence interval for the population standard deviation \(\sigma\):
   \[
   \sqrt{0.0453} \leq \sigma \leq \sqrt{0.3976}
   \]
   \[
   0.213 \leq \sigma \leq 0.631
   \]

### Interpretation:
We are 99% confident that the true population standard deviation of the pH of rainwater in this region lies between 0.213 and 0.631.

---

Would you like further details or have any questions on this process?

Here are 5 questions for you to explore based on this:

1. How do changes in sample size affect the width of the confidence interval?
2. What assumptions must hold for the chi-square distribution to be applicable in this context?
3. How would the interval change if the confidence level were reduced to 95%?
4. Can you compare this interval with one for the mean and discuss the differences?
5. What role do outliers play in computing confidence intervals for standard deviation?

**Tip:** Increasing the confidence level widens the confidence interval but also increases certainty in capturing the true parameter.

The following data represent the pH of rain for a random sample of 12 rain dates in a particular region. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. The sample standard deviation is s = 0.332. Construct and interpret a 99​% confidence interval for the standard deviation pH of rainwater in this region. 4.66, 5.22, 5.10, 4.71, 4.77, 4.86, 5.82, 4.85, 5.10, 4.84, 4.76, 4.64

Learn how to construct and interpret a 99% confidence interval for the standard deviation of pH in rainwater. Using a sample of 12 rain dates and a standard deviation of 0.332, this example explains the use of chi-square distribution and confidence intervals in statistics.

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