Question

The EPA limit on the allowable discharge of suspended solids into rivers and streams is 60 milligrams per liter ( \(\mathrm{mg} /\) l) per day. A study of water samples selected from the discharge at a phosphate mine shows that over a long period, the mean daily discharge of suspended solids is \(48 \mathrm{mg} /\), but day-to-day discharge readings are variable. State inspectors measured the discharge rates of suspended solids for \(n=20\) days and found \(s^{2}=39(\mathrm{mg} / \mathrm{I})^{2}\). Find a \(90 \%\) confidence interval for \(\sigma^{2}\). Interpret your results.

Short answer

Based on a sample size of 20, with a sample variance of 39 $(\text{mg}/\text{l})^2$ and a desired 90% confidence level, we found that the confidence interval for the population variance of daily discharge rates of suspended solids at the phosphate mine is between 24.57 $(\text{mg}/\text{l})^2$ and 73.28 $(\text{mg}/\text{l})^2$. This indicates that there is substantial variability in daily discharge readings, which may require further investigation and monitoring to better understand and manage the situation.

Step-by-step solution

Identify the given information

We are given the following information: - Sample size, \(n = 20\) - Sample variance, \(s^2 = 39 \, (\text{mg}/\text{l})^2\) - Confidence level, \(1 - \alpha = 0.90\)

Calculate the degrees of freedom

To calculate the degrees of freedom, we use the formula: $$ \text{Degrees of freedom} = n - 1 $$ So, we have: $$ \text{Degrees of freedom} = 20 - 1 = 19 $$

Find the chi-square values

We need to find the critical values of the Chi-square distribution for the given degrees of freedom and the level of significance. We will use a Chi-square table or calculator to find the values of \(\chi^2_{0.05, 19}\) and \(\chi^2_{0.95, 19}\). Using a Chi-square table or calculator, we find: $$ \chi^2_{0.05, 19} = 30.144 $$ $$ \chi^2_{0.95, 19} = 10.117 $$

Calculate the confidence interval

Now we plug the values into the formula for the confidence interval of a variance: $$ \frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}} ~ \le ~ \sigma^2 ~ \le ~ \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}} $$ So, our confidence interval becomes: $$ \frac{(20-1)(39)}{30.144} ~ \le ~ \sigma^2 ~ \le ~ \frac{(20-1)(39)}{10.117} $$ This simplifies to: $$ \frac{741}{30.144} ~ \le ~ \sigma^2 ~ \le ~ \frac{741}{10.117} $$ Calculating the left and right limits of the interval, we get: $$ 24.57 ~ \le ~ \sigma^2 ~ \le ~ 73.28 $$ Our 90% confidence interval for the population variance, \(\sigma^2\), is \((24.57, 73.28) \, (\text{mg}/\text{l})^2\).

Interpret the results

We can interpret the results as follows: we are 90% confident that the true variance of the daily discharge rates of suspended solids at the phosphate mine lies between \(24.57 \, (\text{mg}/\text{l})^2\) and \(73.28 \, (\text{mg}/\text{l})^2\). This means that there is still considerable variability in daily discharge readings, which may warrant further investigation and monitoring.

Key concepts

Variance

Variance is a fundamental concept in statistics that measures how widely data points in a set differ from the mean of that set. It's like determining how "spread out" the data is. Variance is denoted as \( s^2 \) for a sample and \( \sigma^2 \) for an entire population.

In our context, the variance of water discharge rates tells us about the variability in the discharge of suspended solids. It's the average of the squared differences from the mean. Here's why variance matters:

• Helps identify data consistency: A small variance means data points are close to the mean, suggesting consistency.
• Detects variability: A larger variance indicates more spread out data, hinting at high variability day-to-day.
• Assists in quality control: In this case, understanding variance can help regulate discharges to ensure environmental standards are met.

Simply put, knowing the variance gives us a better grasp of how the daily discharge numbers are fluctuating around the average, in this case, around 48 mg/l.

Chi-Square Distribution

The chi-square distribution is crucial when you're dealing with variances, especially when estimating them or constructing confidence intervals like in this problem. It's a distribution that arises when you are analyzing variance from a sample.

Key points about the chi-square distribution:

• Asymmetric: The distribution is not symmetric, meaning it's skewed, usually has a peak towards the left, and spreads out rightwards. It starts at zero and moves positively since variance cannot be negative.
• Degrees of Freedom: Controlled by the sample size, degrees of freedom (\( n-1 \) in the formula) determine the shape of the chi-square distribution.
• Critical values: For constructing confidence intervals, you use critical values from this distribution. These are specific points on the curve that separates the distribution into tail values.

In our exercise, you used chi-square critical values to frame the interval around the sample variance. This is what made it possible to estimate the population variance with a given level of confidence.

Sample Size

Sample size is essentially the number of observations or measurements taken in a statistical sample. In our exercise, the sample size \( n \) is 20, meaning inspectors collected discharge data over 20 days.

Here's why sample size is critical:

• Affects Confidence: Larger samples tend to provide more reliable results as they better represent the population.
• Influences Variability: A smaller sample size can introduce more variability, making it harder to estimate the population variance accurately.
• Controls Precision: The sample size influences the degrees of freedom in statistical calculations like those with chi-square, affecting the confidence interval width.

By understanding sample size, you can judge the reliability and precision of your confidence interval. A careful choice of sample size balances between sufficient data collection and resource constraints. Here, a sample size of 20 days was chosen, which is a reasonable number for drawing some general insights yet leaves room for more data to increase reliability.