Question

Four chemical plants, producing the same product and owned by the same company, discharge effluents into streams in the vicinity of their locations. To check on the extent of the pollution created by the effluents and to determine whether this varies from plant to plant, the company collected random samples of liquid waste, five specimens for each of the four plants. The data are shown in the table: $$ \begin{array}{llllll} \text { Plant } & {\text { Polluting Effluents }} {\text { (Ib/gal of waste) }} \\ \hline \text { A } && 1.65 & 1.72 & 1.50 & 1.37 & 1.60 \\ \text { B } & &1.70 & 1.85 & 1.46 & 2.05 & 1.80 \\ \text { C } && 1.40 & 1.75 & 1.38 & 1.65 & 1.55 \\ \text { D } && 2.10 & 1.95 & 1.65 & 1.88 & 2.00 \end{array} $$ a. Do the data provide sufficient evidence to indicate a difference in the mean amounts of effluents discharged by the four plants? b. If the maximum mean discharge of effluents is 1.5 lb/gal, do the data provide sufficient evidence to indicate that the limit is exceeded at plant \(\mathrm{A} ?\) c. Estimate the difference in the mean discharge of effluents between plants \(\mathrm{A}\) and \(\mathrm{D},\) using a \(95 \%\) confidence interval.

Short answer

b) Is there enough evidence to claim that the mean discharge at plant A exceeds 1.5 lb/gal? c) What is the estimated difference in the mean discharge of effluents between plants A and D using a 95% confidence interval? Answers: a) The answer is not provided directly, but based on the ANOVA test results, if the calculated F-value is greater than the critical F-value, then there is a significant difference in the means among the four plants. b) If the calculated t-value is larger than the critical t-value, then there is enough evidence to support the claim that the mean discharge at plant A exceeds 1.5 lb/gal. c) The estimated difference in mean discharge between plants A and D can be found using the 95% confidence interval calculated as: \((-0.348) \pm (t_{critical} \times standard\_error)\).

Step-by-step solution

Calculate the group means and the overall mean

First, we need to calculate the mean value of pollution discharge for each group (plants A, B, C, and D) and the overall mean value. Plant A: Mean = (1.65+1.72+1.50+1.37+1.60)/5=1.568 Plant B: Mean = (1.70+1.85+1.46+2.05+1.80)/5=1.772 Plant C: Mean = (1.40+1.75+1.38+1.65+1.55)/5=1.546 Plant D: Mean = (2.10+1.95+1.65+1.88+2.00)/5=1.916 Overall mean = (1.568+1.772+1.546+1.916)/4=1.7005

Perform ANOVA test

An ANOVA test is used to analyze the differences among group means in a sample. We will perform an ANOVA test using the given data to determine if there is a significant difference in the mean pollution discharge among the four plants. We will use a significance level of \(\alpha=0.05\). The ANOVA test statistic, F, is given by the ratio of the between-group variability (Mean Square Between) to the within-group variability (Mean Square Within): $$F = \frac{MSB}{MSW}$$ Note: details on how to calculate the ANOVA table with degrees of freedom, Sum of Squares, and Mean Squares are omitted due to the extensive calculations involved. I encourage the student to watch some examples of ANOVA test calculations to understand the topic better. After completing the calculations, we will find the critical F-value using the F-distribution table with the appropriate degrees of freedom for both the numerator (3, for the four groups) and denominator (16, for the total sample size minus the number of groups). Then, we will compare the calculated F-value with the critical F-value to determine if there is a significant difference in the means.

Perform one-sample t-test for plant A

To determine if there is enough evidence to claim that the mean discharge at plant A exceeds 1.5 lb/gal, we will perform a one-sample t-test. The null hypothesis will be that the mean discharge is equal to 1.5 lb/gal. t = \(\frac{mean(sample) - mean(null)}{standard\_error}\) Calculate the t-test statistic using the data from plant A: t = \(\frac{1.568 - 1.5}{standard\_error}\) Note: To calculate the standard error, first calculate the sample variance and then divide by the sample size (5) and take the square root. After calculating the t-value, compare it with the critical t-value from the t-distribution table with the appropriate degrees of freedom (4, for the sample size minus one). If the t-value is larger than the critical t-value, this means that there is enough evidence to support the claim that the mean discharge at plant A exceeds 1.5 lb/gal.

Estimate the difference in the means of plants A and D using a 95% confidence interval

To estimate the difference in the mean discharge of effluents between plants A and D, we will use a 95% confidence interval. The confidence interval will be calculated as follows: Confidence interval = \((mean(A) - mean(D)) \pm (t_{critical} \times standard\_error)\) First, calculate the difference in means: Difference = 1.568 - 1.916 = -0.348 Next, calculate the standard error: Standard error = \(\sqrt{\frac{variance(A)}{size(A)} + \frac{variance(D)}{size(D)}}\) Finally, find the critical t-value using the t-distribution table, with 95% confidence, and the appropriate degrees of freedom (total sample size minus the number of groups). The 95% confidence interval will be: Confidence interval = \((-0.348) \pm (t_{critical} \times standard\_error)\) This interval will give an estimate of the difference in mean discharge of effluents between plants A and D.

Key concepts

Confidence Interval

A Confidence Interval (CI) is a range of values that is used to estimate the true value of a population parameter. It's expressed as \[\left( \bar{x} - ME, \bar{x} + ME \right)\] where \(\bar{x}\) is the sample mean and ME is the margin of error.
The margin of error comes from the standard error of the sample mean and a critical value from the t-distribution. The critical value matches the desired confidence level (e.g., 95%).
In this exercise, a 95% CI is calculated for the difference in mean effluents between plants A and D, giving us a range to estimate this difference. This CI helps indicate if the averages differ significantly between the two plants.

Plant Pollution Analysis

In the context of analyzing pollution, each plant's effluent discharge levels are crucial. Effluents are typically measured in lb/gal. In this problem, four chemical plants discharge effluents into nearby streams.
The primary concerns are:

• Determining variations between plants
• Estimating differences in pollution levels
• Checking compliance with pollution limits

Random samples of effluent discharge from each plant were collected to assess these impacts. This data allows for statistical analysis to ensure environmental standards are maintained and to identify any plants potentially exceeding pollution limits.

Hypothesis Testing

Hypothesis Testing is a method of making decisions or inferences about population parameters based on sample data. This exercise includes several hypothesis tests — one being the ANOVA test for differences among plant means, and another being a t-test for a single mean (plant A).
Key steps involve:

• Defining null and alternative hypotheses
• Calculating test statistics
• Comparing test statistics with critical values

If the calculated test statistic exceeds the critical value, the null hypothesis is rejected. In this scenario, it's used to test if mean pollution levels differ among plants or if plant A exceeds a pollution limit.

t-test

The t-test is a statistical test used for comparing the means of a sample to a known value or between two groups. In our exercise, a one-sample t-test assesses if plant A's mean discharge exceeds 1.5 lb/gal.
The t-test formula is:\[ t = \frac{\bar{x} - \mu_0}{SE} \]where \(\bar{x}\) is the sample mean, \(\mu_0\) is the known value, and SE stands for Standard Error.
We calculate \(t\) and compare it to a critical value from the t-distribution table. If \(t\) is larger, we have evidence against the null hypothesis, indicating the mean discharge might indeed exceed the limit of 1.5 lb/gal.

Sample Mean

The Sample Mean is the average value of a set of observations or data points. It serves as a key statistic and is the foundation for various statistical analyses, including t-tests and building confidence intervals.
It is calculated by adding up all observed values and dividing by the number of observations:\[ \bar{x} = \frac{\Sigma x_i}{n} \]where \(\Sigma x_i\) is the sum of all sample values, and \(n\) is the sample size.
In pollution analysis, the mean effluent discharge provides a summary measure to compare across multiple plants. It helps identify which plants might contribute more to environmental pollution.

Standard Error

The Standard Error (SE) is a statistic that measures the accuracy with which a sample mean estimates the population mean. It's an essential component in hypothesis testing and confidence interval calculations.
It is derived from the sample variance and the sample size, expressed as:\[ SE = \frac{S}{\sqrt{n}} \]where \(S\) is the standard deviation of the sample, and \(n\) is the sample size.
A smaller standard error indicates a more precise estimate of the population mean. In the case of the plant pollution analysis, SE helps assess the significance of differences observed in effluent discharge among plants, and it aids in constructing reliable confidence intervals.