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Sewage and industrial pollutants dumped into a body of water can reduce the dissolved oxygen concentration and adversely affect aquatic species. In one study, weekly readings are taken from the same location in a river over a two- month period. $$ \begin{array}{cc} \text { Week Number } & \text { Dissolved } \mathrm{O}_{2}, \text { ppm } \\ \hline 1 & 4.9 \\ 2 & 5.1 \\ 3 & 5.6 \\ 4 & 4.3 \\ 5 & 4.7 \\ 6 & 4.9 \\ 7 & 4.5 \\ 8 & 5.1 \\ \hline \end{array} $$ Some scientists think that \(5.0 \mathrm{ppm}\) is a dissolved \(\mathrm{O}_{2}\) level that is marginal for fish to live. Conduct a statistical test to determine whether the mean dissolved \(\mathrm{O}_{2}\) concentration is less than \(5.0 \mathrm{ppm}\) at the \(95 \%\) confidence level. State clearly the null and alternative hypotheses.

Step by step solution

01

State the Hypotheses

In this problem, we need to determine if the mean dissolved \( \mathrm{O}_{2} \) concentration is less than \( 5.0 \mathrm{ppm} \). Thus, the null hypothesis \( (H_0) \) can be stated as: The mean dissolved oxygen concentration is equal to or greater than \( 5.0 \mathrm{ppm} \). This can be expressed as \( H_0: \mu \geq 5.0 \). The alternative hypothesis \( (H_a) \) can be stated as: The mean dissolved oxygen concentration is less than \( 5.0 \mathrm{ppm} \). This is \( H_a: \mu < 5.0 \).

02

Collect and Describe the Data

The collected data represents weekly dissolved oxygen concentrations in ppm over eight weeks: [4.9, 5.1, 5.6, 4.3, 4.7, 4.9, 4.5, 5.1]. We will use this sample to perform a statistical test.

03

Choose the Right Statistical Test and Significance Level

Since we are comparing the sample mean to a known value with unknown population standard deviation, a one-sample t-test is appropriate. The significance level is given as \( \alpha = 0.05 \) for a 95% confidence level.

04

Calculate the Sample Mean and Standard Deviation

Calculate the sample mean \( \bar{x} \) and standard deviation \( s \):\[\bar{x} = \frac{4.9 + 5.1 + 5.6 + 4.3 + 4.7 + 4.9 + 4.5 + 5.1}{8} = 4.888 \]Use the formula for the standard deviation:\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \].

05

Perform the t-Test

Calculate the t-statistic using the formula:\[t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}\]where \( \mu_0 = 5.0 \), \( n = 8 \). Calculate the t-statistic and compare with the critical t-value from the t-distribution table with 7 degrees of freedom at \( \alpha = 0.05 \). If the t-statistic is less than the critical value, we reject \( H_0 \).

06

Draw Conclusion

Based on calculated t-statistic and comparison with critical value, conclude whether to reject or not reject \( H_0 \). If we reject \( H_0 \), then there is evidence that the mean dissolved oxygen level is less than \( 5.0 \mathrm{ppm} \). If we do not reject \( H_0 \), then we do not have enough evidence to conclude that.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Dissolved Oxygen Concentration

Dissolved oxygen (DO) concentration is a crucial indicator of the health of aquatic ecosystems. It refers to the amount of oxygen gas that is present in water. Aquatic organisms, like fish and insects, rely on sufficient DO to survive and thrive. When pollutants such as sewage and industrial waste enter water bodies, they can significantly decrease DO levels. This can cause stress or even death to aquatic life.

Factors such as water temperature, salinity, and altitude can also influence DO levels. Warmer water holds less oxygen, which can exacerbate the issue during hot months.

• DO levels are measured in parts per million (ppm).
• A DO level of below 5.0 ppm is typically considered low for many aquatic organisms.
• Maintaining healthy DO levels is essential for sustaining aquatic ecosystems.

Monitoring DO levels regularly helps environmentalists and scientists ensure the well-being of aquatic life and assess water quality.

The One-Sample t-Test Explained

A one-sample t-test is a statistical method used when you want to determine if the mean of a single sample is significantly different from a known or hypothesized population mean. In this exercise, we use a one-sample t-test to check if the mean dissolved oxygen concentration in the river is less than 5.0 ppm.

Key steps in a one-sample t-test involve:

• Defining your null hypothesis, usually stating that there is no difference between the sample mean and the population mean.
• Calculating the sample mean and standard deviation.
• Using these calculations to find the t-statistic, which tells you how far, in standard error units, the sample mean is from the known population mean.
• Comparing the t-statistic to a critical value from the t-distribution table, based on the chosen significance level and degrees of freedom.

The test is especially useful when the population standard deviation is unknown, and the sample size is small. It helps determine if observed differences are likely due to chance or represent an actual difference.

The Role of Environmental Chemistry

Environmental chemistry is the study of chemical processes occurring in the environment, impacting air, water, and soil quality. This field is essential for understanding how pollutants affect ecosystems, including the implications for DO levels in water. By analyzing chemical interactions and transformations, environmental chemists can devise strategies to minimize pollution and restore environmental health.

The exercise above involves testing DO concentration, which is a fundamental aspect of water chemistry. Environmental chemists might explore:

• The sources of pollutants reducing DO levels, such as industrial discharges or agricultural runoff.
• The chemical reactions that consume oxygen in water, including microbial decomposition of organic matter.
• Methods to remediate or prevent pollution to improve or maintain DO levels.

Environmental chemistry merges disciplines like biology and geology to address complex ecological issues, providing the tools needed to develop sustainable solutions for environmental protection.