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Dissolved oxygen (DO) refers to the amount of oxygen that is present in water and is crucial for the survival of aquatic life. It is typically measured in parts per million (ppm). A variety of factors can affect DO levels, including water temperature, the presence of plants and algae, and the organic load of water resulting from industrial or sewage discharge. Adequate levels of DO are necessary to sustain fish, invertebrates, and other aquatic organisms. In the context of our problem, the state agency requires a minimum of 5 ppm to support aquatic life. This means that any reading below this threshold could indicate a potentially harmful environment for the river's ecosystem.

The six water specimens taken from the river were tested for their dissolved oxygen content to ensure that they meet the minimum requirement. The readings obtained give us real data that we can analyze using statistical methods such as the one-sample t-test to determine if the DO levels are at an acceptable level.

Hypothesis testing is a statistical method used to make inferences about a population based on sample data. It involves setting up two contradictory hypotheses: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). The null hypothesis typically represents a statement of no effect or no difference, while the alternative hypothesis represents what we are trying to show is statistically significant.

In our exercise, the null hypothesis states that the mean dissolved oxygen content is at least 5 ppm (\(H_0: \rho \text{ }\geq\text{ } 5 \text{ ppm} \)), which is the acceptable level set by the agency. The alternative hypothesis is that the mean dissolved oxygen content is less than 5 ppm (\(H_a: \rho \<\text{ } 5 \text{ ppm} \)), suggesting that the water quality may not support aquatic life. The one-sample t-test will help us determine whether the sample data provides sufficient evidence to reject the null hypothesis in favor of the alternative.

The sample standard deviation is a measure of the amount of variability or dispersion in a set of sample data. It represents the average distance between each data point and the sample mean. A low standard deviation indicates that the data points tend to be very close to the mean, while a high standard deviation indicates that the data points are spread out over a wider range of values.

When conducting a one-sample t-test, the sample standard deviation is used to estimate the standard error of the mean. The standard error gives us an understanding of how much the sample mean is expected to vary from the true population mean. This is crucial for calculating the t-value, which we use to determine how likely it is that our sample came from a population with the hypothesized mean. In our example, the calculated sample standard deviation is approximately 0.142 ppm, which indicates that the readings are relatively close to the sample mean of 4.933 ppm.

The critical t-value is a threshold that determines the cutoff point for rejecting the null hypothesis in a t-test. It represents the value beyond which the likelihood of the sample data occurring by random chance is lower than the significance level (\(\alpha\)), if the null hypothesis were true. The significance level, typically set at 0.05 (5%), helps us control the risk of a Type I error, which is the incorrect rejection of a true null hypothesis.

To find the critical t-value, we use a t-distribution table or an online calculator, which takes into account the degrees of freedom (\text{df\text}) and the significance level. For our one-sample t-test, with \text{5 degrees of freedom and a significance level of 0.05, the critical t-value is approximately -1.476. Comparing this to our calculated t-value of approximately -1.690, we find that our calculated value falls in the critical region, prompting us to reject the null hypothesis. This step is central in hypothesis testing, confirming or denying our initial conjecture about the population mean based on our sample.