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The Department of Parks and Wildlife stocks a large lake with fish every six years. It is determined that a healthy diversity of fish in the lake should consist of \(10 \%\) largemouth bass, \(15 \%\) smallmouth bass, \(10 \%\) striped bass, \(10 \%\) trout, and \(20 \%\) catfish. Therefore each time the lake is stocked, the fish population in the lake is restored to maintain that particular distribution. Every three years, the department conducts a study to see whether the distribution of the fish in the lake has shifted away from the target proportions. In one particular year, a research group from the department observed a sample of 292 fish from the lake with the results given in the table provided. Test, at the \(5 \%\) level of significance, whether there is sufficient evidence in the data to conclude that the fish population distribution has shifted since the last stocking. $$\begin{array}{|c|c|c|} \hline \text { Fish } & \text { Target Distribution } & \text { Fish in Sample } \\ \hline \text { Largemouth Bass } & 0.10 & 14 \\ \hline \text { Smallmouth Bass } & 0.15 & 49 \\ \hline \text { Striped Bass } & 0.10 & 21 \\ \hline \text { Trout } & 0.10 & 22 \\ \hline \text { Catfish } & 0.20 & 75 \\ \hline \text { Other } & 0.35 & 111\\\ \hline \end{array}$$

Step by step solution

01

Define the Hypotheses

We start the process by setting up our null and alternative hypotheses. **Null Hypothesis (H鈧�):** The distribution of fish in the lake follows the target distribution. **Alternative Hypothesis (H鈧�):** The distribution of fish in the lake does not follow the target distribution.

02

Compute Expected Frequencies

Calculate the expected frequencies for each category of fish based on the target distribution and the total sample size (292). The expected frequency for each category can be calculated using:\[\text{Expected Frequency} = \text{Total Sample Size} \times \text{Target Proportion}\]- Largemouth Bass: \(292 \times 0.10 = 29.2\)- Smallmouth Bass: \(292 \times 0.15 = 43.8\)- Striped Bass: \(292 \times 0.10 = 29.2\)- Trout: \(292 \times 0.10 = 29.2\)- Catfish: \(292 \times 0.20 = 58.4\)- Other: \(292 \times 0.35 = 102.2\)

03

Calculate Chi-Square Statistic

The Chi-Square statistic is calculated using the formula:\[\chi^2 = \sum \frac{(\text{Observed} - \text{Expected})^2}{\text{Expected}}\]Applying to each category:- Largemouth Bass: \(\frac{(14 - 29.2)^2}{29.2} = 7.25\)- Smallmouth Bass: \(\frac{(49 - 43.8)^2}{43.8} = 0.62\)- Striped Bass: \(\frac{(21 - 29.2)^2}{29.2} = 2.31\)- Trout: \(\frac{(22 - 29.2)^2}{29.2} = 1.77\)- Catfish: \(\frac{(75 - 58.4)^2}{58.4} = 4.60\)- Other: \(\frac{(111 - 102.2)^2}{102.2} = 0.75\)Adding these values gives us the total Chi-Square statistic: \[\chi^2 = 7.25 + 0.62 + 2.31 + 1.77 + 4.60 + 0.75 = 17.30\]

04

Determine Degrees of Freedom and Critical Value

We need to determine the degrees of freedom (df) to find the critical value from the Chi-Square distribution table.The degrees of freedom are calculated as:\[\text{df} = (\text{Number of Categories} - 1) = 6 - 1 = 5\]Using a Chi-Square distribution table or calculator, we find the critical value for 5 degrees of freedom at \(5\%\) level of significance to be approximately 11.07.

05

Make a Decision

Compare our computed Chi-Square statistic with the critical value:\[\chi^2 = 17.30 > 11.07\] Since the calculated Chi-Square value (17.30) is greater than the critical value (11.07), we reject the null hypothesis.

06

Conclusion

Since we rejected the null hypothesis, this implies that there is sufficient evidence to suggest that the fish population distribution has shifted from the target distribution since the last stocking.

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Key Concepts

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

Null Hypothesis

The null hypothesis serves as a starting point in hypothesis testing and establishes a benchmark for comparison. In the context of the Chi-Square Test, the null hypothesis (\(H_0\)) usually states that there is no significant difference between the observed and expected frequencies. It asserts that any variations are due to random chance rather than genuine differences.

When you state the null hypothesis, it is typically in a predictable format, such as "The distribution of the fish in the lake follows the target distribution." This implies that any deviations in the observed data are not substantial enough to conclude that the distribution has changed.

• If statistical evidence is strong against the null hypothesis, it can be rejected, suggesting that a real difference or relationship exists in the data.
• A key part of hypothesis testing is selecting the right significance level, commonly denoted as \(\alpha\), which impacts your threshold for rejecting the null hypothesis.

Understanding the null hypothesis helps set clear expectations and criteria for analysis in a Chi-Square Test.

Expected Frequency

Expected frequency is crucial in Chi-Square Tests to compare against the actual observed frequency in each category. It is not based on observed data but rather calculated based on the assumption that the null hypothesis is true.

Calculating the expected frequency involves:

• Using the total sample size and the target proportions for each category. For example, if 10% of the fish population should be largemouth bass, the expected frequency is calculated by multiplying the total sample size by the proportion (292 \(\times 0.10 = 29.2\)).
• These numbers provide a baseline for what the population distribution should look like if everything stayed True to the null hypothesis.

Properly calculating expected frequencies allows for a valid comparison between what is observed and what was anticipated, thereby facilitating the computation of the Chi-Square statistic. If the observed frequencies deviate substantially from these expected frequencies, it is likely to impact the decision to accept or reject the null hypothesis.

Degrees of Freedom

The concept of degrees of freedom (df) can initially seem abstract, but it's fundamental for analyzing the Chi-Square statistic's significance. In simple terms, degrees of freedom refer to the number of values in a calculation that are free to vary while still conforming to a statistical model.

For a Chi-Square Test, degrees of freedom are calculated as one less than the number of categories being tested. So, if testing a sample with six categories of fish, as in this exercise, the degrees of freedom is:\[\text{df} = \text{Number of Categories} - 1 = 6 - 1 = 5\]df is essential when determining the critical value from the Chi-Square distribution table. The more degrees of freedom, the closer the Chi-Square distribution approaches a normal distribution.

• Understanding df helps in assessing how the Chi-Square value compares against known statistical distributions.
• It enables you to judge whether the difference between observed and expected data is statistically significant.

Grasping the degrees of freedom concept allows you to correctly interpret Chi-Square statistical results, leading to more reliable conclusions.