91Ó°ÊÓ

Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. (b) Find the value of the chi-square statistic for the sample. Are all the expected frequencies greater than \(5 ?\) What sampling distribution will you use? What are the degrees of freedom? (c) Find or estimate the \(P\) -value of the sample test statistic. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories? (e) Interpret your conclusion in the context of the application. Ecology: Fish The Fish and Game Department stocked Lake Lulu with fish in the following proportions: \(30 \%\) catfish, \(15 \%\) bass, \(40 \%\) bluegill, and \(15 \%\) pike. Five years later it sampled the lake to see if the distribution of fish had changed. It found that the 500 fish in the sample were distributed as follows. \(\begin{array}{cccc}\text { Catfish } & \text { Bass } & \text { Bluegill } & \text { Pike } \\ 120 & 85 & 220 & 75\end{array}\) In the 5 -year interval, did the distribution of fish change at the 0.05 level?

Step by step solution

01

Identify Level of Significance and Hypotheses

The level of significance is given as \( \alpha = 0.05 \). The null hypothesis \( H_0 \) states that the distribution of fish in the lake has not changed: 30% catfish, 15% bass, 40% bluegill, and 15% pike. The alternative hypothesis \( H_a \) states that the distribution of fish has changed.

02

Calculate Expected Frequencies and Chi-Square Statistic

Calculate the expected number of fish in each category using the original proportions and the sample size of 500:- Catfish: \( 500 \times 0.30 = 150 \)- Bass: \( 500 \times 0.15 = 75 \)- Bluegill: \( 500 \times 0.40 = 200 \)- Pike: \( 500 \times 0.15 = 75 \)Next, calculate the chi-square statistic:\[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]where \( O_i \) and \( E_i \) are the observed and expected frequencies.\[\chi^2 = \frac{(120 - 150)^2}{150} + \frac{(85 - 75)^2}{75} + \frac{(220 - 200)^2}{200} + \frac{(75 - 75)^2}{75} = \frac{(-30)^2}{150} + \frac{(10)^2}{75} + \frac{(20)^2}{200} + \frac{(0)^2}{75} = 6 + 1.33 + 2 + 0 = 9.33\]All expected frequencies are greater than 5. The chi-square distribution is appropriate for this test, and the degrees of freedom are \( df = k - 1 = 4 - 1 = 3 \), where \( k \) is the number of categories.

03

Estimate the P-value

Using the chi-square distribution table or a calculator, find the \( P \)-value for \( \chi^2 = 9.33 \) with 3 degrees of freedom. The \( P \)-value is approximately 0.025.

04

Decision Based on P-value

Since the \( P \)-value (0.025) is less than the level of significance (0.05), we reject the null hypothesis.

05

Interpretation of the Conclusion

We conclude that there is enough statistical evidence to suggest that the distribution of fish in Lake Lulu has changed from the original proportions over the five-year period.

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Key Concepts

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

Null Hypothesis

In any statistical test, the null hypothesis is a statement that there is no effect or no difference, and it is the hypothesis that researchers typically aim to test against. In the context of the Chi-Square Test, the null hypothesis represents a scenario where the observed data fit a specified theoretical distribution perfectly.

For the exercise at hand, the null hypothesis ( H_0 ) posits that the distribution of fish in Lake Lulu remains the same as it was initially stocked. Specifically:

• Catfish make up 30%.
• Bass constitute 15%.
• Bluegill account for 40%.
• Pike form the remaining 15%.

It’s crucial because it provides a baseline to compare against the observed data, allowing researchers to determine if any statistically significant deviation exists. If the null hypothesis is rejected, as in this exercise, it suggests that the actual distribution of fish has changed since the original proportions were set.

P-Value

The P-value is a crucial component in determining the outcome of a hypothesis test. It indicates the probability of observing data as extreme as, or more extreme than, those observed under the null hypothesis. A smaller P-value suggests that the observed data is unlikely under the null hypothesis.In this exercise, the calculated P-value is approximately 0.025. This value is derived from the chi-square distribution table, based on the computed chi-square statistic (9.33) and the degrees of freedom.Here’s how to interpret the P-value:

• If \( P \)< the level of significance (often 0.05 or 5%), we reject the null hypothesis.
• If \( P \)> the level of significance, we fail to reject the null hypothesis.

Because the P-value in the current exercise is less than the level of significance (0.05), it leads to rejecting the null hypothesis, indicating that the fish distribution likely changed over the five years.

Degrees of Freedom

Degrees of freedom (df) is a concept used in statistical testing that reflects the number of independent values or quantities which can be assigned to a statistical distribution. It is an essential parameter in the calculations for many tests, including the Chi-Square Test.

In this exercise, we calculate the degrees of freedom based on the number of categories of fish minus one. The formula used is \[ df = k - 1 \]where \(k\) is the number of fish categories, which are catfish, bass, bluegill, and pike. Therefore: \[ df = 4 - 1 = 3 \]The degrees of freedom help determine the correct chi-square distribution to reference when calculating the P-value. Alongside the chi-square statistic, it allows us to find the P-value, which is involved in testing the null hypothesis.

Level of Significance

The level of significance (\(\alpha\)) is a threshold set by researchers before conducting statistical tests to determine whether to reject the null hypothesis. It represents the probability of committing a Type I error, which is falsely rejecting a true null hypothesis.In many standardized tests, including the Chi-Square Test, a level of significance of 0.05 is commonly used. This indicates a 5% risk of concluding that a difference exists when there is no actual difference.For this exercise, the level of significance is set at 0.05. This means that the test will reject the null hypothesis if the P-value is less than 0.05. Since the P-value calculated (0.025) is indeed below this threshold, we reject the null hypothesis. Thus, it suggests that the fish population distribution in Lake Lulu has changed from its originally stocked proportions.