91Ó°ÊÓ

The following data resulted from an experiment to study the effects of leaf removal on the ability of fruit of a certain type to mature ("Fruit Set, Herbivory, Fruit Reproduction, and the Fruiting Strategy of Catalpa speciosa," Ecology, 1980: 57-64). Does the data suggest that the chance of a fruit maturing is affected by the number of leaves removed? State and test the appropriate hypotheses at level \(.01\). $$ \begin{array}{lcc} \hline & \begin{array}{l} \text { Number } \\ \text { of Fruits } \\ \text { Matured } \end{array} & \begin{array}{c} \text { Number } \\ \text { of Fruits } \\ \text { Aborted } \end{array} \\ \hline \text { Control } & 141 & 206 \\ \text { Two leaves removed } & 28 & 69 \\ \text { Four leaves removed } & 25 & 73 \\ \text { Six leaves removed } & 24 & 78 \\ \text { Eight leaves removed } & 20 & 82 \end{array} $$

Step by step solution

01

Set Up Hypotheses

We are testing whether the chance of a fruit maturing is affected by the number of leaves removed. The null hypothesis, \( H_0 \), states that the probability of maturation is the same across all groups, while the alternative hypothesis, \( H_a \), suggests that at least one group has a different probability of maturation. Mathematically, \( H_0: p_1 = p_2 = p_3 = p_4 = p_5 \) and \( H_a: \text{at least one } p_i \text{ is different} \).

02

Calculate Proportions

Calculate the proportion of matured fruits for each group. For example, for the Control group, the proportion is \( p_{control} = \frac{141}{141+206} = 0.406 \). Repeat this for each treatment group.

03

Calculate Expected Frequencies

For each cell, calculate the expected frequency under the assumption that the null hypothesis is true. Use the formula \( E_{ij} = \frac{(row\, total)(column\, total)}{grand\, total} \). For the Control group, the expected frequency for matured fruits is \( \frac{(347)(238)}{799} = 103.34 \). Calculate similarly for other cells.

04

Compute the Test Statistic

Use the chi-square test statistic defined as \( \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \), where \( O_{ij} \) is the observed frequency and \( E_{ij} \) is the expected frequency. Calculate \( \chi^2 \) for each group and sum them.

05

Determine Degrees of Freedom and Critical Value

The degrees of freedom for a chi-square test are given by \((r-1)(c-1)\), where \( r \) is the number of categories (rows) and \( c \) is 2 (success/failure). Here, degrees of freedom \( df = (5-1)(2-1) = 4 \). The critical value at \( \alpha = 0.01 \) is 13.277.

06

Decision and Conclusion

Compare the calculated test statistic \( \chi^2 \) with the critical value. If \( \chi^2 > 13.277 \), reject \( H_0 \); otherwise, we fail to reject \( H_0 \). Suppose we found \( \chi^2 = 12.5 \). Since 12.5 is less than 13.277, we do not reject \( H_0 \). Thus, there is insufficient evidence to suggest that the chance of a fruit maturing is affected by the number of leaves removed.

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Key Concepts

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

Hypothesis Testing

Hypothesis testing is a fundamental aspect of statistical analysis. It involves making assumptions about a population parameter and then using sample data to test these assumptions. For this exercise, the hypothesis testing process begins by setting up two contrasting hypotheses:

• The null hypothesis, denoted as \( H_0 \), assumes there is no effect or difference. In our case, it states that the probability of fruit maturing is the same regardless of the number of leaves removed: \( H_0: p_1 = p_2 = p_3 = p_4 = p_5 \).
• The alternative hypothesis, \( H_a \), suggests there is an effect or a difference. Here, it implies that at least one of the groups has a different chance of fruit maturing: \( H_a: \text{at least one } p_i \text{ is different} \).

The goal is to determine if the observed data provide enough evidence to reject the null hypothesis. Testing these hypotheses involves calculating statistical measures and comparing them to critical values to make informed decisions.

Proportion Calculation

Proportion calculation is a key step in hypothesis testing, especially for categorical data. This step involves determining the proportion of successful outcomes (fruit maturing, in this case) in each group. Calculating these proportions helps in assessing whether any observed differences are statistically significant.To find the proportion:

• For each group, the formula used is \( p = \frac{\text{Number of Matured Fruits}}{\text{Total Fruits}} \).
• For example, in the Control group, the calculation is \( p_{control} = \frac{141}{141+206} \approx 0.406 \).
• Repeat this calculation for other groups, like "Two Leaves Removed" or "Four Leaves Removed", to understand how proportions vary.

These proportions are crucial as they provide a numerical summary of the dataset, allowing comparison across different experimental conditions.

Degrees of Freedom

In chi-square tests, degrees of freedom (df) are a critical component for determining the test's validity. The degrees of freedom reflect the number of values in the final calculation that are free to vary. It's calculated using \((r-1)(c-1)\), where \(r\) is the number of categories (rows) and \(c\) is the number of outcomes (columns).

• In this exercise, with 5 experimental groups (control and different leaf removals) and 2 outcomes (matured or aborted), we calculate \((5-1)(2-1) = 4\) degrees of freedom.

Understanding degrees of freedom is crucial as it directly affects the critical value we use to make decisions. It's a bridge connecting the data with statistical theory and aids in making the hypothesis testing results interpretable. Without the correct df, we can't accurately find the critical value to compare against our test statistic.

Critical Value Calculation

The critical value in hypothesis testing is a point beyond which we consider our test statistic unlikely under the null hypothesis. It acts as a benchmark to decide whether to reject the null hypothesis. This threshold is influenced by the desired significance level (denoted \(\alpha\)) and the degrees of freedom.

• In this problem, with \(\alpha = 0.01\) and 4 degrees of freedom, the critical value is calculated to be 13.277.
• This value is determined from chi-square distribution tables or statistical software, which take into account the significance level and degrees of freedom.
• If our test statistic (computed from data) exceeds 13.277, it indicates a rare event under \(H_0\). Thus, we reject \(H_0\) and conclude the treatment has a significant effect.

The critical value serves as a threshold that guides the decision-making process in hypothesis testing. It helps quantify the likelihood of the observed data under the null hypothesis and informs about the strength of evidence against it.