Question

The two claws of the lobster (Homarus americanus) are identical in the juvenile stages. By adulthood, however, the two claws normally have differentiated into a stout claw called a "crusher" and a slender claw called a "cutter." In a study of the differentiation process, 26 juvenile lobsters were reared in smooth plastic trays and 18 were reared in trays containing oyster chips (which they could use to exercise their claws). Another 23 lobsters were reared in trays containing only one oyster chip. The claw configurations of all the lobsters as adults are summarized in the table. \({ }^{35}\) $$ \begin{array}{|lccc|} \hline && {\text { Claw configuration }} \\ \hline\text { Treatment } & \begin{array}{l} \text { Right } \\ \text { crusher, } \\ \text { Left } \\ \text { cutter } \end{array} & \begin{array}{l} \text { Right } \\ \text { cutter, } \\ \text { Left } \\ \text { crusher } \end{array} & \begin{array}{l} \text { Right } \\ \text { cutter, } \\ \text { Left } \\ \text { cutter } \end{array} \\ \hline \text { Oyster chips } & 8 & 9 & 1 \\ \text { Smooth plastic } & 2 & 4 & 20 \\ \text { One oyster chip } & 7 & 9 & 7 \\ \hline \end{array} $$ (a) The value of the contingency-table chi-square statistic for these data is \(\chi_{s}^{2}=24.35 .\) Carry out the chi-square test at \(\alpha=0.01\) (b) Verify the value of \(\chi_{s}^{2}\) given in part (a). (c) Construct a table showing the percentage distribution of claw configurations for each of the three treatments. (d) Interpret the table from part (c): In what way is claw configuration related to treatment? (For example, if you wanted a lobster with two cutter claws, which treatment would you choose and why?)

Short answer

The chi-square test indicates whether there's an association between the treatment type and the claw configuration outcome. With the given \( \chi^2 \) value of \(24.35\) and a significance level of \(0.01\), we would either reject or fail to reject the null hypothesis based on a comparison to a critical value. To get a lobster with two cutter claws, choose the treatment with the highest percentage of such an outcome, which can be determined using the percentage distribution table.

Step-by-step solution

State the hypotheses

The null hypothesis \(H_0\) states that there is no association between the type of treatment and the claw configuration of the lobsters. The alternative hypothesis \(H_a\) states that there is an association between the treatment and the claw configuration.

Calculate the expected frequencies

To perform the chi-square test, we first need to calculate the expected frequencies for each cell in the contingency table. The expected frequency for each cell is calculated by multiplying the sum of the row by the sum of the column, then dividing by the total sample size. The expected frequency is given by \( E_{ij} = (Row_i \times Column_j) / Total \).

Compute chi-square statistic

Compute the value of the chi-square statistic using the formula \( \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \) where \(O_{ij}\) is the observed frequency and \(E_{ij}\) is the expected frequency for the \(i\)th row and \(j\)th column. Sum the results over all cells to obtain \( \chi^2 \) value.

Determine the critical chi-square value

For the given significance level of \( \alpha = 0.01 \) and the degrees of freedom \(df\) which is \( (rows - 1) \times (columns - 1) \), determine the critical chi-square value from the chi-square distribution table.

Make the decision

If the calculated \( \chi^2 \) is greater than the critical value from the chi-square distribution table, reject the null hypothesis \(H_0\). Otherwise, fail to reject the null hypothesis \(H_0\).

Verify the given \( \chi^2 \) value

Using the observed and expected frequencies, go through the calculations as described in Step 3 to confirm the given \( \chi^2 \) value of \(24.35\).

Construct the percentage distribution table

For each cell in the original contingency table, calculate the percentage by dividing the cell value by the total number of lobsters in its treatment group and then multiply by 100.

Interpret the percentage distribution table

Analyze the percentage distribution table to determine which treatment results in the highest percentage of a particular claw configuration. For example, to find the treatment that results in the highest percentage of lobsters with two cutter claws, identify the row with the highest value in the relevant column.

Key concepts

Contingency Table

A contingency table is a type of data presentation commonly used in statistics to show the frequency distribution of variables. It provides a simple way to visualize and analyze the relationship between two categorical variables. Each cell within the table represents the frequency count for the corresponding combination of the categories. In the case of the lobster claw study, the contingency table is used to summarize the number of lobsters with each claw configuration within each treatment group (oyster chips, smooth plastic, and one oyster chip).

When performing a chi-square test, the contingency table is the foundation of calculating observed frequencies, which can then be compared to expected frequencies to determine if there is a significant association between the variables being examined. To improve comprehension, it is essential to understand how to read and interpret contingency tables, as they are recurrent in the context of statistical hypothesis testing.

Lobster Claw Differentiation

Lobster claw differentiation refers to the process by which juvenile lobsters develop into adults with two distinct types of claws: a 'crusher' and a 'cutter.' While they are initially identical, environmental factors and genetic predisposition influence their development. The study mentioned in the exercise investigates this phenomenon under varying conditions, which provides valuable data on how external stimuli may affect physical development in lobsters. This specific application of the chi-square test can help determine if the treatment (type of surface the lobsters are reared on) has any significant effect on the likelihood of developing a particular configuration of claws.

Understanding the biology behind claw differentiation can enhance students' grasp of why the chi-square test is being applied to such data and underscores the relevance of statistical methods in biological research.

Statistical Hypothesis Testing

Statistical hypothesis testing is a method used to decide if there is significant evidence to reject a null hypothesis, based on data from a sample. In the context of the chi-square test, the null hypothesis typically states that there is no association between the variables being tested. For the lobster claw study, the null hypothesis posits that claw configuration is independent of the treatment.

To determine whether the observed data supports this hypothesis, we compare the calculated chi-square statistic against the critical value at a predetermined significance level. If the test statistic exceeds the critical value, it suggests that the pattern observed in the data is unlikely to occur by chance alone, leading to the rejection of the null hypothesis. It's vital for students to understand that hypothesis testing is a cornerstone of empirical research, enabling scientists to draw conclusions from data with a known risk of being incorrect (the significance level).

Expected Frequencies

Expected frequencies are hypothetical counts calculated under the assumption that the null hypothesis is true. They provide a benchmark to which the observed frequencies are compared when conducting the chi-square test. To calculate expected frequencies, one must use the marginal totals of the rows and columns from the contingency table.

In the lobster study, expected frequencies are calculated by multiplying the sum of lobsters for each treatment by the sum of lobsters with each claw configuration, then dividing by the total number of lobsters observed. These calculations allow us to evaluate whether the differences between expected and observed frequencies are statistically significant, or simply due to random variation. As part of their learning process, students need to comprehend how these expected frequencies tie into the overall framework of hypothesis testing and why they are critical for determining the p-values, which ultimately guide decisions about hypotheses.