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The article "Susceptibility of Mice to Audiogenic Seizure Is Increased by Handling Their Dams During Gestation" (Science, 1976: 427-428) reports on research into the effect of different injection treatments on the frequencies of audiogenic seizures. \begin{tabular}{lc|c|c|c} & \multicolumn{1}{c}{ No } & \multicolumn{1}{c}{ Wild } & Clonic & Tonic \\ Treatment & Response & \multicolumn{1}{c}{ Running } & \multicolumn{1}{c}{ Seizure } & \multicolumn{1}{c}{ Seizure } \\ \cline { 2 - 5 } Thienylalanine & 21 & 7 & 24 & 44 \\ \cline { 2 - 5 } Solvent & 15 & 14 & 20 & 54 \\ \cline { 2 - 5 } Sham & 23 & 10 & 23 & 48 \\ \cline { 2 - 5 } Unhandled & 47 & 13 & 28 & 32 \\ \cline { 2 - 5 } & & & & \end{tabular} Does the data suggest that the true percentages in the different response categories depend on the nature of the injection treatment? State and test the appropriate hypotheses using \(\alpha=.005\).

Step by step solution

01

Define Hypotheses

We need to use a chi-square test for independence to investigate if the type of injection treatment affects the seizure response. Our null hypothesis \(H_0\) states that the proportions of responses are independent of the treatment. The alternative hypothesis \(H_a\) claims that the proportions depend on the treatment.

02

Construct the Contingency Table

Construct a table with observed frequencies as follows:\[\begin{array}{lcccc}\text{Treatment} & \text{No Response} & \text{Wild Running} & \text{Clonic Seizure} & \text{Tonic Seizure} \\hline\text{Thienylalanine} & 21 & 7 & 24 & 44 \\text{Solvent} & 15 & 14 & 20 & 54 \\text{Sham} & 23 & 10 & 23 & 48 \\text{Unhandled} & 47 & 13 & 28 & 32\end{array}\]

03

Calculate Expected Frequencies

To calculate the expected frequency for each cell, use the formula \(E = \frac{(\text{Row Total}) \times (\text{Column Total})}{\text{Grand Total}}\). Compute expected frequencies for each cell using the totals obtained from the observed frequencies.

04

Calculate the Chi-Square Test Statistic

Use the formula \[\chi^2 = \sum \frac{(O - E)^2}{E}\]where \(O\) is the observed frequency and \(E\) is the expected frequency. Sum this calculation for all cells in the table.

05

Determine Degrees of Freedom

The degrees of freedom \(df\) for a chi-square test are calculated as \(( ext{Number of Rows} - 1) \times ( ext{Number of Columns} - 1)\). For this 4 \(\times\) 4 table, \(df = (4-1) \times (4-1) = 9\).

06

Compare with Critical Value

Use chi-square distribution tables to find the critical value for \(df = 9\) at \(\alpha = 0.005\). Compare the calculated \(\chi^2\) statistic to this critical value. If the statistic exceeds the critical value, reject the null hypothesis.

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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 powerful statistical method used to make decisions or inferences about a population based on sample data. The objective is to determine whether there is enough evidence to reject a hypothesis about a population parameter. In this context, the hypothesis test involves checking if the type of injection treatment affects the frequency of seizures in mice.

The first step is to establish two hypotheses:

• The null hypothesis (\( H_0 \)) states that the seizure response is independent of the injection treatment. This implies that any observed differences in seizure rates across treatments are due to random chance.
• The alternative hypothesis (\( H_a \)) suggests that the seizure response depends on the type of injection treatment, meaning that the treatment has a significant impact on the seizure rates.

The significance level (\( \alpha \)) is set at 0.005. This means there is a 0.5% risk of rejecting the null hypothesis if it is actually true, indicating the test's strictness. A key part of hypothesis testing is calculating a test statistic, such as the chi-square statistic, and comparing it to a critical value to make a decision.

Contingency Table

A contingency table is a matrix used to display the frequency distribution of variables. Here, it helps visualize the relationship between different treatments and their corresponding seizure responses in mice.

The table consists of:

• Rows representing different types of treatments: Thienylalanine, Solvent, Sham, and Unhandled.
• Columns representing various seizure responses: No Response, Wild Running, Clonic Seizure, and Tonic Seizure.

Each cell in the contingency table shows how many times each combination of treatment and response occurred in the sample. For instance, in the table provided, we see that 21 mice showed no response under the Thienylalanine treatment, while 44 exhibited Tonic Seizures.

The contingency table is critical for performing the chi-square test, as it highlights the observed frequencies. These observed counts are later compared to expected frequencies to assess whether the distribution of responses is independent of the treatments.

Expected Frequencies

Expected frequencies are a key component in the chi-square test for independence. They represent the counts we would expect in each cell of a contingency table if the null hypothesis were true — that is, if there were no relationship between treatment and seizure response.

To compute the expected frequency for each cell, we use the formula:\[E = \frac{(\text{Row Total}) \times (\text{Column Total})}{\text{Grand Total}}\]With this formula:

• The Row Total is the sum of observed frequencies for each treatment.
• The Column Total is the sum of observed frequencies for each response category.
• The Grand Total is the total number of observations across all treatments and responses.

By calculating these expected frequencies, we can perform the chi-square test by comparing each expected frequency to its corresponding observed frequency. Any significant differences between observed and expected frequencies might suggest a dependency between treatment types and seizure responses. The process involves summing the squared differences between observed and expected frequencies, divided by the expected frequency, across all cells.