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Criminologists have long debated whether there is a relationship between weather conditions and the incidence of violent crime. The author of the article "Is There a Season for Homicide?" (Criminology, 1988: 287-296) classified 1361 homicides according to season, resulting in the accompanying data. Test the null hypothesis of equal proportions using \(\alpha=.01\) by using the chi-squared table to say as much as possible about the \(P\)-value. \begin{tabular}{llll} Winter & Spring & Summer & Fall \\ \hline 328 & 334 & 372 & 327 \end{tabular}

Short Answer

Expert verified
Fail to reject the null hypothesis; no significant seasonal difference in homicide proportions.

Step by step solution

01

Define Hypotheses

First, define the null and alternative hypotheses. The null hypothesis \((H_0)\) states that the proportions of homicides are equal across the four seasons, while the alternative hypothesis \((H_1)\) states that the proportions are not equal.
02

Calculate Expected Frequencies

Calculate the expected frequency for each season under the null hypothesis. Since the null hypothesis assumes equal proportions, each expected frequency is the total number of homicides divided by the number of seasons. Total number of homicides = 328 + 334 + 372 + 327 = 1361.Expected frequency for each season = \(\frac{1361}{4} = 340.25\).
03

Perform Chi-Square Test

Use the formula for the chi-square statistic:\[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]Where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency for each category (season).\[\chi^2 = \frac{(328 - 340.25)^2}{340.25} + \frac{(334 - 340.25)^2}{340.25} + \frac{(372 - 340.25)^2}{340.25} + \frac{(327 - 340.25)^2}{340.25}\]\[\chi^2 = \frac{(-12.25)^2}{340.25} + \frac{(-6.25)^2}{340.25} + \frac{(31.75)^2}{340.25} + \frac{(-13.25)^2}{340.25}\]Calculate each term and sum them to find the \(\chi^2\) statistic.
04

Calculate Chi-Square Statistic

Continuing from Step 3:\[\chi^2 = \frac{150.06}{340.25} + \frac{39.06}{340.25} + \frac{1008.06}{340.25} + \frac{175.56}{340.25}\]\[\chi^2 = 0.441 + 0.115 + 2.962 + 0.516\]\[\chi^2 \approx 4.034\]
05

Determine Degrees of Freedom

For a chi-square test, the degrees of freedom (df) is calculated as the number of categories minus one. Here, we have four categories (seasons), so \(df = 4 - 1 = 3\).
06

Determine Critical Value and Compare Test Statistic

Using the chi-square distribution table, with \(\alpha = 0.01\) and \(df = 3\), find the critical value. The critical value is approximately 11.345. Compare the calculated \(\chi^2\) statistic (4.034) to this critical value.
07

Conclusion and P-Value Interpretation

Since the calculated \(\chi^2\) (4.034) is less than the critical value (11.345) at \(\alpha = 0.01\), we fail to reject the null hypothesis. This implies that there is insufficient evidence to conclude that the proportions of homicides differ by season. With regard to the \(P\)-value, since 4.034 is much lower than 11.345, the \(P\)-value is greater than 0.01, indicating a lack of statistical significance.

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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 methodical approach used in statistics to determine if a hypothesis about a data set is reasonable or not. It begins with a null hypothesis (represented as \(H_0\)), which is a statement of no effect or no difference that you aim to test. In our example about homicides in different seasons, \(H_0\) posits that the number of homicides is the same across all the seasons. On the other side is the alternative hypothesis \(H_1\), asserting that the homicides count is not equal across seasons.
All hypothesis tests follow a similar framework:
  • Define null and alternative hypotheses
  • Decide on the significance level (often represented as \(\alpha\))
  • Collect data and calculate test statistics
  • Make a decision: reject or fail to reject the null hypothesis based on test results and critical values or \(P\)-value.
Understanding the framework of hypothesis testing helps you draw inferences about populations from sample data.
Degrees of Freedom
Degrees of freedom (df) describes the number of values in a calculation that are free to vary. It's a crucial concept in statistics when you perform tests like the chi-square test. For a chi-square test, degrees of freedom are calculated based on the number of categories being analyzed. In our example with four seasonal categories, the degrees of freedom is calculated as the number of categories minus one: \(df = 4 - 1 = 3\).
Why does it matter? Because the degrees of freedom affect the shape of the chi-square distribution, which in turn affects how you interpret the test statistic and make decisions. More categories generally mean higher degrees of freedom, which allows for a more stable estimate of population variance from sample variance.
The degrees of freedom inform which row to reference in chi-square distribution tables when determining critical values for hypothesis testing.
Statistical Significance
Statistical significance is a measure of whether observed data is likely to occur by random chance or if it's reflective of a true effect. In hypothesis testing, we compare the test statistic to a critical value determined by the chosen significance level, \(\alpha\).
  • When a test statistic exceeds the critical value, the data is said to be statistically significant.
  • If the statistic is below the threshold, there's no significant effect.
In our exercise, the significance level was set at 0.01, meaning we'd only reject the null hypothesis if there was only a 1% or less probability of observing the data if the null hypothesis was true. Our chi-square statistic of 4.034 did not exceed the critical value of 11.345 for 3 degrees of freedom. Thus, we concluded there was no statistical significance in the homicide data across seasons.
P-Value Interpretation
The \(P\)-value is a probability measure used to interpret the results of hypothesis tests. It represents the probability of observing a test statistic as extreme as, or more extreme than, the observed, under the null hypothesis. In simple terms, it's about assessing the strength of the evidence against the null hypothesis.
  • A small \(P\)-value (typically ≤ \(\alpha\)) indicates strong evidence against \(H_0\), leading to its rejection.
  • A large \(P\)-value (> \(\alpha\)) suggests weak evidence against \(H_0\), so you don't reject it.
In our scenario, the chi-square test statistic was 4.034, corresponding to a \(P\)-value larger than 0.01. This higher \(P\)-value implies that the observed seasonal differences in homicide counts could easily have occurred by chance, meaning we fail to reject the null hypothesis of equal proportions.

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Most popular questions from this chapter

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