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In a 2016 article published in the Journal of American College Health, Heller et al. surveyed a sample of students at an urban community college. Students' ages and frequency of alcohol use per month are recorded in the following table. Because some of the expected counts are less than 5, we should combine some groups. For this question, combine the frequencies \(10-29\) days and Every day into one group. Label this group \(10+\) days and show your new table. Then test the new table to see whether there is an association between age group and alcohol use using a significance level of \(0.05\). Assume this is a random sample of students from this college. $$\begin{array}{lcccc}\hline & {\text { Alcohol Use }} \\\\\text { Age } & \text { None } & \text { 1-9 days } & \text { 10-29 days } & \text { Every day } \\ \hline 18-20 & 182 & 100 & 27 & 4 \\ \hline 21-24 & 142 & 109 & 35 & 4 \\\25-29 & 49 & 41 & 5 & 2 \\\\\hline 30+ & 76 & 32 & 8 & 2 \\ \hline\end{array}$$

Short Answer

Expert verified
After computing the chi-square test, if the test statistic is greater than the critical value at the 0.05 level of significance, this means that there is evidence to suggest that there is an association between age group and alcohol use. If the test statistic is not greater than the critical value, then there is no significant association between age group and alcohol use.

Step by step solution

01

Consolidating frequencies

First, we combine the frequencies for the third column '10-29 days' and the fourth column 'Everyday' into a single column labelled '10+ days'. The new table would look as follows, where the figures represent the number of students: \[\begin{array}{lcccc}\hline & {\text { Alcohol Use }} \\{\text { Age }} & \text { None } & \text { 1-9 days } & \text { 10+ days } \\\hline 18-20 & 182 & 100 & 31 \\\hline 21-24 & 142 & 109 & 39 \\25-29 & 49 & 41 & 7 \\\hline 30+ & 76 & 32 & 10 \\\hline\end{array} \]
02

Calculating expected values

Now we need to calculate the expected counts for each cell in our table. This can be done by multiplying the row total by the column total and then dividing by the grand total. Expected counts are used to determine what we would expect if there were no relationship between the two variables.
03

Computing Chi-Square Test statistic

After calculating the expected values, we apply the chi-square formula \( (O - E)^2 / E \) where O is the observed value in the cell and E is the expected value. We add up all these values to find our chi-square test statistic.
04

Testing for significance

We compare our test statistic to a chi-square distribution with (k - 1)(r - 1) degrees of freedom, where k and r are the number of rows and columns respectively. If our test statistic is greater than the value on the chi-square distribution for our level of significance (0.05), then we reject the null hypothesis that there is no association between age group and alcohol use.

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Key Concepts

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

Statistical Significance
When working with statistical tests, it's critical to understand what it means for a result to be 'statistically significant'. This concept relates to the likelihood that the results of our analysis are not due to random chance. For example, in our context, if we find that there is a statistically significant association between age group and alcohol use, it suggests that the differences observed in the data are unlikely to have occurred by accident.

Statistical significance is usually determined using a p-value, which measures the probability of obtaining a result at least as extreme as the one we have if there were no actual association. In general, if this p-value is less than our chosen significance level (usually 0.05 or 5%), we have enough evidence to reject the null hypothesis—in this case, that there is no association between the two variables. By conducting a chi-square test and comparing our test statistic to a critical value from the chi-square distribution, we can decide whether our results are statistically significant at our chosen significance level.
Expected Counts in Statistics
Expected counts play a pivotal role in categorical data analysis, especially when conducting a chi-square test for association. These counts represent the number of observations that we would expect in each category if there were no association between the variables being tested—in our case, age group and frequency of alcohol use.

To calculate the expected count for a particular cell in a contingency table, you multiply the row total for that cell by the column total, and then divide by the grand total of all observations. Expected counts are a fundamental part of the chi-square test because they provide the baseline to which we compare our observed counts. If the observed counts deviate significantly from the expected counts, it suggests an association between the variables. For our analysis, it's vital to ensure that these expected counts are sufficiently large—usually at least 5—to ensure the validity of the chi-square test.
Consolidating Frequency Data
In datasets with multiple categories, frequency data can sometimes require consolidation for proper analysis. When some categories have a small number of observations—like in our exercise where some expected counts are less than 5—it's necessary to merge these categories to ensure the validity of statistical tests. This not only helps in satisfying the assumptions of the test, such as the chi-square test but also helps in simplifying the data and making patterns more discernible.

Consolidating frequency data involves combining categories with similar characteristics or those that do not meet certain statistical criteria. In our exercise, we combined the '10-29 days' and 'Every day' categories into a single '10+ days' category. This ensures that the expected counts are above the threshold needed for the chi-square test and balances the distribution of data for more reliable analysis. As a best practice, researchers should plan their data collection strategies to minimize the need for such consolidation, but in practical scenarios, re-categorization like this is often unavoidable.

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Most popular questions from this chapter

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