Binge Drinking The American College Health Association -...

Binge Drinking The American College Health Association - National College Health Assessment survey $,{ }^{17}$ introduced on page 60 , was administered at 44 colleges and universities in Fall 2011 with more than 27,000 students participating in the survey. Students in the ACHA-NCHA survey were asked "Within the last two weeks, how many times have you had five or more drinks of alcohol at a sitting?" The results are given in Table 7.31 . Is there a significant difference in drinking habits depending on gender? Show all details of the test. If there is an association, use the observed and expected counts to give an informative conclusion in context. $$ \begin{array}{c|rr|r} \hline & \text { Male } & \text { Female } & \text { Total } \\ \hline 0 & 5,402 & 13,310 & 18,712 \\ 1-2 & 2,147 & 3,678 & 5,825 \\ 3-4 & 912 & 966 & 1,878 \\ 5+ & 495 & 358 & 853 \\ \hline \text { Total } & 8,956 & 18,312 & 27,268 \\ \hline \end{array} $$

Binge Drinking The American College Health Association - National College Health Assessment survey $,{ }^{17}$ introduced on page 60 , was administered at 44 colleges and universities in Fall 2011 with more than 27,000 students participating in the survey. Students in the ACHA-NCHA survey were asked "Within the last two weeks, how many times have you had five or more drinks of alcohol at a sitting?" The results are given in Table 7.31 . Is there a significant difference in drinking habits depending on gender? Show all details of the test. If there is an association, use the observed and expected counts to give an informative conclusion in context.
$$
\begin{array}{c|rr|r}
\hline & \text { Male } & \text { Female } & \text { Total } \\
\hline 0 & 5,402 & 13,310 & 18,712 \\
1-2 & 2,147 & 3,678 & 5,825 \\
3-4 & 912 & 966 & 1,878 \\
5+ & 495 & 358 & 853 \\
\hline \text { Total } & 8,956 & 18,312 & 27,268 \\
\hline
\end{array}
$$

Key Concepts

P-value

The p-value is the probability of obtaining a chi-square statistic at least as extreme as the one observed, assuming the null hypothesis is true. A low p-value indicates that the observed data is unlikely under the null hypothesis, providing evidence to reject it in favor of the alternative hypothesis.

Interpreting Statistical Results in Context

Beyond numerical calculations, interpreting the results in context involves explaining what a statistically significant or non-significant result means in practical terms. This includes discussing the implications of observed differences between categories and considering factors such as sample size and real-world impact.

Degrees of Freedom

Degrees of freedom refer to the number of values that are free to vary when estimating statistical parameters. In the chi-square test for independence, they are calculated based on the number of categories in each variable, generally as (number of rows minus 1) multiplied by (number of columns minus 1).

Hypothesis Testing

Hypothesis testing is a method used to decide whether the observations from a study deviate sufficiently from what would be expected under a null hypothesis. Typically, the null hypothesis states that no association exists between the variables, while the alternative hypothesis asserts that an association does exist.

Expected Counts

Expected counts represent the frequencies that would be anticipated in each cell of the contingency table under the assumption that there is no association (independence) between the variables. They are computed using the marginal totals of the table and are essential for calculating the chi-square statistic.

Contingency Table

A contingency table is a matrix format that displays the frequency distribution of variables. It organizes data into rows and columns, representing the categories of the variables being analyzed, which facilitates the calculation of both observed and expected counts in statistical tests.

Chi-square Test for Independence

This test is used to determine whether there is a statistically significant association between two categorical variables. It compares the observed frequencies in each category of a contingency table with the frequencies that would be expected if the variables were indeed independent.

Transcript

00:01 Here we have a question that involves college students, gender, and drinking habits.

00:05 A study of college students examined the link between gender and drinking habits.

00:10 So students were asked the number of times over the previous two weeks in which they'd had five or more drinks in one sitting.

00:16 And the results are shown in the table below for both males and females.

00:20 So we are to test for an association between drinking habits and gender.

00:25 And you can see here that the questions involved categorizing the number of drinks as 0, 1 to 2, 3 to 4, and 5 or more.

00:36 And again, we have two groups, males, and females.

00:41 So first thing to do is write down our hypotheses.

00:45 So as we're stating our hypotheses, the null hypothesis states that there is no association between gender and drinking habits.

00:55 And then the alternative hypothesis states that there is an association between gender and drinking habits.

01:03 Now, our table shows us in the black typeface here, the observed counts.

01:10 We need to calculate the expected counts ourselves.

01:13 And remember how to find the expected counts.

01:16 What we'll do is take the row total times the column total and divide by the overall total.

01:25 So, for example, in the first cell, the row total is 18 ,712, the column total 8 ,956, the overall total 27 ,268.

01:39 So you can see that those three numbers are used in the calculation right here to get 6 ,146 as the expected count.

01:48 So i did that for all eight cells in the table, and then in parentheses in green wrote the expected counts for each of those cells.

01:58 The next step is to use the expected counts along with the observed counts for each cell to calculate the kai squared statistic.

02:09 So what do we do here? for each cell, we'll take our observed count, and starting with the first cell, the observed count, was 5 .3 .3 .5 .3.

02:20 402, will subtract the expected count.

02:23 We'll square that difference and divide by the expected count.

02:27 We're going to have eight terms, one for each cell, that do the same thing, squared difference of observed minus expected divided by expected.

02:37 So i didn't write all eight of them out here, but you can see that pattern to calculate the kai square statistic...

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