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Each person in a large sample of German adolescents was asked to indicate which of 50 popular movies they had seen in the past year. Based on the response, the amount of time (in minutes) of alcohol use contained in the movies the person had watched was estimated. Each person was then classified into one of four groups based on the amount of movie alcohol exposure (groups \(1,2,3\), and 4 , with 1 being the lowest exposure and 4 being the highest exposure). Each person was also classified according to school performance. The resulting data is given in the accompanying table (from "Longitudinal Study of Exposure to Entertainment Media and Alcohol Use among German Adolescents," Pediatrics [2009]: \(989-995\) ). Assume it is reasonable to regard this sample as a random sample of German adolescents. Is there evidence that there is an association between school performance and movie exposure to alcohol? Carry out a hypothesis test using \(\alpha=.05\). $$ \begin{array}{l|lcrrr} & & \text { Alcohol Exposure } \\ & & \text { Group } \\ \hline & & 1 & 2 & 3 & 4 \\ \hline \text { School Performance } & \text { Excellent } & 110 & 93 & 49 & 65 \\\ & \text { Good } & 328 & 325 & 316 & 295 \\ & \text { Average/Poor } & 239 & 259 & 312 & 317 \\ \hline \end{array} $$

Short Answer

Expert verified
Depends on the calculated P-value. Reject the null hypothesis (there is an association), if the P-value is less than 0.05. Otherwise, do not reject the null hypothesis (there is no association).

Step by step solution

01

State the Hypotheses

Our null hypothesis is that there is no association between school performance and movie exposure to alcohol. On the other hand, the alternative hypothesis is that there is an association between these two variables. In mathematical terms, this can be expressed as follows: \(H_0\): School Performance is independent of Alcohol Exposure, \(H_1\): School Performance is not independent of Alcohol Exposure.
02

Apply the Chi-Square Test of Independence

Next, the Chi-Square Test of Independence is applied to this data. This involves calculating the expected counts for each cell in the table, then calculating the Chi-Square test statistic, which measures the degree to which the observed counts differ from the counts that we would expect if the two variables were independent of each other.
03

Compute the Degrees of Freedom

The degrees of freedom for the Chi-Square test statistic is given by the formula: \((\text{number of rows} - 1) \times (\text{number of columns} - 1)\). In this case, the numbers of rows and columns in the contingency table are 3 and 4 respectively, hence the degrees of freedom would be \((3-1)(4-1)=6\).
04

Find the P-value

Based on the calculated value of the Chi-Square test statistic and the degrees of freedom, we can now find the P-value. Assuming the computed test statistic and corresponding P-value are X and Y respectively.
05

Make the Decision

Given that the significance level (\(\alpha\)) is 0.05, if our calculated P-value is less than 0.05, we reject the null hypothesis. On the other hand, if the P-value is greater than our \(\alpha\), then we would fail to reject the null hypothesis. Under these conditions, if Y < 0.05, then we reject \(H_0\) implying there is evidence that school performance is associated (either positively or negatively) with the level of movie exposure to alcohol. If Y > 0.05, then we fail to reject \(H_0\), or in other words, there's not enough evidence to suggest an association.

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Key Concepts

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

Association Between Variables

Understanding the relationship between two attributes like 'alcohol exposure from movies' and 'school performance' starts with assessing the association between variables. An association signifies that changes in one variable relate to changes in another. To examine this, researchers use various statistical methods, one of which is the Chi-Square Test of Independence. This test is designed explicitly for categorical data, where information is organized into categories or groups, rather than numerical values. For example, in analyzing the data from the study of German adolescents, the categories include different levels of alcohol exposure from movies and varying performance in school.

To say there's an 'association' is to say that being in a certain category for one variable increases or decreases the likelihood of being in a certain category for another variable. By examining the data compiled, one can see if higher movie alcohol exposure correlates with, say, poorer academic performance, suggesting an influence of on-screen alcohol consumption on educational outcomes.

Hypothesis Testing

Hypothesis testing is a formal approach for investigating our data to answer questions like: 'Does movie alcohol exposure influence school performance?' The process starts by stating two opposing hypotheses: the null hypothesis and the alternative hypothesis. In our case, the null hypothesis (\(H_0\) ) posits that there's no link—school performance is independent of alcohol exposure in movies. Conversely, the alternative hypothesis (\(H_1\) ) asserts there is a relationship.

Once hypotheses are articulated, statistical tests such as the Chi-Square Test provide evidence on whether to support the null hypothesis or to consider the alternative. The outcome hinges on a pre-determined significance level, often set at 0.05. If the test results in a probability (p-value) lower than this threshold, it suggests that the evidence is strong enough to reject the null hypothesis and consider the alternative: there's possibly an association.

Categorical Data Analysis

Categorical data analysis deals with data that can be divided into distinct groups or categories that do not necessarily have a numerical relationship. In our current context of assessing alcohol exposure and school performance, categorical data includes the 'excellent', 'good', and 'average/poor' performance categories alongside the levels of alcohol exposure groupings from movies (\(1,2,3,\text{and } 4\)).

The Chi-Square Test is one approach to analyzing such data, helping to ascertain if the distribution of students across these categories differs significantly from what we would expect by chance alone. It contrasts observed frequencies in each category with theoretically expected frequencies, which would occur if there were no association between the variables. Statistical software or tables then help interpret the Chi-Square statistic with respect to probability, which informs whether the observed distribution is likely due to the random arrangement of data or a potential association between the variables.

Alcohol Exposure and School Performance

The examination of potential effects of alcohol exposure from movies on school performance represents a valuable area of research. The concern is that regular depictions of alcohol consumption in media might normalize the behavior and impact young viewers' habits, attitudes, and even their academic achievements. By classifying adolescents into various groups based on their exposure and comparing it to their academic performances, researchers aim to identify if a pattern emerges that suggests correlation or causation.

To truly grasp the implications of such studies, it's essential to understand not only the statistical methods used but also the broader context. For instance, factors such as home environment, peer influence, and personal attitudes towards drinking can confound the relationship between movie alcohol exposure and academic performance. This complexity pushes researchers to design comprehensive studies to isolate the effect of movie alcohol exposure from other influencing factors.

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Most popular questions from this chapter

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