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Chapter 12: Problem 20

The authors of the paper "Movie Character Smoking and Adolescent Smoking: Who Matters More, Good Guys or Bad Guys?" (Pediatrics [2009]: 135-141) classified characters who were depicted smoking in movies released between 2000 and \(2005 .\) The smoking characters were classified according to sex and whether the character type was positive, negative or neutral. The resulting data is given in the accompanying table. Assume that it is reasonable to consider this sample of smoking movie characters as representative of smoking movie characters. Do the data provide evidence of an association between sex and character type for movie characters who smoke? Use \(\alpha=.05\). $$ \begin{array}{lccc} & & \text { Character Type } \\ \hline \text { Sex } & \text { Positive } & \text { Negative } & \text { Neutral } \\ \hline \text { Male } & 255 & 106 & 130 \\ \text { Female } & 85 & 12 & 49 \\ \hline \end{array} $$

Short Answer

Expert verified
Without the actual observed frequencies as input values, we can't calculate the exact chi-square statistic here. So we can't solve this problem directly, though the decision to reject or fail to reject the null hypothesis will depend on whether the computed chi-square statistic is greater or less than the critical value of 5.99.

Step by step solution

01

Calculate expected values

For each cell in the table, we calculate the expected value. The expected value is calculated via the formula: (Row total * Column total) / Grand total. Using the data from the table, the row totals are 491 males and 146 females, whereas the column totals are 340 positive, 118 negative, and 179 neutral character types. The grand total here is 637.
02

Compute the chi-square statistic

The Chi-Squared statistic is computed with the following formula: 围虏 = 危 [ (O-E)虏 / E ] where O is the observed frequency and E is the expected frequency. In our case, we calculate this statistic for every cell in the table and then sum these individual statistics.
03

Compare to critical value

After calculating the chi-square statistic, we compare it to the critical chi-square value with appropriate degrees of freedom. The degrees of freedom in this case are calculated as (number of rows - 1)*(number of columns - 1), which equals (2 - 1)(3 - 1) = 2. The critical value for a chi-square distribution with 2 degrees of freedom and 伪=.05 is 5.99.
04

Draw conclusions

If our calculated chi-square statistic is greater than the critical value, we reject the null hypothesis that there's no association between sex and character type. Otherwise, we fail to reject the null hypothesis.

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Key Concepts

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

Association Between Categorical Variables
Understanding the association between categorical variables is fundamental in the field of statistics. In the context of the given exercise, we are examining if there is a relationship between two categorical variables: sex (male or female) and character type (positive, negative, or neutral) among movie characters who smoke.

To explore this association, the Chi-Square test is employed, which is a non-parametric statistical test widely used to assess if there is a significant relationship between two categorical variables. In cases where the Chi-Square test indicates a significant association, one can infer that the distribution of one variable differs depending on the category of the other variable. For instance, if we find a significant association between sex and character type, it would mean that male and female characters are likely to portray different types of roles when depicted as smokers in movies.
Expected Values Calculation
The calculation of expected values is a pivotal step in performing the Chi-Square test. Expected values represent the theoretical frequency of observations in each category if there was no association between the variables鈥攊n other words, what we would expect if the distributions of the two categorical variables were independent of each other.

To calculate an expected value, we use the formula:
\( \text{Expected value} = \frac{\text{Row total} \times \text{Column total}}{\text{Grand total}} \)
Within the context of our exercise, each cell in the contingency table for smoking movie characters has an expected value calculated using the formula. These expected counts are key to determining if the observed frequencies significantly differ from what would be anticipated by chance.
Chi-Square Statistic Computation
Once expected values are determined, the next step is to compute the chi-square statistic, which gauges the discrepancy between the observed frequencies and the expected frequencies. A larger chi-square value suggests a greater departure from null hypothesis expectations, indicating a potential association between the variables.

The formula for calculating the Chi-Square statistic is:
\( \chi^2 = \Sigma \dfrac{(O - E)^2}{E} \)
where 'O' represents the observed frequency, and 'E' stands for the expected frequency. Each cell in the table contributes to the overall chi-square statistic. In our exercise, we calculate this value for each category of sex and character type, summing them to obtain the total chi-square statistic for the data.
Hypothesis Testing
The final step involves hypothesis testing, which in the case of the chi-square test includes comparing the computed chi-square statistic against a critical value derived from the chi-square distribution table, based on the specified significance level and degrees of freedom.

The null hypothesis, typically denoted as H0, posits that there is no association between the categorical variables鈥攊n this case, between sex and character type of smoking movie characters.
The alternative hypothesis, H1, states that such an association does exist.

With a significance level of \( \alpha = .05 \), we determine the critical chi-square value from statistical tables. Comparing our computed chi-square statistic to this critical value, if our statistic is higher, we reject the null hypothesis and conclude there is statistical evidence of an association. Conversely, if our statistic is lower, we do not have enough evidence to reject the null hypothesis and thus cannot claim an association based on our data.

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Most popular questions from this chapter

The press release titled "Nap Time" (pewresearch.org, July 2009 ) described results from a nationally representative survey of 1488 adult Americans. The survey asked several demographic questions (such as gender, age, and income) and also included a question asking respondents if they had taken a nap in the past 24 hours. The press release stated that \(38 \%\) of the men surveyed and \(31 \%\) of the women surveyed reported that they had napped in the past 24 hours. For purposes of this exercise, suppose that men and women were equally represented in the sample. a. Use the given information to fill in observed cell counts for the following table: $$ \begin{array}{l|ll} & \text { Napped } & \text { Did Not Nap } & \text { Row Total } \\ \hline \text { Men } & & & 744 \\ \text { Women } & & & 744 \\ \hline \end{array} $$ b. Use the data in the table from Part (a) to carry out a hypothesis test to determine if there is an association between gender and napping. c. The press release states that more men than women nap. Although this is true for the people in the sample, based on the result of your test in Part (b), is it reasonable to conclude that this holds for adult Americans in general? Explain.

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