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According to the 2015 High School Youth Risk Behavior Survey, \(41.5 \%\) of high school students reported they had texted or emailed while driving a car or other vehicle. Suppose you randomly sample 80 high school students and ask if they have texted or emailed while driving. Suppose 38 say yes and 42 say no. Calculate the observed value of the chi-square statistic for testing the hypothesis that \(41.5 \%\) of high school students engage in this behavior.

Short Answer

Expert verified
The observed value of the chi-square statistic for the hypothesis is approximately \(1.178\).

Step by step solution

01

Identify sample size and observed frequencies

The sample size is \(80\) students, with \(38\) students admitting to texting while driving and \(42\) students saying no.
02

Identify expected frequencies

According to the High School Youth Risk Behavior Survey, expected proportion who text or email while driving is \(41.5 \% = 0.415\). This means out of \(80\) students, we would expect \(0.415 * 80 = 33.2\) students to have texted while driving. The remaining \(80 - 33.2 = 46.8\) students are expected to say no.
03

Compute Chi-square statistic

The chi-square statistic is given by the formula: \[\chi^{2}= \sum \frac{(O-E)^{2}}{E}\] Where \(O\) is the observed frequency and \(E\) is the expected frequency. Substituting the observed and expected frequencies, we get:\[\chi^{2} = \frac{(38-33.2)^{2}}{33.2} + \frac{(42-46.8)^{2}}{46.8} \approx 0.688 + 0.49 = 1.178 \]
04

Conclusion

The observed value of the chi-square statistic to test the hypothesis that \(41.5 \%\) of high school students text or email while driving is approximately \(1.178\).

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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 statistical method used to determine whether there is enough evidence in a sample of data to infer that a certain condition is true for the entire population. In the context of the chi-square statistic, hypothesis testing is often used to compare observed frequencies with expected frequencies to see if there are significant differences.

In our exercise, the null hypothesis is that the true proportion of students who text or email while driving is indeed the reported 41.5%. The alternative hypothesis is that the actual proportion is different from this percentage. The chi-square statistic helps us to test these hypotheses. If the chi-square value is high enough, it can indicate that our observed frequencies are unlikely to have occurred based on random chance alone, leading to the rejection of the null hypothesis.
Observed Frequency
Observed frequency refers to the count of occurrences or responses that are actually observed in the collected data. It represents the real-world evidence collected from experiments or surveys.

In the provided example, the observed frequencies are the actual responses from the 80 high school students. The frequency reported was that 38 students admitted to texting while driving and 42 stated they do not engage in this behavior. These are the real-world counts that will be compared against what we would expect to see, according to the hypothesis we are testing.
Expected Frequency
Expected frequency is the theorized count of occurrences or responses that we would expect to find if the null hypothesis is true. This is based on probabilistic models and assumptions about the population.

In the exercise, the expected frequency of students who text or email while driving is calculated using the proportion given by the 2015 High School Youth Risk Behavior Survey. Assuming the null hypothesis is true, we would expect 41.5% of the 80 students, which is approximately 33.2 students, to say they text while driving. The expected frequency for students not engaging in this behavior would then be the remainder, which is about 46.8 students.
Statistical Significance
Statistical significance is a determination of whether the observed results in a study are unlikely to have occurred by random chance. It is usually evaluated by a p-value, which is calculated using a test statistic like the chi-square statistic. When the p-value is below a predetermined threshold (commonly 0.05), the result is considered statistically significant, suggesting that the observed data is sufficiently unlikely under the null hypothesis.

In the chi-square test, the statistic calculated provides an idea of how much the observed frequencies deviate from the expected frequencies. The larger the chi-square value, the greater the evidence against the null hypothesis. In our exercise, the resulting chi-square value would be compared to a critical value from a chi-square distribution to determine statistical significance. The p-value would be derived from this comparison. The approximate chi-square statistic of 1.178 by itself does not indicate if the result is significant; this determination would depend on the degrees of freedom and the aforementioned comparison to the critical chi-square values.

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Most popular questions from this chapter

A 2018 Gallup poll asked college graduates if they agreed that the courses they took in college were relevant to their work and daily lives. The respondents were also classified by their field of study. If we wanted to test whether there was an association between response to the question and the field of study of the respondent, should we do a test of independence or homogeneity?

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According to the website MedicalNewsToday.com, coronary artery disease accounts for about \(40 \%\) of deaths in the United States. Many people believe this is due to modern-day factors such as high-calorie fast food and lack of exercise. However, a study published in the Journal of the American Medical Association in November 2009 (www.medicalnewstoday .com) reported on 16 mummies from the Egyptian National Museum of Antiquities in Cairo. The mummies were examined, and 9 of them had hardening of the arteries, which seems to suggest that hardening of the arteries is not a new problem. a. Calculate the expected number of mummies with artery disease (assuming the rate is the same as in the modern day). Then calculate the expected number of mummies without artery disease (the rest). b. Calculate the observed value of the chi-square statistic for these mummies.

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