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91Ó°ÊÓ

According to the article "Which Adults Do Underage Youth Ask for Cigarettes?" (American Journal of \(P u b\) lic Health \([1999]: 1561-1564), 43.6 \%\) of the 14918 - to 19 -year-olds in a random sample have been asked to buy cigarettes for an underage smoker. a. Is there convincing evidence that fewer than half of 18 to 19 -year-olds have been approached to buy cigarettes by an underage smoker? b. The article went on to state that of the 110 nonsmoking 18- to 19 -year- olds, only \(38.2 \%\) had been approached to buy cigarettes for an underage smoker. Is there evidence that less than half of nonsmoking 18 - to 19 -year- olds have been approached to buy cigarettes?

Short Answer

Expert verified
Yes, there is convincing evidence that fewer than half of 18-19 year-olds and fewer than half of non-smoking 18-19 year-olds have been approached to buy cigarettes for underage smokers, as suggested by the low P-values (< 0.05) obtained from hypothesis tests.

Step by step solution

01

State the Hypotheses

For the first part of exercise, the null and alternative hypotheses are as follows: \nNull Hypothesis (\(H_0\)): \(p = 0.5\), meaning half of 18-19 year olds have been approached to buy cigarettes.\nAlternative Hypothesis (\(H_A\)): \(p < 0.5\), meaning less than half of 18-19 year olds have been approached to buy cigarettes. \nIn the second part, the hypotheses for non-smokers are similar to the above, with the alternative hypothesis stating that less than half of non-smoking 18-19 year-olds have been approached.
02

Test Statistic and P-value

The test statistic (Z score) is calculated using the formula:\n \(Z = (p̂ - p) / \(\sqrt{pq/n}\) \), where \(p = 0.5\)\nFor the first part of the exercise, plug \(p̂ = 0.436\), \(q = 1-p = 0.5 \), and \(n = 14918\) into the formula to get the Z score. Calculating it gives \(Z = -9.853\). \nFor the second part, plug \(p̂ = 0.382\), \(q = 0.5\), and \(n = 110\) into the formula to find the Z score: \(Z = -1.79\). \nNext, find the P-value by looking up the Z scores in a standard normal (Z) table or using statistical software. The P-value for the first part is closer to 0 and essentially zero. For the second part, the P-value is 0.037.
03

Draw Conclusions

If the P-value is less than the significance level (usually 0.05), we reject the null hypothesis. For the first part of the exercise, the P-value is less than 0.05. Therefore, there is convincing evidence that fewer than half of 18-19 year olds have been approached to buy cigarettes. For the second part, the P-value is also less than 0.05 - providing evidence that less than half of non-smoking 18-19 year-olds have been approached to buy cigarettes. However, as the sample size is small, these results should be interpreted with caution.

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Key Concepts

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

Null Hypothesis
Understanding the null hypothesis is essential when conducting hypothesis testing. It's a starting point that suggests there is no effect or no difference in the population; in other words, it assumes the status quo. For example, in our exercise, the null hypothesis (\(H_0\)) posits that half (\(50\text{%}\)) of 18 to 19-year-olds have been asked to buy cigarettes for an underage smoker. Similarly, for nonsmoking 18 to 19-year-olds, it maintains that half have been approached with the same request. It is crucial because it represents the hypothesis that the researcher aims to test against the alternative and is often expressed in a form that allows for mathematical manipulation and calculation.
Alternative Hypothesis
The alternative hypothesis (\(H_A\text{ or }H_1\text{ in other contexts}\)) represents what the researcher really wants to prove. It suggests that there is a change, a difference, or an effect in the population. In contrast to the null, it's the hypothesis that there is something noteworthy or scientifically interesting occurring in our data. For instance, in the provided exercise, the alternative hypothesis is that less than half of the targeted age group – both smokers and non-smokers – have been approached to buy cigarettes for someone underage (\(p < 0.5\)). This hypothesis is what we are looking to find evidence for—it runs counter to the null and is tested through the collection and analysis of data.
Significance Level
The significance level, commonly denoted by \(\alpha\), is the threshold against which we measure the p-value of our test statistic to decide whether we can reject the null hypothesis. It's essentially the probability of making a type I error, which is rejecting the null hypothesis when it's actually true. The most common significance level used is 0.05, indicating a 5% risk of concluding that a difference exists when there is no actual difference. The significance level is pre-determined before carrying out the hypothesis test. If the computed p-value is less than \(\alpha\), we reject the null hypothesis, as is the case in our exercise with both smokers and non-smokers being less likely than not to be asked to buy cigarettes.
Test Statistic
A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It's used to decide whether to reject the null hypothesis. It is calculated under the assumption that the null hypothesis is true. The form of the test statistic depends on the type of test being performed but it is typically a z-score (Z) in large samples when the population variance is known or could be approximated well. In this exercise, we calculate the Z score by using the observed proportion \(p̂\), the hypothesized population proportion \(p\), the number of observations \(n\), and the standard deviation of the sampling distribution. If the test statistic falls into a critical area, which is determined by the significance level, the null hypothesis is rejected.
P-value
The p-value, or probability value, is the probability of obtaining test results at least as extreme as the ones observed during the test, assuming that the null hypothesis is true. It tells us how 'surprising' our data is. A small p-value (\(\le 0.05\text{ for a }5\text{%} \text{significance level}\)) indicates strong evidence against the null hypothesis, so we reject it. In our exercise, the p-values calculated from the Z scores were essentially zero for the first part and 0.037 for the second, both of which are less than the typical \(\alpha = 0.05\) level. This suggests that our data is indeed surprising under the null hypothesis and supports the alternative hypothesis that fewer than half have been approached to buy cigarettes.

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Most popular questions from this chapter

The desired percentage of silicon dioxide in a certain type of cement is \(5.0 \%\). A random sample of \(n=36\) specimens gave a sample average percentage of \(\bar{x}=5.21\). Let \(\mu\) be the true average percentage of silicon dioxide in this type of cement, and suppose that \(\sigma\) is known to be 0.38. Test \(H_{0}: \mu=5\) versus \(H_{a}: \mu \neq 5\) using a significance level of .01.

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