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Chapter 12: Problem 7

The paper "Cigarette Tar Yields in Relation to Mortality from Lung Cancer in the Cancer Prevention Study II Prospective Cohort" (British Medical journal [2004]: 72-79) included the accompanying data on the tar level of cigarettes smoked for a sample of male smokers who subsequently died of lung cancer. $$ \begin{array}{lc} \text { Tar Level } & \text { Frequency } \\ \hline 0-7 \mathrm{mg} & 103 \\ 8-14 \mathrm{mg} & 378 \\ 15-21 \mathrm{mg} & 563 \\ \geq 22 \mathrm{mg} & 150 \\ \hline \end{array} $$ Assume it is reasonable to regard the sample as representative of male smokers who die of lung cancer. Is there convincing evidence that the proportion of male smoker lung cancer deaths is not the same for the four given tar level categories?

Short Answer

Expert verified
Without specific numerical calculations, it is not possible to conclude whether there is convincing evidence that the proportion of male smoker lung cancer deaths is not the same for the four given tar level categories. However, if p-value < 0.05 from the Chi-Square test, then there is statistically significant evidence to reject the null hypothesis in favor of the alternative, implying the proportions are different across the tar level categories.

Step by step solution

01

Define Null and Alternative Hypotheses

The null hypothesis \(H_0\) is that there is no association between tar levels and the proportion of male smokers who die of lung cancer, implying that the proportion remains the same across all tar level categories. The alternative hypothesis \(H_A\) is that there is an association between tar levels and the proportion of male lung cancer deaths, indicating the proportion is different across the tar level categories.
02

Calculate Observed and Expected Frequencies

We are given the observed frequencies in the exercise. To calculate the expected frequencies, we calculate the total number of deaths which is 103 + 378 + 563 + 150 = 1194. Under \(H_0\), we assume each tar level has the same proportion of deaths. Therefore, the expected frequency for each category is the total number of deaths divided by the number of categories, which is 1194 / 4 = 298.5.
03

Compute Chi-Square Test Statistic

The Chi-Square test statistic calculates the sum of the squared difference between observed and expected frequencies divided by the expected frequency for all levels. It is given as \(\chi^2 = \Sigma [ (O_i - E_i)^2 / E_i ] \), where \(O_i\) and \(E_i\) represent the observed and expected frequency for the i-th category. Therefore, we have:\[\chi^2 = [(103 - 298.5)^2 / 298.5] + [(378 - 298.5)^2 / 298.5] + [(563 - 298.5)^2 / 298.5] + [(150 - 298.5)^2 / 298.5]\]Calculating these values will give us our Chi-Square statistic.
04

Determine p-value

Next, we use the Chi-Square distribution with 3 degrees of freedom (4 categories - 1) to find the area to the right of the calculated Chi-Square value. This area is the p-value. If the p-value is less than 0.05, then there is statistically significant evidence to reject the null hypothesis in favor of the alternative.

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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 that helps us decide whether to accept or reject a hypothesis based on sample data. In this exercise, we are particularly interested in comparing proportions across different groups. Here's what happens step by step:

1. **Establish Hypotheses**: We initially assume a null hypothesis. For our exercise, the null hypothesis \( H_0 \) is that there is no difference in the proportion of lung cancer deaths among the four different cigarette tar levels. In contrast, the alternative hypothesis \( H_A \) asserts that there is a significant difference in these proportions.

2. **Collect Data**: We use the given data, which includes the number of deaths in each tar level category.

3. **Perform the Test**: By calculating a test statistic, like the Chi-Square statistic (detailed in later steps), we can determine the likelihood of the observed data under the null hypothesis.

4. **Decision Making**: If this probability (or p-value) is low, typically less than 0.05, we reject the null hypothesis and accept that there is a significant difference in proportions.
Expected Frequencies
In hypothesis testing, especially when using the Chi-Square Test, expected frequencies are crucial. They represent the frequencies we would expect in each category if the null hypothesis were true. Let's break it down:

- **Calculate Total Cases**: First, sum up all observed frequencies, which represents the total number of cases, in this case, the total number of deaths from lung cancer across all tar levels. This amounts to 1194.

- **Equitable Distribution**: If the null hypothesis is true (no proportional differences), we would expect the deaths to be evenly distributed across the 4 tar level categories.

- **Determine Expected Frequencies**: Therefore, the expected frequency for each category = Total number of cases / Number of categories. For our problem, that's \( 1194 / 4 = 298.5 \).

Expected frequencies give us a baseline against which we can compare the observed frequencies.
Observed Frequencies
Observed frequencies are the actual numbers gathered from data that tell us how often something happens. In our exercise, these frequencies are the counted number of lung cancer deaths at each tar level:

- 0-7 mg: 103
- 8-14 mg: 378
- 15-21 mg: 563
- 鈮� 22 mg: 150

These observed frequencies drive our analysis. They help us determine if there's a significant difference from what we expect under the null hypothesis. We compare these observed values to the expected frequencies using a Chi-Square Test. This comparison highlights any disparities and aids in concluding whether these differences are due to chance or a significant factor, like the tar level of cigarettes.

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Most popular questions from this chapter

The paper "Sociochemosensory and Emotional Functions" (Psychological Science [2009]: \(1118-1124\) ) describes an interesting experiment to determine if college students can identify their roommates by smell. Forty-four female college students participated as subjects in the experiment. Each subject was presented with a set of three t-shirts that were identical in appearance. Each of the three t-shirts had been slept in for at least 7 hours by a person who had not used any scented products (like scented deodorant, soap, or shampoo) for at least 48 hours prior to sleeping in the shirt. One of the three shirts had been worn by the subject's roommate. The subject was asked to identify the shirt worn by her roommate. This process was then repeated with another three shirts, and the number of times out of the two trials that the subject correctly identified the shirt worn by her roommate was recorded. The resulting data is given in the accompanying table. $$ \begin{array}{l|ccc} \hline \text { Number of Correct Identifications } & 0 & 1 & 2 \\ \hline \text { Observed Count } & 21 & 10 & 13 \\ \hline \end{array} $$ a. Can a person identify her roommate by smell? If not, the data from the experiment should be consistent with what we would have expected to see if subjects were just guessing on each trial. That is, we would expect that the probability of selecting the correct shirt would be \(1 / 3\) on each of the two trials. It would then be reasonable to regard the number of correct identifications as a binomial variable with \(n=2\) and \(p=1 / 3\). Use this binomial distribution to compute the proportions of the time we would expect to see 0,1, and 2 correct identifications if subjects are just guessing. b. Use the three proportions computed in Part (a) to carry out a test to determine if the numbers of correct identifications by the students in this study are significantly different than what would have been expected by guessing. Use \(\alpha=.05 .\) (Note: One of the expected counts is just a bit less than \(5 .\) For purposes of this exercise, assume that it is \(\mathrm{OK}\) to proceed with a goodness-of-fit test.)

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