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This is a continuation of Exercise 20 in Chapter \(12 .\) A case-control study in Berlin, reported by Kohlmeier et al. (1992) and by Hand et al. (1994), asked 239 lung cancer patients and 429 controls (matched to the cases by age and sex) whether they had kept a pet bird during adulthood. of the 239 lung cancer cases, 98 said yes. of the 429 controls, 101 said yes. a. State the null and alternative hypotheses for this situation. b. Construct a contingency table for the data. c. Calculate the expected counts. d. Calculate the value of the chi-square statistic. e. Make a conclusion about statistical significance using a level of \(0.05 .\) State the conclusion in the context of this situation.

Short Answer

Expert verified
Significant association between keeping a pet bird and lung cancer.

Step by step solution

01

Formulate Hypotheses

First, articulate the null and alternative hypotheses: - Null Hypothesis ( H_0 ): There is no association between keeping a pet bird during adulthood and lung cancer cases. - Alternative Hypothesis ( H_a ): There is an association between keeping a pet bird during adulthood and lung cancer cases.
02

Construct Contingency Table

Create a contingency table using the provided data: | | Lung Cancer | Control | Total | |------------------|-------------|---------|-------| | Kept a pet bird | 98 | 101 | 199 | | Did not keep a bird | 141 | 328 | 469 | | Total | 239 | 429 | 668 |
03

Calculate Expected Counts

Using the formula for expected counts, \( E_{ij} = \frac{(Row_i \; Total) \times (Column_j \; Total)}{Grand \; Total} \), calculate the expected counts:- Expected count for 'Kept a pet bird' and 'Lung Cancer': \(\frac{(199 \times 239)}{668} \approx 71.24\)- Expected count for 'Kept a pet bird' and 'Control': \(\frac{(199 \times 429)}{668} \approx 127.76\)- Expected count for 'Did not keep a bird' and 'Lung Cancer': \(\frac{(469 \times 239)}{668} \approx 167.76\)- Expected count for 'Did not keep a bird' and 'Control': \(\frac{(469 \times 429)}{668} \approx 301.24\)
04

Calculate Chi-Square Statistic

Use the formula for the chi-square statistic,\[ \chi^2 = \sum \frac{(Observed - Expected)^2}{Expected} \], to calculate:- \( \chi^2 = \frac{(98 - 71.24)^2}{71.24} + \frac{(101 - 127.76)^2}{127.76} + \frac{(141 - 167.76)^2}{167.76} + \frac{(328 - 301.24)^2}{301.24} \)- Calculate each component and sum them: \( \chi^2 \approx 13.16 \)
05

Interpret the Result

Using a significance level of 0.05 and degrees of freedom \((df = 1)\), compare the calculated chi-square statistic with the critical value from the chi-square distribution table.- The critical value for \(df = 1\) at \(\alpha = 0.05\) is 3.841.- Since \(\chi^2 \approx 13.16 \) is greater than 3.841, reject the null hypothesis.
06

State the Conclusion

In the context of this study, there is statistically significant evidence at the 0.05 level to suggest an association between keeping a pet bird during adulthood and being a lung cancer case.

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Key Concepts

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

Null Hypothesis
The null hypothesis is a fundamental concept in statistics, especially when conducting tests like the chi-square test. In the context of this study, the null hypothesis, often denoted as \(H_0\), assumes there is no association between two factors. Here, it means believing that there is no link between keeping a pet bird and the occurrence of lung cancer in individuals.
This assumption is crucial because it provides a baseline for comparison when evaluating data. By starting with the idea that there's no effect or association, researchers can use statistical methods to test whether observed data provide enough evidence to discard this assumption.
It acts like a "status quo" statement that requires substantial evidence to be overturned. If the null hypothesis is not rejected, it suggests that the study did not find significant evidence of an association, although it does not prove that no association exists.
Alternative Hypothesis
Contrasting with the null hypothesis, the alternative hypothesis represents a different possibility. It is labeled as \(H_a\) and suggests that there is some kind of association or effect. In this specific study, the alternative hypothesis posits that keeping a pet bird during adulthood is associated with an increased likelihood of developing lung cancer.
The alternative hypothesis is what researchers typically hope to support. It is formulated to challenge the null hypothesis and is accepted if the statistical evidence strongly contradicts the null hypothesis.
When a test statistic shows results far from what the null hypothesis would predict, the alternative hypothesis gains weight. This is what occurred in this study, where the results were enough to reject the null hypothesis.
Contingency Table
A contingency table is a table used to display the frequency distribution of variables. It helps in organizing data to see the relationship between different variables. In this study, a contingency table shows the responses from people with lung cancer and the control group regarding pet bird ownership.
The table has rows and columns corresponding to the categories being compared. For instance, one row represents those who kept birds and another for those who didn’t. Corresponding columns indicate lung cancer patients and healthy controls, allowing a clear visual comparison.
This table is important for calculating the expected counts and providing a groundwork for the chi-square test. It simplifies the data and presents it in a structured manner to facilitate further analysis.
Expected Counts
Expected counts are calculated to understand what the data distribution would look like if there was no association between the variables, according to the null hypothesis. Using the formula \( E_{ij} = \frac{(Row_i \; Total) \times (Column_j \; Total)}{Grand \; Total} \), expected counts provide a baseline comparison to actual observed counts.
In this study, expected counts indicate how many people in each scenario (owned a bird, didn’t own a bird) would be expected in both groups (lung cancer cases and controls) if there was truly no association between bird ownership and lung cancer.
  • An important part of chi-square calculations, they allow researchers to quantify how much the observed data deviates from what would be expected under the null hypothesis. Large deviations may suggest evidence against the null hypothesis, supporting the alternative hypothesis.
Understanding expected counts helps make sense of why certain data might differ noticeably from assumptions, leading to significant findings in studies such as this one.

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Most popular questions from this chapter

"Research shows women harder hit by. hangovers" and the accompanying Original Source \(2 .\) In the study, 472 men and 758 women, all of whom were college students and alcohol drinkers, were asked about whether they had experienced each of 13 hangover symptoms in the previous year. What population do you think is represented by the sample for this study? Explain.

If there is a relationship between two variables in a population, which is more likely to result in a statistically significant relationship in a sample from that population \(-\) a small sample, a large sample, or are they equivalent? Explain.

The chi-square test described in this chapter can be used for tables with more than two rows and/or columns. Use software (such as Excel), a calculator, or a website to find the \(p\) -value for each of the following chi-square statistics calculated from a table of the specified number of rows and columns. You may round off your answer to three decimal places. In each case, specify the degrees of freedom you used to find the \(p\) -value. (Hint: Remember that in general, degrees of freedom for a table with \(r\) rows and \(c\) columns are \(\mathrm{df}=(r-1)(c-1) .\) ) Then make a conclusion about whether you would reject the null hypothesis for a test with level \(0.05\). a. Two rows and three columns; chi-square statistic \(=7.4\) b. Three rows and three columns; chi-square statistic \(=11.15\) c. Four rows and three columns; chi-square statistic \(=7.88\) d. Four rows and two columns; chi-square statistic \(=12.20\)

Explain whether each of the following is possible. A relationship exists in the observed sample but not in the population from which the sample was drawn. A relationship does not exist in the observed sample but does exist in the population from which the sample was drawn. A relationship does not exist in the observed sample, but an analysis of the sample shows that there is a statistically significant relationship, so it is inferred that there is a relationship in the population.

For each of the following situations, would a chi-square test based on a \(2 \times 2\) table using a level of 0.01 be statistically significant? Justify your answer. a. chi-square statistic \(=1.42\) b. chi-square statistic \(=14.2\) c. \(p\) -value \(=0.02\) d. \(p\) -value \(=0.15\)

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