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Aspirin and heart attacks for women A study in the New England Journal of Medicine compared cardiovascular events for treatments of low-dose aspirin or placebo among 39,876 healthy female health eare providers for anaverage duration of about 10 years. Results indicated that women receiving aspirin and those receiving placebo did not differ for rates of a first major cardiovascular event, death from cardiovascular causes, or fatal or nonfatal heart attacks. However, women receiving aspirin had lower rates of stroke than those receiving placebo (data from N. Engl.J.Med., vol. \(352,2005, \mathrm{pp} .1293-1304) .\) \begin{tabular}{lccc} \hline \multicolumn{3}{c} { Women's Aspirin Study Data } \\ \hline Group & Mini-Stroke & Stroke & No Strokes \\ \hline Placebo & 240 & 259 & 19443 \\ Aspirin & 185 & 219 & 19530 \\ \hline \end{tabular} a. Use software to test independence. Show (i) assump- tions, (ii) hypotheses, (iii) test statistic, (iv) P-value, (v) conclusion in the context of this study. b. Describe the association by finding and interpreting the relative risk for the stroke category.

Short Answer

Expert verified
Aspirin use is associated with a reduced risk of strokes, but overall there's no independence between treatments and strokes according to the chi-square test.

Step by step solution

01

Calculate the Contingency Table Totals

First, we calculate the row totals, column totals, and grand total for the data. This will help us in further calculations for the chi-square test. Placebo Totals: - Mini-Stroke: 240 - Stroke: 259 - No Strokes: 19443 - Total = 240 + 259 + 19443 = 19942 Aspirin Totals: - Mini-Stroke: 185 - Stroke: 219 - No Strokes: 19530 - Total = 185 + 219 + 19530 = 19934 Column Totals: - Mini-Stroke = 240 + 185 = 425 - Stroke = 259 + 219 = 478 - No Strokes = 19443 + 19530 = 38973 Grand Total = 19942 + 19934 = 39876.
02

State Hypotheses and Assumptions

We need to define the hypotheses we are testing: - **Null hypothesis (H鈧):** Aspirin is independent of stroke incidence (no association). - **Alternative hypothesis (H鈧):** Aspirin is not independent of stroke incidence (there is an association). The assumptions for the chi-square test include: 1. The sample should be large enough, with an expected frequency greater than 5 for each cell. 2. Observations should be independent.
03

Calculate Expected Frequencies

We calculate the expected frequencies for each cell using the formula: \[ E_{ij} = \frac{{(Row \, Total) imes (Column \, Total)}}{{Total \, Sample \, Size}} \]For **Placebo Mini-Stroke**: \[ E = \frac{19942 imes 425}{39876} \approx 212.89 \]For **Aspirin Mini-Stroke**: \[ E = \frac{19934 imes 425}{39876} \approx 212.11 \]Continue similarly for other cells using the same formula.
04

Calculate Chi-Square Test Statistic

The chi-square test statistic is calculated using the formula:\[ \chi^2 = \sum \frac{{(O_{ij} - E_{ij})^2}}{E_{ij}} \]Substitute each observed (O) and expected (E) value:For all categories: sum the squared differences divided by expected frequency.For example:\[ \chi^2_{mini-stroke,placebo} = \frac{(240 - 212.89)^2}{212.89} + ... \] Calculate similar for all cells and sum them up.
05

Find P-value and Conclude

With the computed chi-square statistic from the previous step, find the corresponding P-value using the chi-square distribution with Degrees of freedom = (Number of rows - 1) * (Number of columns - 1) = 2 * 2 = 4. Assume a significance level of 0.05. Compare the P-value to the significance level to conclude: - If P-value 鈮 0.05, reject the null hypothesis (there is an association). - If P-value > 0.05, fail to reject the null hypothesis (no association).
06

Calculate Relative Risk for Stroke

Relative risk (RR) helps us understand the risk of strokes due to a daily aspirin regimen:\[ RR = \frac{E_{aspirin,stroke}/N_{aspirin}}{E_{placebo,stroke}/N_{placebo}} = \frac{219/19934}{259/19942} \approx \frac{0.011}{0.013} \approx 0.846 \]Interpretation: Women taking aspirin have 0.846 times the risk of stroke compared to those taking placebo, indicating a lower relative risk associated with aspirin use.

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Key Concepts

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

Chi-Square Test
A chi-square test is a statistical method used to determine if there's a significant association between two categorical variables. It's especially useful in experiments with data arranged in contingency tables. In this context of the aspirin study, we want to test if taking aspirin or placebo impacts the occurrence of strokes among women. To begin, we define the assumptions required for the chi-square test:
  • The sample size must be sufficiently large, ideally with each expected frequency over 5.
  • Each observation should be independent of the others.
The chi-square test involves calculating expected frequencies for each situation based on the assumption that there's no association. We do this using the formula: \[ E_{ij} = \frac{{(\text{Row Total}) \times (\text{Column Total})}}{\text{Grand Total}} \] This calculation helps us understand what the data distribution would look like if there were no association between aspirin use and stroke occurrences.Once the expected frequencies are known, the chi-square statistic \( \chi^2 \) is calculated. This sum evaluates the differences between observed and expected frequencies, allowing us to judge whether these differences are too large to occur by chance alone. By comparing the calculated \( \chi^2 \) to a critical value from the chi-square distribution at a given significance level (like 0.05), we decide whether to reject or fail to reject the null hypothesis that suggests no association exists.
Hypothesis Testing
Hypothesis testing is a procedure in statistics that allows us to make inferences about population parameters based on sample data. You begin by proposing a null hypothesis \( H_0 \), which represents a statement of no effect or association, and an alternative hypothesis \( H_1 \) that suggests an effect or association exists. For our aspirin study, these hypotheses are:
  • Null hypothesis (\( H_0 \)): Aspirin intake is independent of stroke incidence.
  • Alternative hypothesis (\( H_1 \)): Aspirin intake is not independent of stroke incidence.
The significance level, commonly set at 0.05, is the threshold for deciding whether to reject \( H_0 \). A hypothesis test, like the chi-square test, computes a test statistic that informs us about the sample鈥檚 data in relation to the population. With this, we calculate a p-value, quantifying the probability of obtaining results as extreme as the observed, assuming \( H_0 \) is true.If this p-value is less than or equal to the significance level, we reject the null hypothesis in favor of the alternative hypothesis. This suggests there's statistically significant evidence of a real effect or association, such as aspirin having a relationship with stroke occurrence. If it鈥檚 more, we conclude that the data do not provide enough evidence to say the effect exists and thus do not reject the null hypothesis.
Relative Risk
Relative risk (RR) is a measure used to determine the strength of the association between an exposure (like taking aspirin) and an outcome (like having a stroke). It compares the risk of a certain event happening in the exposed group with the risk in the non-exposed group. In the aspirin study, we compute it to understand how much aspirin impacts stroke risk.The formula used is \[ RR = \frac{E_{\text{aspirin, stroke}}/N_{\text{aspirin}}}{E_{\text{placebo, stroke}}/N_{\text{placebo}}} \]This ratio tells us how much more or less likely strokes are among aspirin users compared to placebo takers. An RR of < 1 implies reduced risk; for instance, an RR of 0.846 indicates a lower risk in the aspirin group, meaning they have fewer strokes relative to the placebo group. Conversely, an RR > 1 would suggest increased risk.While RR can guide health decisions, it's crucial to consider the absolute risk as it provides context. This is especially helpful in understanding the real-world impact or public health implications, balancing the benefit of aspirin against possible side effects or risks.

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Most popular questions from this chapter

Standardized residuals for \(2 \times 2\) tables The table that follows shows the standardized residuals in parentheses for GSS data about the statement, "Women should take care of running their homes and leave running the country up to men." The absolute value of the standardized residual is 13.2 in every cell. For chi-squared tests with \(2 \times 2\) tables, since \(d f=1,\) only one nonredundant piece of information exists about whether an association exists. If observed count \(\geq\) expected count in one cell, observed count \(<\) expected count in the other cell in that row or column. Explain why this is true, using the fact that observed and expected counts have the same row and column totals. (In fact, in \(2 \times 2\) tables, all four standardized residuals have absolute value equal to the square root of the \(X^{2}\) test statistic.) \begin{tabular}{lcc} \hline Year & Agree & Disagree \\ \hline 1974 & 509(13.2) & 924(-13.2) \\ 1998 & 280(-13.2) & 1534(13.2) \\ \hline \end{tabular}

Keeping old dogs mentally sharp In an experiment with beagles ages \(7-11,\) the dogs attempted to learn how to find a treat under a certain black-colored block and then relearn that task with a white-colored block. The control group of dogs received standard care and diet. The diet and exercise group were given dog food fortified with vegetables and citrus pulp and vitamin \(E\) and C supplements plus extra exercise and social play. All 12 dogs in the diet and exercise group were able to solve the entire task, but only 2 of the 8 dogs in the control group could do so. (Background material from \(\mathrm{N}\). W. Milgram et al., Neurobiology of Aging, vol. \(26,2005,\) pp. \(77-90 .)\) a. Show how to summarize the results in a contingency table. b. Conduct all steps of Fisher's exact test of the hypothesis that whether a dog can solve the task is independent of the treatment group. Use the two- sided alternative hypothesis. Interpret. c. Why is it improper to conduct the chi-squared test for these data?

Smelling and mortality A recent study (Pinto et al., Olfactory Dysfunction Predicts 5-Year Mortality in Older Adults. PLoS ONE \(9(10): \mathrm{e} 107541,2014)\) mentions that anosmic (those with almost no sense of smell) older adults had more than three times the odds of death over a 5-year span compared to normosmic (those with normal smell) individuals. Does this imply that anosmic older adults were more than three times as likely to die over the next 5 years than normosmic ones? Explain.

In the GSS, subjects who were married were asked about the happiness of their marriage, the variable coded as HAPMAR. a. Go to the GSS website sda.berkeley.edu/GSS/, click GSS with no weight as the default, and construct a contingency table for 2012 relating family income (measured as in Table 11.1 ) to marital happiness: Enter FINRELA(r:4;3;2) as the row variable (where \(4=\) "above average," \(3=\) "average, \("\) and \(2=\) "below average") and HAPMAR(r:3;2;1) as the column variable \((3=\) not \(t o 0,2=\) pretty \(,\) and \(1=\) very happy \()\) As the selection filter, enter YEAR(2012). Under Output Options, put a check in the row box (instead of in the column box) for the Percentaging option and put a check in the Summary Statistics box further below. Click on Run the Table. b. Construct a table or graph that shows the conditional distributions of marital happiness, given family income. How would you describe the association? c. Compare the conditional distributions to those in Table \(11.2 .\) For a given family income, what tends to be higher, general happiness or marital happiness for those who are married? Explain.

True or false: Statistical but not practical significance Even when the sample conditional distributions in a contingency table are only slightly different, when the sample size is very large it is possible to have a large \(X^{2}\) statistic and a very small P-value for testing \(\mathrm{H}_{0}\) independence.

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