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Chapter 10: Problem 46

In a 2018 article published in The Lancet, Kappos et al. studied the effect of the drug siponimod in treating patients with secondary progressive multiple sclerosis (SPMS) using a double-blind, randomized, controlled study. Of the 1099 patients given the drug, 198 experienced a severe adverse outcome. Of the 546 patients given the placebo, 82 experienced a severe adverse outcome. a. Find the percentage in each group that suffered a severe adverse outcome. b. Create a two-way table with the treatment labels (drug/placebo) across the top. c. Test the hypothesis that treatment and severe adverse outcome are associated using a significance level of \(0.05\).

Short Answer

Expert verified
The percentage in drug and placebo group that had a severe adverse outcome are 18.02% and 15.02% respectively. Based on the p-value from the Chi-Square Test of Independence, decision on whether to reject or fail to reject the null hypothesis would be made.

Step by step solution

01

Finding the percentage

First, calculate the percentage of patients who had a severe adverse outcome in both groups. In the drug group, the percentage is calculated as 198 out of 1099, given by \[\text{{Percentage (Drug Group)}} = \left(\frac{{198}}{{1099}}\right) \times 100 = 18.02\%\] In the placebo group, the percentage is calculated as 82 out of 546, which is \[\text{{Percentage (Placebo Group)}} = \left(\frac{{82}}{{546}}\right) \times 100 = 15.02\%\]
02

Creating a two-way table

Next, create a two-way table reflecting the number of patients in each group with and without severe adverse outcomes. \[ \begin{{array}}{{ccc}} & \text{{Drug Group}} & \text{{Placebo Group}} \ \text{{Severe Adverse Outcome}} & 198 & 82 \ \text{{No Severe Adverse Outcome}} & 901 & 464 \ \text{{Total}} & 1099 & 546\ \end{{array}} \]
03

Hypothesis test

To test whether the treatment and severe adverse outcomes are associated, run a chi-square test of independence. Here, the null hypothesis \(H_0\) is that treatment and severe adverse outcomes are independent, while the alternative hypothesis \(H_1\) is that they are not independent. By using the stats software, generate the test statistic and p-value for the chi-square test. If the p-value is less than the significance level \(0.05\), reject the null hypothesis.

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Key Concepts

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

Percentage Calculation
Calculating percentages is a crucial statistical tool that helps us understand proportions between different categories. It's a value that represents a part per hundred of a given quantity. In this exercise, we're interested in determining what percent of patients experienced a severe adverse outcome for both the drug and placebo groups.

To calculate this, we use the formula:
  • Percentage = \( \left( \frac{\text{Number of cases}}{\text{Total number of participants}} \right) \times 100 \)
By applying this formula to the drug group, where 198 out of 1099 patients had adverse outcomes, we get:
  • \(18.02\%\) of patients in the drug group suffered a severe adverse outcome.
  • Similarly, for the placebo group, \(15.02\%\) out of 546 patients experienced severe outcomes.
This step involves division and multiplication to convert the ratio into a percentage, giving a clear and understandable measure of comparison between the two groups.
Two-Way Table
A two-way table is a simple yet powerful tool to organize and display data that involves two categorical variables, such as "treatment type" and "adverse outcomes". It's beneficial for seeing relationships between the variables and making comparisons.

In our scenario, the rows of the table represent different outcome categories (severe and no adverse outcomes), while the columns depict the treatment type (drug or placebo). This setup allows for easy comparison at a glance.

The table helps us summarize data:
  • 198 patients who had a severe adverse outcome from the drug.
  • 82 patients who had the same outcome from the placebo.
  • Additionally, we note 901 (drug) and 464 (placebo) patients did not experience severe outcomes.
This organization aids statistically in hypothesis testing, like the chi-square test, by clearly laying out how many belong to each group.
Chi-Square Test
The chi-square test of independence is a statistical method used to determine whether there's a significant association between two categorical variables. In this scenario, these are treatment types (drug vs. placebo) and whether patients suffered severe adverse outcomes.

We begin this test by positing a null hypothesis \(H_0\), suggesting that treatment and outcomes are independent, meaning there's no real connection between the two variables. The alternative hypothesis \(H_1\) suggests they are related.

The test calculates a chi-square statistic, which measures the extent of difference between expected and observed frequencies in our two-way table. Along with this, the test gives a p-value:
  • If the p-value is less than the significance level, often set at 0.05, it points to the null hypothesis being unlikely, suggesting a relationship between treatment and adverse outcomes.
  • A high p-value indicates weak evidence against the null hypothesis, meaning any association could merely be by chance.
Thus, the chi-square test assists us in determining if there's a statistically significant connection in the dataset.

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Most popular questions from this chapter

There is a theory that relative finger length depends on testosterone level. The table shows a summary of the outcomes of an observational study that one of the authors carried out to determine whether men or women were more likely to have a ring finger that appeared longer than their index finger. Identify both of the variables, and state whether they are numerical or categorical. If numerical, state whether they are continuous or discrete. $$ \begin{array}{|lcc|} \hline & \text { Men } & \text { Women } \\ \hline \text { Ring Finger Longer } & 23 & 13 \\ \hline \text { Ring Finger Not Longer } & 4 & 14 \\ \hline \end{array} $$

In a 2016 article published in the Journal of American College Health, Heller et al. surveyed a sample of students at an urban community college. Students' ages and frequency of alcohol use per month are recorded in the following table. Because some of the expected counts are less than 5, we should combine some groups. For this question, combine the frequencies \(10-29\) days and Every day into one group. Label this group \(10+\) days and show your new table. Then test the new table to see whether there is an association between age group and alcohol use using a significance level of \(0.05\). Assume this is a random sample of students from this college. $$ \begin{array}{lcccc} \hline & & \text { Alcohol Use } \\ \text { Age } & \text { None } & \text { 1-9 days } & \text { 10-29 days } & \text { Every day } \\ \hline 18-20 & 182 & 100 & 27 & 4 \\ \hline 21-24 & 142 & 109 & 35 & 4 \\ \hline 25-29 & 49 & 41 & 5 & 2 \\ \hline 30+ & 76 & 32 & 8 & 2 \end{array} $$

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?

The Perry Preschool Project discussed in exercises \(10.39\) to \(10.41\) found that 8 of the 58 students who attended preschool had at least one felony arrest by age 40 and that 31 of the 65 students who did not attend preschool had at least one felony arrest (Schweinhart et al. 2005 ). a. Compare the percentages descriptively. What does this comparison suggest? b. Create a two-way table from the data and do a chi-square test on it, using a significance level of \(0.05 .\) Test the hypothesis that preschool attendance is associated with being arrested. c. Do a two-proportion \(z\) -test. Your alternative hypothesis should be that preschool attendance lowers the chances of arrest. d. What advantage does the two-proportion \(z\) -test have over the chi-square test?

In the study described in exercise \(10.55\). researchers (Du Toits et al., 2015) also studied infants with eczema, egg allergies, or both who also had a preexisting sensitivity to peanut extract. These infants were also randomly assigned to either consume or avoid peanuts until 60 months of age. The numbers in each group developing a peanut allergy by 60 months of age are shown in the following table. $$ \begin{array}{lcc} & \text { Treatment Group } \\ \hline \begin{array}{l} \text { Peanut allergy at age } \\ \text { 60 mos. } \end{array} & \text { Consume peanuts } & \text { Avoid peanuts } \\ \hline \text { Yes } & 5 & 18 \\ \text { No } & 43 & 33 \end{array} $$ a. Compare the percentages in each group that developed a peanut allergy by age 60 months. b. Test the hypothesis that treatment group and peanut allergy are associated using the chi-square statistic. Use a significance level of \(0.05\). c. Do a Fisher's Exact Test for the data with the same significance level. Report the two-tailed p-value and your conclusion. (Use technology to run the test.) d. Compare the p-values for parts \(\mathrm{b}\) and \(\mathrm{c}\). Which is more accurate? Explain.
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