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Chapter 10: Problem 7

A group of patients with a binge-eating disorder were randomly assigned to take either the experimental drug fluvoxamine or a placebo in a 9-week long double-blind clinical trial. At the end of the trial the condition of each patient was classified into one of four categories: no response, moderate response, marked response, or remission. The following table shows a cross classification of the data. \({ }^{38}\) Is there statistically significant evidence, at the 0.10 level, to conclude that there is an association between treatment group (fluvoxamine versus placebo) and condition? $$ \begin{array}{|lccccc|} \hline & \text { No response } & \text { Moderate response } & \text { Marked response } & \text { Remission } & \text { Total } \\ \hline \text { Fluvoxamine } & 15 & 7 & 3 & 15 & 40 \\ \text { Placebo } & 22 & 7 & 3 & 11 & 43 \\ \text { Total } & 37 & 14 & 6 & 26 & \\ \hline \end{array} $$

Short Answer

Expert verified
To determine if there is an association between the treatment and the condition, perform a Chi-square test, calculate the degrees of freedom, find the critical value or p-value, and conclude whether to reject the null hypothesis at the 0.10 significance level.

Step by step solution

01

- State the Hypotheses

State the null hypothesis (H0) which claims that there is no association between the treatment group and the condition of the patients. That is, the treatment outcome is independent of whether the patient received fluvoxamine or a placebo. State the alternative hypothesis (H1) which claims that there is an association between the treatment group and the condition of the patients.
02

- Conduct a Chi-Square Test

We will use a Chi-square test to determine if there is a significant association between treatment group and condition. Calculate the expected counts for each cell in the table assuming the null hypothesis is true. Then, use the observed counts and expected counts to compute the test statistic.
03

- Determine the Degrees of Freedom

The degrees of freedom for a Chi-square test are calculated as (number of rows - 1) times (number of columns - 1). In our case, (2 - 1) * (4 - 1) = 3.
04

- Find the Critical Value or P-Value

Consult the Chi-square distribution table with 3 degrees of freedom to find the critical value for a 0.10 significance level, or use a statistical software to find the p-value directly.
05

- Make a Decision

Compare the test statistic to the critical value or the p-value to the significance level. If the test statistic is greater than the critical value or the p-value is less than the significance level, reject the null hypothesis.
06

- Draw a Conclusion

Based on the decision made in the previous step, conclude whether or not there is statistically significant evidence at the 0.10 level to support an association between treatment group and patient condition.

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Key Concepts

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

Statistical Significance
Understanding statistical significance is crucial when analyzing data in research. It tells us if the outcome of a study is likely due to a specific factor or simply a random occurrence. In the context of our fluvoxamine clinical trial example, we're trying to determine if the variances observed in patient conditions are associated with the treatment (either fluvoxamine or placebo) or just random chance.

When researchers set a level of significance (in this case, 0.10), they're specifying the probability of rejecting the null hypothesis when it's actually true. Think of this significance level as a strictness setting on how we make decisions from our data - the lower it is, the more stringent and confident we are that the results aren't by chance. If our test yields a p-value lower than the significance level, we have enough evidence to suggest a relationship between the variables studied - here, the treatment and patient response outcomes.
Null Hypothesis
The null hypothesis (H_0), often denoted as H_0, is a default statement that there is no effect or no association between two measured phenomena. In clinical trials, such as the one comparing fluvoxamine to a placebo, the null hypothesis posits that the drug has no effect on the outcome of patients' conditions compared to the placebo.

It serves as a starting point for statistical testing, where the aim is to either provide evidence against this hypothesis or fail to find enough evidence to dispute it. The null hypothesis is helpful because it sets a clear criterion for decision-making: if there is sufficient evidence to reject it, researchers may accept the alternative hypothesis which suggests that there is some effect or association present in the data.
Degrees of Freedom
Degrees of freedom are an essential component in many statistical tests, including the Chi-square test, because they define the number of values in the final calculation that are free to vary. It's a bit like determining how much 'wiggle room' your statistics have. To calculate the degrees of freedom (DF) in a Chi-square test, you take the number of categories in one variable and subtract one, then do the same for the other variable, and finally multiply these two numbers together.

The formula is given by DF = (number of rows - 1) - (number of columns - 1), which in our exercise with the 2x4 table (2 treatment groups and 4 condition categories), equates to (2-1)-(4-1) = 3 degrees of freedom. This number is crucial for determining the correct critical values from the Chi-square distribution to make decisions about our null hypothesis.

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Most popular questions from this chapter

Consider a study to investigate a certain suspected disease-causing agent. One thousand people are to be chosen at random from the population; each individual is to be classified as diseased or not diseased and as exposed or not exposed to the agent. The results are to be cast in the following contingency table: Let EY and EN denote exposure and nonexposure and let DY and DN denote presence and absence of the disease. Express each of the following statements in terms of conditional probabilities. (Note that "a majority" means "more than half.") (a) The disease is more common among exposed than among nonexposed people. (b) Exposure is more common among diseased people than among nondiseased people. (c) Exposure is more common among diseased people than is nonexposure. (d) A majority of diseased people are exposed. (e) A majority of exposed people are diseased. (f) Exposed people are more likely to be diseased than are nonexposed people. (g) Exposed people are more likely to be diseased than to be nondiseased

Most salamanders of the species \(P\). cinereus are red striped, but some individuals are all red. The all-red form is thought to be a mimic of the salamander \(N\). viridescens, which is toxic to birds. In order to test whether the mimic form actually survives more successfully, 163 striped and 41 red individuals of \(P\). cinereus were exposed to predation by a natural bird population. After 2 hours, 65 of the striped and 23 of the red individuals were still alive. \({ }^{4}\) Use a chi-square test to assess the evidence that the mimic form survives more successfully. Use a directional alternative and let \(\alpha=0.05 .\) (a) State the null hypothesis in words. (b) State the null hypothesis in symbols. (c) Compute the sample survival proportions for each group and display the values in a table similar to Table 10.1 .2 (d) Find the value of the test statistic and the \(P\) -value. (e) State the conclusion of the test in the context of this setting.

Children with juvenile arthritis were randomly assigned to receive the drug tocilizumab or a placebo for 12 weeks. In the tocilizumab group, 64 out of 75 patients showed marked improvement versus 9 out of 37 for the placebo group. \(^{52}\) An appropriate \(95 \%\) confidence interval is (0.43,0.75) . Write a sentence that interprets this confidence interval, in context. Use cause- effect language if appropriate or say why no causal statement can be made.

For each of the following tables, calculate (i) the relative risk and (ii) the odds ratio. $$\begin{aligned}&\text { (a) }\\\&\begin{array}{|rr|}\hline 25 & 23 \\\492 & 614 \\\\\hline\end{array}\end{aligned}$$ $$\begin{aligned}&\text { (b) }\\\&\begin{array}{|cr|}\hline 12 & 8 \\\93 & 84 \\\\\hline\end{array}\end{aligned}$$

Many over-the-counter decongestants and appetite suppressants contain the ingredient phenylpropanolamine. A study was conducted to investigate whether this ingredient is associated with strokes. The study found that 6 of 702 stroke victims had used an appetite suppressant containing phenylpropanolamine, compared to only 1 of 1,376 subjects in a control group. The following table summarizes these data. (a) Calculate the sample value of the odds ratio. (b) Construct a \(95 \%\) confidence interval for the population value of the odds ratio. (c) Upon hearing of these data, some scientists called the study "inconclusive" because the numbers of users of appetite suppressants containing phenylpropanolamine (7 total: 6 in one group and 1 in the other) are so small. What is your response to these scientists?
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