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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 conducting Fisher's exact test with the following fictitious table of data. Let the null hypothesis be that treatment and response are independent, and let the alternative be the directional hypothesis that treatment \(\mathrm{B}\) is better than treatment \(\mathrm{A}\). List the tables of possible outcomes that more strongly support \(H_{A}\). $$ \begin{array}{|ll|cc|c|} \hline & & {\text { Treatment }} & \\ & & \text { A } & \text { B } & \text { Total } \\ \hline \text { Outcome } & \text { Die } & 4 & 2 & 6 \\ & \text { Live } & 10 & 14 & 24 \\ & \text { Total } & 14 & 16 & 30 \\ \hline \end{array} $$

In an experiment to treat patients with "generalized anxiety disorder," the drug hydroxyzine was given to 71 patients, and 30 of them improved. A group of 70 patients were given a placebo, and 20 of them improved. \(^{51}\) Let \(p_{1}\) and \(p_{2}\) represent the probabilities of improvement using hydroxyzine and the placebo, respectively. Construct a \(95 \%\) confidence interval for \(\left(p_{1}-p_{2}\right)\).

Phenytoin is a standard anticonvulsant drug that unfortunately has many toxic side effects. A study was undertaken to compare phenytoin with valproate, another drug in the treatment of epilepsy. Patients were randomly allocated to receive either phenytoin or valproate for 12 months. Of 20 patients receiving valproate, 6 were free of seizures for the 12 months, while 6 of 17 patients receiving phenytoin were seizure free. (a) Consider a chi-square test to compare the seizure-free response rates for the two drugs using a nondirectional alternative (i) State the null and alternative hypotheses in symbols. (ii) What is the value of the test statistic? (iii) The \(P\) -value for the test is \(0.73 .\) If \(\alpha=0.10,\) what is your conclusion regarding the hypotheses in (ii)? (b) Do your conclusions in part (a)(iii) provide evidence that valproate and phenytoin are equally effective in preventing seizures? Discuss.

The growth factor pleiotrophin is associated with cancer progression in humans. In an attempt to monitor the growth of tumors, doctors measured serum pleiotrophin levels in patients with pancreatic cancer and in a control group of patients. They found that only 2 of 28 control patients had serum levels more than two standard deviations above the control group mean, whereas 20 of 41 cancer patients had serum levels this high. \(^{27}\) Use Fisher's exact test to determine whether a discrepancy this large \((2\) of 28 versus 20 of 41\()\) is likely to happen by chance. Use a directional alternative and let \(\alpha=0.05\)

An experiment involving subjects with schizophrenia compared "personal therapy" to "family therapy." Only 2 out of 23 subjects assigned to the personal therapy group suffered psychotic relapses in the first year of the study, compared to 8 of the 24 subjects assigned to the family therapy group. \(^{29}\) Is this sufficient evidence to conclude, at the 0.05 level of significance, that the two types of therapies are not equally effective? Conduct Fisher's exact test using a nondirectional alternative.

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