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Chapter 10: Problem 7

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.

Short Answer

Expert verified
The null hypothesis, which states there is no difference in effectiveness between the two drugs, is not rejected based on the P-value of 0.73 with an alpha level of 0.10. This result does not conclusively prove that the drugs are equally effective; it only shows that there is not sufficient evidence to claim a difference in effectiveness from the data.

Step by step solution

01

State the hypotheses

The null hypothesis (\( H_0 \) states that there is no difference in seizure-free response rates between the two drugs. The alternative hypothesis (\( H_1 \) states that there is a difference in seizure-free response rates between the two drugs.- Null hypothesis: \( H_0: p_{valproate} = p_{phenytoin} \)- Alternative hypothesis: \( H_1: p_{valproate} eq p_{phenytoin} \)
02

Calculate the test statistic

To calculate the chi-square test statistic, use the formula:\[ \chi^2 = \sum \left( \frac{(O - E)^2}{E} \right) \] where \(O\) is the observed frequency and \(E\) is the expected frequency under the null hypothesis. The expected frequencies are calculated based on the assumption that both treatments have equal seizure-free rates.For valproate: \(O_{valproate} = 6\), \(E_{valproate} = \frac{20}{37} \times 12\)For phenytoin: \(O_{phenytoin} = 6\), \(E_{phenytoin} = \frac{17}{37} \times 12\)Insert these values into the formula to calculate the chi-square test statistic.
03

Compare the P-value to alpha

With a given \( P \)-value of 0.73 and a significance level of \( \alpha = 0.10 \), compare the two to make a decision about the null hypothesis. If the \( P \)-value is less than\( \alpha \), the null hypothesis is rejected. If the \( P \)-value is greater than or equal to \( \alpha \), the null hypothesis is not rejected.
04

Formulate the conclusion

Since the \( P \)-value of 0.73 is greater than the significance level of 0.10, you do not reject the null hypothesis. There isn't enough evidence to suggest there is a difference in seizure-free response rates between valproate and phenytoin.
05

Discuss the effectiveness comparison

The conclusion from part (a)(iii) indicates that there is not enough statistical evidence to claim a difference in the effectiveness between valproate and phenytoin in preventing seizures. However, this does not necessarily mean that both drugs are equally effective, as factors such as study design, sample size, and power of the test could affect the conclusion.

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Key Concepts

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

Anticonvulsant Drug Comparison
Anticonvulsant drugs like phenytoin and valproate are critical in managing epilepsy, a neurological condition characterized by recurring seizures. Determining the efficacy of these drugs is essential for improving treatment outcomes for patients. Researchers compare anticonvulsant drugs by looking at various measures such as seizure-free response rates, which reflect the proportion of patients who remain seizure-free over a certain period while on medication.

In studies comparing anticonvulsant drugs, it is crucial to ensure that the comparison is fair and unbiased. This involves random allocation of patients to different treatment groups, consistent dosing protocols, and careful monitoring of outcomes. By utilizing statistical methods, such as the chi-square test, we can objectively evaluate and compare the performance of the drugs based on data collected from these studies.
Seizure-Free Response Rates
Seizure-free response rates are a key indicator in epilepsy treatment as they directly measure the primary goal of therapy, which is to prevent seizures. In clinical trials, seizure-free response rates are expressed as the percentage of patients who did not experience any seizures during the study period. When comparing two anticonvulsant drugs, the seizure-free rates are juxtaposed to assess their relative effectiveness.

Higher seizure-free rates suggest better drug performance, but it is also important to consider the margin of error and the sample size. A small sample size might not provide a clear picture, and results could be due to chance. Therefore, researchers must use proper statistical testing to draw reliable conclusions about drug efficacy from seizure-free response data.
Statistical Hypothesis Testing
Statistical hypothesis testing is a framework used to determine whether there is enough evidence to reject a null hypothesis in favor of an alternative hypothesis. In the context of anticonvulsant drug comparison, the null hypothesis typically states that there is no difference in seizure-free response rates between the two drugs. The alternative hypothesis posits that a difference does exist.

To test these hypotheses, a chi-square test, which compares expected frequencies under the null hypothesis to observed frequencies, can be used. The test statistic resulting from the chi-square formula reflects how far the observed data deviate from what would be expected if the null hypothesis were true. If the deviation is large enough, the null hypothesis can be rejected, suggesting a significant difference in drug efficacy.
P-value Interpretation
The P-value is a probability measure that helps researchers decide whether to reject the null hypothesis. It represents the likelihood of observing a test statistic as extreme as, or more extreme than, the one obtained if the null hypothesis were true. In simpler terms, a small P-value indicates that the observed data are unlikely under the assumption that there is no difference between groups.

In the given exercise, with a P-value of 0.73, one can interpret that there is a 73% chance of seeing the observed difference, or a greater one, between the two drugs if, in reality, there is no difference in their effectiveness. Since the P-value is much higher than the typical cutoff of 0.10 (or 10%), we fail to reject the null hypothesis. This suggests that the data does not provide sufficient evidence to say that there is a difference in effectiveness between the two anticonvulsant drugs.

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Most popular questions from this chapter

Men with prostate cancer were randomly assigned to undergo surgery \((n=347)\) or "watchful waiting" (no surgery, \(n=348\) ). Over the next several years there were 83 deaths in the first group and 106 deaths in the second group. The results are given in the table. \({ }^{18}\). (a) Let \(D\) and \(A\) represent died and alive, respectively, and let \(S\) and \(W W\) represent surgery and watchful waiting. Calculate \(\operatorname{Pr}\\{\mathrm{D} \mid \mathrm{S}\\}\) and \(\operatorname{Pr}\\{\mathrm{D} \mid \mathrm{WW}\\}\). (b) The value of the contingency-table chi-square statistic for these data is \(\chi_{s}^{2}=3.75 .\) Test for a relationship between the treatment and survival. Use a nondirectional alternative and let \(\alpha=0.05 .\)

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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.

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