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Chapter 10: Problem 9

Patients with pleural infections (fluid buildup in the chest) were randomly assigned to placebo, the treatment "tPA" (tissue plasminogen activator), the treatment "DNase" (deoxyribonuclease), or a combination of tPA and DNase in a double-blind clinical trial. Some of the patients needed to be given penicillin. The following table shows a cross-classification of the data. \({ }^{40}\) Is there evidence of an association between treatment group and use of penicillin? $$ \begin{array}{|lrrrrr|} \hline & \text { Placebo } & \text { tPA } & \text { DNase } & \text { tPA+DNase } & \text { Total } \\ \hline \text { Penicillin } & 4 & 2 & 3 & 3 & 12 \\ \text { No penicillin } & 51 & 50 & 48 & 49 & 198 \\ \text { Total } & 55 & 52 & 51 & 52 & 210 \\ \hline \end{array} $$ Here is computer output for a chi-square test to compare the four groups. \(x\) -squared \(=0.59, d f=3, p\) -value \(=0.90\) (a) State the null and alternative hypotheses in context. (b) Compute the expected frequency of the Penicillin/ Placebo cell under the null hypothesis. (c) If \(\alpha=0.10\), what is your conclusion regarding \(H_{0}\) ?

Short Answer

Expert verified
The null hypothesis is that there is no association between treatment group and penicillin usage, which cannot be rejected given the p-value of 0.90. The expected number of patients who received penicillin and a placebo is (12 * 55) / 210.

Step by step solution

01

State the Hypotheses

In context, the null hypothesis (\(H_0\)) states that there is no association between the treatment group and the use of penicillin. The alternative hypothesis (\(H_a\)) states that there is an association between the treatment group and the use of penicillin.
02

Compute Expected Frequency

To calculate the expected frequency for the Penicillin/Placebo cell, use the formula: Expected frequency = (Row total * Column total) / Grand total. So, here it is: Expected frequency for Penicillin/Placebo = (12 * 55) / 210.
03

Analyze the Chi-Square Test Output

With a chi-square test statistic of 0.59 and a p-value of 0.90, we compare the p-value to the significance level \(\alpha = 0.10\). Since the p-value (0.90) is greater than \(\alpha\), we fail to reject the null hypothesis, suggesting there is not enough evidence to conclude an association between treatment group and penicillin usage.

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Key Concepts

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

Null and Alternative Hypotheses
Understanding the null and alternative hypotheses is a fundamental part of any statistical test, including the chi-square test used in clinical trials. The null hypothesis (\(H_0\)) is the default assumption that there is no effect or no difference; it represents a statement of 鈥渘o effect鈥� or 鈥渘o association.鈥� In the context of the given exercise, the null hypothesis posits that there is no association between the type of treatment given to patients with pleural infections and whether they required penicillin.

The alternative hypothesis (\(H_a\text{ or }H_1\)), on the other hand, is what the researcher aims to prove. It suggests that there is an effect, a difference, or an association that is statistically significant. In our clinical trial example, the alternative hypothesis claims that the use of penicillin is indeed related to the type of treatment鈥攑lacebo, tPA, DNase, or a combination of tPA and DNase.

It's crucial for students to understand these concepts because they frame the entire hypothesis testing process and determine the direction of statistical analysis. To examine the hypotheses accurately, researchers use data from the study and apply the chi-square test to determine whether the observed frequencies in the data significantly deviate from the expected frequencies under the null hypothesis.
Expected Frequency Calculation
The expected frequency is a cornerstone of the chi-square test and requires a clear understanding for accurate statistical analysis in clinical trials. It refers to the frequency of observations that we would expect to find in a particular category if the null hypothesis were true鈥攎eaning if there are no associations in our data set.

To calculate the expected frequency as shown in the exercise, we use the formula: \[ \text{Expected frequency} = \frac{(\text{Row total} \times \text{Column total})}{\text{Grand total}} \]
This formula helps determine what numbers we would expect if there were no association between the treatment groups and the usage of penicillin. For the Penicillin/Placebo cell in our example, this calculates to:\[ \text{Expected frequency} = \frac{(12 \times 55)}{210} \]
It's important for students to grasp that this calculation is essential for every cell in their contingency table to perform the chi-square test. These expected frequencies are then compared to the observed frequencies within the data, allowing us to investigate any significant differences that might indicate associations or effects that could be of clinical importance.
Associations in Clinical Data
Associations in clinical data are connections or relationships that may exist between different variables鈥攕uch as treatment types and outcomes like medication usage. Discovering these associations can provide valuable insights into the effectiveness of treatments or possible side effects.

In the context of a chi-square test, we are particularly interested in the strength and significance of these associations. The chi-square test assesses whether the distribution of observed frequencies across different categories deviates from what would be expected under the null hypothesis of no association. In simpler terms, it lets us question if there鈥檚 a pattern or a relationship, or if our results could just as likely occur by random chance.

The exercise provided offers a clear example: researchers wanted to know if the type of treatment for pleural infections was linked to the need for penicillin. With a high p-value (0.90) in this case, suggesting a high probability that results as extreme as the observed data could occur under the null hypothesis, we do not find evidence of a significant association. When interpreted correctly, these statistical tests can guide critical decisions in healthcare and inform future research. It鈥檚 essential for students to understand how to evaluate such associations correctly to make informed decisions based on clinical data.

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Most popular questions from this chapter

A randomized trial of clinically obese women examined the efficacy and safety of a partial meal replacement diet using nutritionally fortified meal- replacement shakes. Fourteen clinically obese women received the nutritional shakes and healthy diet counseling during their pregnancy. Another nine clinically obese women only received healthy diet counseling (control). Among the meal-replacement group, two women had preterm deliveries (i.e., before 37 weeks' gestation), while none in the control group did. \({ }^{25}\) (a) Does the use of the meal-replacement shakes cause excessive risk of preterm birth? Fisher's exact test with a directional alternative gives a \(P\) -value of \(0.3597 .\) Interpret the conclusions of this test. (b) Given the results above, do the results provide compelling evidence that the meal-replacements are safe with respect to causing preterm birth? (c) Provide an argument for why a directional test could be preferred over a nondirectional test given the context of this research (d) Provide an argument for why a nondirectional test could be preferred over a directional test given the context of this research

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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.

Do women respond to men's solicitations more readily during the fertile phase of their menstrual cycles? In a study of this question, each of two hundred 18 - to 25-year-old women who were walking alone in a city were approached by an attractive 20-year-old man who solicited the woman's telephone number. Previous research suggested that during the fertile phase of her menstrual cycle a woman would be more receptive to this kind of request than at other times. Of 60 women who were in the fertile phase of their cycles, 13 gave out their phone numbers and 47 refused. The corresponding numbers for the 140 women not in the fertile phase of their cycles were 11 and \(129 .^{14}\) The data are summarized in the following table. Consider a chi-square test to determine whether the difference in success rates provides significant evidence in favor of an appropriate directional alternative. Here is computer output for a chi-square test that used a nondirectional alternative. $$ X \text { -squared }=7.585, d f=1, \text { p-value }=0.0059 $$ (a) State the null and appropriate directional alternative hypotheses in context. (b) Compute the sample proportions and the expected frequencies. (c) If \(\alpha=0.02\), what is your conclusion regarding \(H_{0}\) ?

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
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