Question

To determine the effectiveness of a drug for arthritis, a researcher studied two groups of 200 arthritic patients. One group was inoculated with the drug; the other received a placebo (an inoculation that appears to contain the drug but actually is nonactive). After a period of time, each person in the study was asked to state whether his or her arthritic condition had improved. $$ \begin{array}{lcc} \hline & \text { Treated } & \text { Untreated } \\ \hline \text { Improved } & 117 & 74 \\ \text { Not Improved } & 83 & 126 \\ \hline \end{array} $$ You want to know whether these data indicate that the drug was effective in improving the condition of arthritic patients. a. Use the chi-square test of homogeneity to compare the proportions improved in the populations of treated and untreated subjects. Test at the \(5 \%\) level of significance. b. Test the equality of the two binomial proportions using the two-sample \(z\) -test of Section 9.5 . Verify that the squared value of the test statistic \(z^{2}=X^{2}\) from part a. Are your conclusions the same as in part a?

Short answer

In this problem, we used both the chi-square test of homogeneity and the two-sample z-test to determine if a drug is effective in improving the condition of arthritic patients. Based on the chi-square test statistic (17.578) and the squared value of the z-test statistic (17.568), we reject the null hypothesis and conclude that the drug is effective in improving the condition of arthritic patients. Both tests show a significant difference between the proportions of treated and untreated subjects who reported improvement.

Step-by-step solution

(Step 1: Calculate total number of patients in each row and column)

We will add the values in each row and column to get the total numbers. The table should now look like this: $$ \begin{array}{lcccr} \hline & \text { Treated } & \text { Untreated } & \text { Total } \\ \hline \text { Improved } & 117 & 74 & 191 \\ \text { Not Improved } & 83 & 126 & 209 \\ \hline & 200 & 200 & 400 \\ \end{array} $$

(Step 2: Calculate the expected values)

The expected values are calculated by multiplying the row total by the column total, and then dividing by the grand total. We want to compare the actual values to the expected values to see if they are significantly different. $$ E_{11} = \frac{\text{Total Improved} \times \text{Total Treated}}{\text{Grand Total}} = \frac{191 \times 200}{400} = 95.5 $$ $$ E_{12} = \frac{\text{Total Improved} \times \text{Total Untreated}}{\text{Grand Total}} = \frac{191 \times 200}{400} = 95.5 $$ $$ E_{21} = \frac{\text{Total Not Improved} \times \text{Total Treated}}{\text{Grand Total}} = \frac{209 \times 200}{400} = 104.5 $$ $$ E_{22} = \frac{\text{Total Not Improved} \times \text{Total Untreated}}{\text{Grand Total}} = \frac{209 \times 200}{400} = 104.5 $$

(Step 3: Calculate the Chi-Square Test Statistic)

Now we will calculate the chi-square test statistic \(X^{2}\) using the formula: $$ X^{2} = \sum \frac{(O-E)^{2}}{E} $$ where O is the observed value and E is the expected value. Plugging in the values, we get: $$ X^{2} = \frac{(117-95.5)^{2}}{95.5} + \frac{(74-95.5)^{2}}{95.5} + \frac{(83-104.5)^{2}}{104.5} + \frac{(126-104.5)^{2}}{104.5} \approx 17.578 $$

(Step 4: Determine critical value and p-value)

For a chi-square test with 1 degree of freedom and a significance level (alpha) of 0.05, we find the critical value to be \(\chi^{2}_{0.05,1} = 3.841\). Since \(X^{2} = 17.578 > 3.841\), we reject the null hypothesis and conclude that the drug is effective in improving the condition of arthritic patients as there is a significant difference between the proportions of treated and untreated subjects who reported improvement.

(Step 5: Calculate the two-sample z-test)

We now test the equality of the two binomial proportions using the two-sample z-test by first calculating the pooled proportion, \(\hat{p}\), as follows: $$ \hat{p} = \frac{117 + 74}{200 + 200} = \frac{191}{400} = 0.4775 $$ Now, we can calculate the z-score as follows: $$ z = \frac{\frac{117}{200} - \frac{74}{200}}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{200}+\frac{1}{200})}} \approx 4.191 $$ The significance level is 0.05, so the critical z-value is \(1.96\) for a two-tailed test. Since \(|4.191| > 1.96\), we reject the null hypothesis.

(Step 6: Verify squared z-value)

Finally, we will verify that the squared value of the test statistic \(z^{2}=X^{2}\) from part a. $$ z^{2} = (4.191)^{2} \approx 17.568 $$ This value is approximately equal to the chi-square test statistic \(X^{2}\) calculated earlier. Based on both tests (chi-square and two-sample z-test), we reject the null hypothesis and conclude that the drug is effective in improving the condition of arthritic patients as there is a significant difference between the proportions of treated and untreated subjects who reported improvement.

Key concepts

Understanding Probability and Statistics in Chi-Square Test of Homogeneity

Probability and statistics are fundamental concepts used in a variety of scientific and business applications, including hypothesis testing in medical research.

Take, for example, the chi-square test of homogeneity. This test is a non-parametric statistical method used to determine whether the frequency distribution of certain events observed in a sample is consistent across different populations or groups. In essence, it assesses whether the proportions across different categories are the same for various groups.

In the exercise, the chi-square test is applied to data from a medical study comparing the effect of a drug versus a placebo on arthritic patients. To understand whether the drug was effective, the comparison of proportions of patients reporting improvement is essential. The test hinges on calculating expected values based on the null hypothesis that there is no difference between the treated and untreated groups. Discrepancies between observed and expected proportions are then measured using the chi-square statistic.

By comparing the calculated statistic with a critical value from the chi-square distribution, a decision about the null hypothesis can be made. A chi-square value larger than the critical value signals that there is a statistically significant difference in proportions, suggesting that the drug may have had an effect.

Binomial Proportions' Role in Statistical Analysis

Binomial proportions are probabilities that arise from processes where there are two possible outcomes—often termed 'success' and 'failure'. The analysis of binomial proportions is central to the interpretation of binary data, such as 'improved' or 'not improved' conditions in a clinical trial.

In our example, each patient represents a binomial trial where the outcome 'improved' can be considered as a success. The proportion of successes in a sample is used to estimate the success probability in the population. The chi-square test of homogeneity, employed to determine if there is a difference in success proportions between the treated and untreated groups, implicitly compares binomial proportions.

It is critical to understand that the validity of these comparisons assumes that each patient’s outcome is independent of others, and that the probability of improvement remains constant across patients. When these assumptions hold, binomial proportions provide a robust tool to infer population trends based on sample data, which is pivotal to forming conclusions in experiments such as drug efficacy trials.

Two-Sample Z-Test for Comparing Binomial Proportions

The two-sample z-test is another statistical tool used to assess whether the proportions of successes in two independent samples differ significantly. It is appropriate when dealing with large sample sizes and when the sampling distribution of the proportion can be approximated by a normal distribution, based on the Central Limit Theorem.

In our exercise, the two-sample z-test is applied to the proportion of improved patients in the treated and untreated groups. A pooled estimate of the proportion is first calculated to establish a common probability of success under the assumption that there is no effect from the treatment. The z-score, or test statistic, is then computed to measure the standard deviations by which the difference between the two sample proportions deviates from zero.

If the absolute value of the z-score is greater than the critical value from the normal distribution, the result is considered statistically significant. This means that it is unlikely for such an extreme statistic to arise under the null hypothesis of equal proportions. The exercise demonstrates consistency between the chi-square test result and the two-sample z-test result, affirming the conclusion about the drug's effectiveness.

Both these tests provide essential methodologies in analyzing categorical data and are fundamental in verifying the efficacy of a treatment in clinical trials.