Amelie has developed a new drug against the flu and wanted to test its effectiveness. She recruited 50 people and divided them into two groups. She gave the medicine to the first group (34 people) and provided a placebo to the second (16). Here are the results.
In the treatment group, 15 out of 34 (44%) recovered in 1 day.
In the control group, 4 out of 16 (25%) recovered in 1 day.
Amelie thinks her medicine is effective because of the big defence (44 – 25 = 19%) in performance and the larger sample size for the treatment group. Do you agree?
We must do a statistical test to conclude. The data is:
| Sample | Events | Trials |
| Treatment | 15 | 34 |
| Control | 4 | 16 |
The Null Hypothesis, H0: No impact of medicine; recovery proportion on drug equals that of placebo.
The Alternative Hypothesis, H1: Medicine improves the condition; recovery proportion on drug greater than placebo.
We use the ‘prop.test()’ function in R.
prop.test(x = c(15, 4), n = c(34, 16), alternative = "greater")We used the one-tailed test to determine whether the treatment has improved the condition compared to the control.
2-sample test for equality of proportions with continuity correction
data: c(15, 4) out of c(34, 16)
X-squared = 0.97389, df = 1, p-value = 0.1619
alternative hypothesis: greater
95 percent confidence interval:
-0.0813273 1.0000000
sample estimates:
prop 1 prop 2
0.4411765 0.2500000
The p-value of 0.1619 (more than the significance value of 0.05) is not enough to reject the null hypothesis.
What would have happened if she had recruited more people in the placebo group and found results at the same proportions?
| Sample | Events | Trials |
| Treatment | 15 | 34 |
| Control | 16 | 64 |
prop.test(x = c(15, 16), n = c(34, 64), alternative = "greater") 2-sample test for equality of proportions with continuity correction
data: c(15, 16) out of c(34, 64)
X-squared = 2.9205, df = 1, p-value = 0.04373
alternative hypothesis: greater
95 percent confidence interval:
0.002691967 1.000000000
sample estimates:
prop 1 prop 2
0.4411765 0.2500000 Here, the results are significant (p-value = 0.04373 < 0.05) to reject the null hypothesis in favour of the alternate (that the drug works).
What about recruiting more to the treatment group?
| Sample | Events | Trials |
| Treatment | 150 | 340 |
| Control | 4 | 16 |
prop.test(x = c(150, 4), n = c(340, 16), alternative = "greater") 2-sample test for equality of proportions with continuity correction
data: c(150, 4) out of c(340, 16)
X-squared = 1.5631, df = 1, p-value = 0.1056
alternative hypothesis: greater
95 percent confidence interval:
-0.02503096 1.00000000
sample estimates:
prop 1 prop 2
0.4411765 0.2500000 p-value = 0.1056; the null hypothesis stays!
Or, had many individuals in both groups?
| Sample | Events | Trials |
| Treatment | 150 | 340 |
| Control | 40 | 160 |
prop.test(x = c(150, 40), n = c(340, 160), alternative = "greater") 2-sample test for equality of proportions with continuity correction
data: c(150, 40) out of c(340, 160)
X-squared = 16.076, df = 1, p-value = 3.042e-05
alternative hypothesis: greater
95 percent confidence interval:
0.1149402 1.0000000
sample estimates:
prop 1 prop 2
0.4411765 0.2500000 Both the groups have plenty of samples, and now the difference (44 – 25 = 22%) overwhelmingly supports the impact of the medicine.
