Answer to Question #182838 in Statistics and Probability for royston

Question

A new drug cures 9 of 200 patients suffering from a type of cancer, for which the historical cure rate is 2%. Perform a test to check on the significance of this result, at both 5% and 1% levels of significance. Based on your conclusion, comments on the efficacy of the treatment using this new drug.

Answer 1

One Proportion z-test

$p_0 = 2 \% = 0.02 \\ H_0 : p = 0.02 \\ H_1 : p > 0.02 \\ n = 200$

The number of successes:

x = 9

The sample proportion:

$\hat{p} = \frac{x}{n}= \frac{9}{200}= 0.045$

The test statistics:

$z = \frac{\hat{p} -p_0}{ \sqrt{ \frac{p_0(1-p_0)}{n} }} \\ = \frac{0.045 -0.02}{ \sqrt{ \frac{0.02 \times (1-0.02)}{200} }} \\ = 2.5254 \\ P(z> 2.5254) = 1 -P(Z<2.5254) = 0.0058$

At the 5 % significance level (α=0.05), we reject the H₀, since p-value < α. We have sufficient evidence to conclude that the new drug has a better efficacy than the historical cure rate in curing the type of cancer at 5 % significance level.

At the 1 % significance level (α=0.01), we reject the H₀, since p-value < α. We have sufficient evidence to conclude that the new drug has a better efficacy than the historical cure rate in curing the type of cancer at 1 % significance level.