Using Wald Test with unknown distribution

Question

I am stuck on the following task:

They conducted an experiment on different medicines. For placebo there are $80$ patients and $45$ complications, for medicine A, there are $75$ patients and $26$ complications. Test success of medicine A against placebo on significance level $0.05$.

I have read a wiki article and I don't understand how to formulate the hypothesis.

Maybe it is better to use chi-squared distibution. What is better to use here?

My attempt:

Let's consider $\alpha = \frac{complications}{total}$.

Placebo parameter is $35/80$ and medicine A parameter is $49/75$.

$1.959$ is threshold for $N(0,1)$ with $0.95$ significance.

$$\frac{\theta - \theta_0}{se(\theta)} = ?$$

I suppose that I need put difference to the numerator. But I do not understand what to put to standard error and how to interpret the result.

Answer 1

The parameter we estimate is the ratio of patients with complications.

We have a parameter for placebo, which is $35/80$, and want to test a hypothesis that the other distribution (for other drug) is the same as placebo (which means that this medicine is just as effective as placebo).

A formula for the test:

$$\frac{\theta - \theta_0}{s / \sqrt n}.$$

$s$ is a mean sum of squared differences between sample elements (which are zeros and ones) and their mean (which for drug A is $26/75$).

$n$ is $75$.

We get the value 8.25 for the test value. It is more that 1.9 so test is not passed and the drug is placebo.

Answer 2

Here is output from Minitab statistical software comparing complication rates for Placebo (Sample 1) and Drug A (Sample 2). Of course, the smaller rate is preferable.

Test and CI for Two Proportions 

Sample   X   N  Sample p
1       45  80  0.562500
2       26  75  0.346667

Difference = p (1) - p (2)
Estimate for difference:  0.215833
95% CI for difference:  (0.0628052, 0.368862)
Test for difference = 0 (vs ≠ 0):  Z = 2.76  P-Value = 0.006

Fisher’s exact test: P-Value = 0.010

The Wald test is an 'asymptotic' test depending on a normal approximation for large data. Sample sizes here are small for a drug trial. Thus, I would prefer Fisher's Exact Test. However, both two-sided tests reject the null hypothesis that placebo and drug have the same complication rate. A one-sided test with the alternative that the drug has the smaller complication rate has P-value 0.003 for the first test and 0.005 for Fisher's Exact test.

Note: Formulas for the first test differ. Some texts base the standard error on separately estimated variances and some on the combined rate $(45 + 26)/(90 + 75),$ suggested by the null hypothesis. The output shown is for separately estimated variances. (For the 'combined' standard error the results are slightly different, but not enough different to change the conclusion to reject.) If this is for a class, you need to use exactly the formula in your text or notes.