Testing Hypothesis (vaccine)

[Agent Smith]

TL;DR Summary

Vaccine efficacy hypothesis using a method that may/may not be appropriate

Placebo/control group has 800 people. They aren't given the vaccine. 60 of them develop the disease
Treatment group has 1000 people. They're given the vaccine. 15 develop the disease

It seems there's a formula viz $\frac{p_1 - p_2}{\sqrt{p (1 - p)}\left(\frac{1}{n_1} + \frac {1}{n_2}\right)}$ to do this.

However I tried something different. Would like to know if it's valid.

For $p_1$ = proportion of diseased in unvaccinated and $p_2$ = proportion of diseased in vaccinated, my hypotheses are:
$H_0: p_2 =p_1$
$H_a: p_2 < p_1$

Is this ok?

 

[Dale]

It is equivalent. Your $H_0$ can be tested by seeing if the above formula is significantly different from 0.

 

[Agent Smith]

@Dale when I do a Chi-square test, my $\chi^2 \approx 40$ for a degree of freedom = 1, that gives a P-value << 0.05,

BUT

when I use the method in the OP, the P-value = 0.0113. It is < 0.05, but not << 0.5.

How come?

 

[Dale]

I don’t know. I think they should be the same, but I have to admit that I have not calculated this test by hand in almost 30 years

 

[Agent Smith]

@Dale can you break the formula down for me? We seem to be "combining" the frequencies with p.
$p = \frac{15 + 60}{1000 + 800}$ and then using it compute a standard deviation $\sqrt {p (1 - p)}$. What does $\left(\frac{1}{n_1} + \frac{1}{n_2}\right)$ do?

A person has assigned values for $p_1$ and $p_2$ as follows:

Can I do $p_1 = 60/800$ and $p_2 = 15/1000$?

 

[Agent Smith]

In the original question, it says "variances unknown". What if variances are known?
All the relevant variances/standard deviations seem computable from givens.

 

[Gavran]

@Dale when I do a Chi-square test, my $\chi^2 \approx 40$ for a degree of freedom = 1, that gives a P-value << 0.05,

This is okay.

BUT

when I use the method in the OP, the P-value = 0.0113. It is < 0.05, but not << 0.5.

How come?

I have got $6,3$ for $z$ value which gives $P$ much less than $0,05$.

View attachment 353603

@Dale can you break the formula down for me? We seem to be "combining" the frequencies with p.
$p = \frac{15 + 60}{1000 + 800}$ and then using it compute a standard deviation $\sqrt {p (1 - p)}$. What does $\left(\frac{1}{n_1} + \frac{1}{n_2}\right)$ do?

$\begin{align} z&=\frac{p_1-p_2}{\sqrt{p(1-p)(\frac{1}{n_1}+\frac{1}{n_2})}}\nonumber\\ &=\frac{p_1-p_2}{\sqrt{\frac{p(1-p)}{n_1}+\frac{p(1-p)}{n_2}}}\nonumber \end{align}$

where $$ \sigma_1^2=\frac{p(1-p)}{n_1} $$ is the first population variance (800 people, 60 of them develop the disease ) and $$ \sigma_2^2=\frac{p(1-p)}{n_2} $$ is the second population variance (1000 people, 15 of them develop the disease).

So $p_1-p_2$ has a normal distribution $\mathcal{N}(0,\sigma_1^2+\sigma_2^2)$.

A person has assigned values for $p_1$ and $p_2$ as follows:
View attachment 353604

Can I do $p_1 = 60/800$ and $p_2 = 15/1000$?

Yes you can and now you will have: $H_0:p_1-p_2=0$ and $H_A:p_1-p_2<0$. In this case $z$ value will be $-6,3$ and $P$ will be the same as for the case in the original post.