I want to make an hypothesis test about two proportions $p_{1},p_{2}$ and the condition is that $p_1 \geq kp_2$, in particular $p_1 \geq 2p_2$. So I made this hypothesis $$H_0:p_1 \geq 2p_2; H_1:p_1 \lt 2p_2 $$

I've seen examples of hypothesis tests using difference of proportions and using a Z test, but I don't know if it is correct. I have something like $(\hat{p_1}-\hat{p_2})-(p_1 - p_2)$, but I don't know what to do because $p_1 - p_2 = p_2$. I also have calculated proportions $\hat{p_1},\hat{p_2}$ and a significance $\alpha$. How can I use this info to calculate a Z score?

  • 1
    $\begingroup$ Let me provide you with a non$-$rigorous answer to your question when $H_0:p_1=2p_2$ and $H_a:p_1 \neq 2p_2$. Let $\hat{P}_1,\hat{P}_2$ be random sample proportions of successes generated from simple random samples (chosen independently from each other) of sizes $n_1,n_2$. From the central limit theorem we have $\hat{P}_1\approx N\Bigg(p_1,\frac{p_1(1-p_1)}{n_1}\Bigg)$ and $\hat{P}_2\approx N\Bigg(p_2,\frac{p_2(1-p_2)}{n_2}\Bigg)$ so therefore $$\hat{P}_1-2\hat{P_2}\approx N\Bigg(p_1-2p_2,\frac{p_1(1-p_1)}{n_1}+\frac{4p_2(1-p_2)}{n_2}\Bigg)$$ $\endgroup$ Apr 18 at 3:08
  • 1
    $\begingroup$ Since we're assuming $p_1=2p_2$ we have $$\frac{\hat{P}_1-2\hat{P_2}}{\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{4p_2(1-p_2)}{n_2}}}\approx N(0,1)$$ If $\hat{p}_1,\hat{p}_2$ are our observed sample proportions of successes we can use these observations to estimate $p_1,p_2$ in the denominator of the preceding expression and develop a suitable (and analogous) $z-$score to be $$z=\frac{\hat{p}_1-2\hat{p}_2}{\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\frac{4\hat{p}_2(1-\hat{p}_2)}{n_2}}}$$ $\endgroup$ Apr 18 at 3:08
  • $\begingroup$ @MatthewPilling, By task $p_{1} = 2p_{2}$. Seems clear linear relationship $corr(p_{1} , p_{2} ) = 1$. In this case $\sigma^2_{\bar{P_{1}} - \bar{P_{2}}} = $ what you write $+2cov(\bar{P_{1}}, \bar{P_{2}})$. Am I wrong? I have similar question. I say about $p_{2}$ that it is consant and let prove that $p_{1} =2p_{2}$ (better see my suggestion (question 28.04.2021)) $\endgroup$ Apr 28 at 18:47
  • $\begingroup$ It should be $-2\text{cov}(\hat{P}_1,\hat{P}_2)$. Also, $\text{corr}(p_1,p_2)$ is meaningless since $p_1,p_2$ are parameters, not random variables. Lastly, $\text{cov}(\hat{P}_1,\hat{P}_2)=0$ since we chose our samples independently from one another. $\endgroup$ Apr 28 at 18:48
  • $\begingroup$ @MatthewPilling Thanks $\endgroup$ Apr 28 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.