Binomial testing

Question

Suppose in any given year, the probability of getting a success is $0.8$. Suppose we have $10$ years of data and we have observed $3$ failures. How can I test if these three failures are statistically significant at say the 90% confidence interval? I have vague memories of high school hypothesis testing but cannot seem to recall it!

Answer 1

This is a population proportion Hypothesis testing. You'd want to test if $$\mathcal{H}_0: \hat{p} = 0.7 = 1- \frac{3}{10}$$ is accepted or not, given that your true $p = 0.8$. The way to do so here is first to get your standard error, $$SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.8(1-0.8)}{10}} = 0.1265$$ Then you get your test statistic $$z = \frac{\hat{p} - p}{SE} = \frac{0.7 - 0.8}{\sqrt{\frac{0.8(1-0.8)}{10}}} = - \frac{0.1}{ 0.1265} = - 0.7906$$ If $\vert z \vert > z_{\frac{\alpha}{2}} $, reject $\mathcal{H}_0$ (here $z_{\frac{\alpha}{2}}$ is the critical value on both negative and positive tails). But in your case for $\alpha = 1 - 0.9 = 0.1$ and by referring to the z-table, you get: $$z_{\frac{\alpha}{2}} = 1.645 > \vert z \vert$$ So we accept $\mathcal{H}_0$.