Question

Let y be a random variable following b(300, p, binomial distribution with (unknown probability of success p ? (0, 1 and 300 bernoulli trials. if the observed value of y is y = 75, find an approximate 90 percent confidence interval for p.

Answer 1

$y=75\implies\hat p=\dfrac{75}{300}=0.25$

For a $(1-\alpha)\times100\%$ confidence level, the confidence interval will be given by

$\hat p\pm Z_{\alpha/2}\sqrt{\dfrac{\hat p(1-\hat p)}n}$

where $Z_{\alpha/2}=Z_{0.05}\approx1.645$, since $\mathbb P(|Z|<Z_{0.05})\approx0.90$. So the interval is

$0.25\pm1.645\sqrt{\dfrac{0.25\times0.75}{300}}\approx(0.2089,0.2911)$