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=(75)/(300)=0.25`

For a
`(1-\alpha)*100\%` confidence level, the confidence interval will be given by


`\hat p\pm Z_(\alpha/2)\sqrt{\frac{\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{(0.25*0.75)/(300)}\approx(0.2089,0.2911)`