Confidence interval problem for tires One tire manufacturer found that after $5,000$ miles, $y=32$ of $n=200$ steel-belted tires selected at random were defective. Find an approximate $99$ percent confidence interval for $p$, the proportion of defective tires in the total production.
In order to use the formula, I would need to know $N, S^2,$ and $\alpha.$
We know that $\alpha=0.01$, and $N = 200$, but what would be $S^2$?
 A: Note: Your $Y\sim \text{Bin}(n=200,p=\displaystyle \frac{32}{200})$. So your $S^2=n*p*(1-p)$.
A: i was looking at a wrong formula, in then end i had to do this
(32/200)+-2.57*Sqroot(((32/200)*1-(32/200))/200)=[.1,.22]
i had to use confidence interval for proportions
A: We have that the true proportion can be estimated as $\hat{p} = \frac{Y}{n}$ where $Y$ is a random variable having two possible outcomes "defective" or "not defective", hence $Y\sim Bin(n, p)$. Nevertheless, since $n$ is large and we have that $n\hat{p}\geq 10$ and $n(1-\hat{p})\geq 10$ it is useful to use the following approximation:
$$Y \sim N(np , np(1-p))$$
otherwise the computations are really tedious. Moreover, this is a standard widely used approximation.
Hence
$$\frac{Y-np}{\sqrt{np(1-p)}}\sim N(0,1).$$
Now observe that deriving a confidence interval for $p$ means choosing $\alpha\in (0,1)$ and imposing $1-\alpha = P(-z_{\alpha/2} < Z < z_{\alpha/2})$ and isolating $Y$ then dividing by $n$ to get $p=\frac{Y}{n}$, but this is really difficult. Hence one "reapproximates" $Y$ as follows: by approximating $p$ by $\hat{p}$:
$$\frac{Y-n\hat{p}}{\sqrt{n\hat{p}(1-\hat{p})}}\approx N(0,1).$$
Now diving by $n$ we get
$$Z= \frac{p-\hat{p}}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}}\approx N(0,1)$$
and now from this it is very easy to construct a confidence interval: choose $\alpha\in (0,1)$,
\begin{align*}
1-\alpha = P(-z_{\alpha/2} < Z < z_{\alpha/2}) &\approx P((-z_{\alpha/2} < \frac{p-\hat{p}}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}}< z_{\alpha/2}) \\
&= P( \hat{p}-z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} < p < \hat{p}+z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}).
\end{align*}
Altogether:
$$\left( \hat{p}-z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} , \, \hat{p}+z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \right)$$
is an approximated $\alpha$-confidence interval for $p$.
Using your data: n=200, y=32 and $\alpha=0.01$ then
$$\left( 0.16-z_{0.005}\sqrt{\frac{0.16(1-0.16)}{200}} , \, 0.16+z_{0.005}\sqrt{\frac{0.16(1-0.16)}{200}} \right)$$
