Let $Y$ be $b(300,p)$. If the observed value of $Y$ is $y=75$, find the approximate $90\%$ confidence interval for $p$. I tried solving this problem by what I've learned about confidence intervals for proportions, but something goes wrong at some point because I'm not matching the textbook answer.
For this problem, since $Y$ is not normally distributed, let's consider the following general pivot random variable:
$$
Z = \frac{\overline{X}-\mu}{\frac{\sqrt{S^2/n}}{n}}
$$
where $\overline{X}$ is the sample mean (mle of $\mu$), $\mu$ is the mean (the parameter we wish to estimate), $S^2$ is the sample variance and $n$ is the sample size. By the central limit theorem and by the independence of the samples $Y_1, \ldots, Y_n$, $Z$ approximately follows a standard normal distribution, i.e. $N(0,1)$.
A $(1-\alpha)100\%$ confidence interval for $\mu$ is one such that
$$
\text{Pr}\left(\overline{X} - z_{\alpha/2}{\frac{\sqrt{S^2/n}}{n}} < \mu < \overline{X} + z_{\alpha/2} \frac{\sqrt{S^2/n}}{n}\right) = 1-\alpha
$$
where $0 < \alpha < 1$.
In our case,
\begin{align}
\alpha &= 0.10 \\
z_{\alpha/2} &= 1.645 \\
n &= 300 \\
\overline{X} &= \widehat{p} = 75/300 = 1/4 \\
\\
S^2 &= \sigma^2 = 300\widehat{p}(1-\widehat{p})
\end{align}
With these values in mind, our confidence interval for $p$ is:
\begin{align}
\left(1/4 - 1.645\frac{\sqrt{300(1/4)(1-1/4)/300}}{300}, 1/4 + 1.645\frac{\sqrt{300(1/4)(1-1/4)/300}}{300}\right) &= 
(0.2476, 0.2524)
\end{align}
My textbook has the following confidence interval:
$$
(0.2089,0.2911)
$$
What did I do wrong?
 A: Population standard deviation $\sigma$ is unknown, so the textbook is to use t-distribution, with degree of freedom $=300-1=299$, so you find t-score $=1.6504$
Plug in and you will get the consistent answer with your textbook.
$$0.25 - 1.650\frac{\sqrt{300\cdot(0.25)\cdot(1-0.25)}}{300}=0.209$$
$$0.25 + 1.650\frac{\sqrt{300\cdot(0.25)\cdot(1-0.25)}}{300}=0.291$$
A: I need to write this answer  , because i thought that you must learn some basics. First of all , when we work over the confidence interval of proportions , we use z-table .However, to be able to use the z-table , we must have large sample case!! Then , what is the large sample case ? The large sample case is the case that $\hat{p}\times\hat{q}\times n >9$ , or $\hat{p}\times n>5$ and $\hat{q}\times n>5$.
In our question , $\hat{p}:75/300 =1/4 $, $\hat{q}:3/4$ ,$n:300$ . So , $(1/4) \times (3/4) \times 300 =56.25$.As a result ,we can use z-table.
Then , lets write the formula for confidence interval for $\hat{p}$ such that $$\hat{p}-z_{\alpha/2}\sqrt{\frac{\hat{p}\hat{q}}{n}}<p<\hat{p}+z_{\alpha/2}\sqrt{\frac{\hat{p}\hat{q}}{n}}$$
where $z_{0.05}=1.645$
$$0.25-1.645 \times (0.025)<p<0.25+1.645 \times0.025$$
$$0.208875< p<0.291125$$
Addentum: The answer of @MathFail gave the correct result ,as well. The resaon behind it is that the z-table and t-table getting closer and closer for the large sample cases. However , the best way is to use z-table .
