# 统计学|Statistics代写 STAT 516

We are being asked to construct a $100(1-\alpha)$ confidence interval estimate, with $\alpha=0.10$ in part (a) and $\alpha=0.01$ in part (b). Now
$$z_{0.05}=1.645 \text { and } z_{0.005}=2.576$$
and so the 90 percent confidence interval estimator is
$$\bar{X} \pm 1.645 \frac{\sigma}{\sqrt{n}}$$

and the 99 percent confidence interval estimator is
$$\bar{X} \pm 2.576 \frac{\sigma}{\sqrt{n}}$$
For the data of Example 8.5, $n=10, \bar{X}=19.3$, and $\sigma=3$. Therefore, the 90 and 99 percent confidence interval estimates for $\mu$ are, respectively,
$$19.3 \pm 1.645 \frac{3}{\sqrt{10}}=19.3 \pm 1.56$$
and
$$19.3 \pm 2.576 \frac{3}{\sqrt{10}}=19.3 \pm 2.44$$

## STAT516 COURSE NOTES ：

$$\bar{X} \pm 1.96 \frac{\sigma}{\sqrt{n}}$$
Since the length of this interval is
$$\text { Length of interval }=2(1.96) \frac{\sigma}{\sqrt{n}}=3.92 \frac{\sigma}{\sqrt{n}}$$
we must choose $n$ so that
$$\frac{3.92 \sigma}{\sqrt{n}} \leq b$$
or, equivalently,
$$\sqrt{n} \geq \frac{3.92 \sigma}{b}$$
Upon squaring both sides we see that the sample size $n$ must be chosen so that
$$n \geq\left(\frac{3.92 \sigma}{b}\right)^{2}$$