# 309553032_林育駿_hw07

###### tags: `Biostatistics`

## 1

### (1)
$0.2=Pr(Z>k)=1-Pr(Z\leq k)$
$\iff Pr(Z\leq k)=0.8$
| 0.040 | k | 0.055 |
| ------ | ------ | ------ |
| 0.7995 | 0.8000 | 0.8023 |
$k = 0.841785$
(Solve by linear interpolation, the same for the following)
### (2)
$0.1=Pr(Z>k)=1-Pr(Z\leq k)$
$\iff Pr(Z\leq k)=0.9$
| 1.280 | k | 1.290 |
| ------ | ------ | ------ |
| 0.8997 | 0.9000 | 0.9015 |
$k = 1.281666$
### (3)
$0.05=Pr(Z>k)=1-Pr(Z\leq k)$
$\iff Pr(Z\leq k)=0.95$
| 1.640 | k | 1.650 |
| ------ | ------ | ------ |
| 0.9495 | 0.9500 | 0.9505 |
$k = 1.644999$
### (4)
$Pr(Z\leq k)=0.2$
$\implies Pr(Z\gt (-k))=0.2$ ( similarity of Gaussian
$\implies -k=0.841785$ ( (1)
$\implies k = -0.841785$
### (5)
$Pr(Z\leq k)=0.1$
$\implies Pr(Z\gt (-k))=0.1$ ( similarity of Gaussian
$\implies -k=1.281666$ ( (2)
$\implies k = -1.281666$
### (6)
$Pr(Z\leq k)=0.05$
$\implies Pr(Z\gt (-k))=0.05$ ( similarity of Gaussian
$\implies -k=1.644999$ ( (3)
$\implies k = -1.644999$

## 2
### (a)
$\bar{X}\sim N(\mu,\frac{\sigma^2}{n})=N(91.1,4.571428)$
### (b)
margin of error: $z_{0.025}\times \frac{\sigma}{\sqrt{n}}=1.96\times \frac{8}{\sqrt{14}}=4.190656$

confidence interval:
[86-4.190656,86+4.190656] = [81.809344, 90.190656] (mm)

critical values:
81.809344, 90.190656 (mm)

### (c\)
confidence interval:
[91.1-4.190656,91.1+4.190656] = [86.909344, 95.290656] (mm)

## 3
### (a)
margin of error: $z_{0.05}\times \frac{\sigma}{\sqrt{n}}=1.645\times \frac{8}{\sqrt{14}}=3.517157$ (mm)
confidence interval:
[91.1-3.517157, 91.1+3.517157] = [87.582843, 94.617157] (mm)
### (b)
margin of error = $z_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}$
而 percentage of confidence interval = $1-\alpha$
由
可知在固定$\sigma$與$n$時，信心區間越小，margin of error越短。