hw4.Rmd

---
title: "Homework #4"
author: "Ming Chen & Wenqiang Feng"
date: "2/2/2017"
output: html_document
---

## You can also get the [PDF format](./pdf/HW4.pdf) of Wenqiang's Homework.

## Q 5.66
### Q 5.66(a)

* Evaluate whether the sample is from a normal distribution
* From the normal quantile plot, we can see that the data points appear to be realtively close to the fitted line, which suggests that the normality of the population is plausible.

```{R, fig.width=5, fig.height=5}
dissolution = c(19.5,19.7,19.7,20.4,19.2,19.5,19.6,20.8,19.9,
                19.2,20.1,19.8,20.4,19.8,19.6,19.5,19.3,19.7,
                19.5,20.6,20.4,19.9,20.0,19.8)
qqnorm(dissolution)
qqline(dissolution)
```

### Q 5.66(b)

* Mean = `mean(dissolution)` = `r mean(dissolution)`
* The 99% CI for unknown $\mu$ and $\sigma$ is given by $\bar{y} \pm t_{\alpha/2}s/\sqrt{n}$, where
    + $\bar{y}$ = `mean(dissolution)` = `r mean(dissolution)`
    + $t_{\alpha/2}$ = $t_{0.01/2, df=23}$ = `r -qt(0.01/2, df=23)`
    + s = `r sd(dissolution)`
    + n = 24
* so, 99% CI = $19.83 \pm 2.81(0.432)\sqrt{23}$ = [19.58, 20.08]

### Q 5.66(c)
* Let $H_{0} = \mu\geq\mu_{0}, H_{a} = \mu<\mu_{0}$. Reject $H_{0}$ if $t\geq t_{\alpha}$
* From the right-tailed t-test, p-value = 0.0325 > 0.01, given that $\alpha = 0.01$
* Conclusion: we cann't reject the $H_{0}$ hypothesis

```{R}
t.test(dissolution, mu=20, alternative = "less", conf.level = 0.99)
```


## Q 5.68

* Let $H_{0} = \mu\leq\mu_{0}, H_{a} = \mu>\mu_{0}$. Reject $H_{0}$ if $t\geq t_{\alpha}$
* From the right-tailed t-test, p-value = 0.1485 > 0.05, given that $\alpha = 0.05$
* Conclusion: we can't reject the $H_{0}$ hypothesis

```{R}
time = c(28,25,27,31,10,26,30,15,55,12,24,32,28,42,38)
t.test(time, mu=25, alternative = "greater")
```

## Q 5.72

### Q 5.72(a)

* 95% CI: [26.26906, 34.75951]
* Explanation: randomely select a healthy army inductee, the probablity that his exercise capacity will fall into [26.26906, 34.75951] is 95%.

```{R}
capacities = c(23,19,36,12,41,43,19,28,14,44,15,46,36,25,
               35,25,29,17,51,33,47,42,45,23,29,18,14,48,
               21,49,27,39,44,18,13)
CI95 = t.test(capacities)$conf.int
CI95
```

### Q 5.72(b)
* 99% CI: [24.81485, 36.21372]
* The 99% confidence interval becomes wider than 95% confidence interval 
```{R}
CI99 = t.test(capacities, conf.level=0.99)$conf.int
CI99
```