Statistal Inference

class: center, middle, inverse, title-slide

# Statistal Inference
## RSS Module 3 Exam Paper 2011

---

### Fertilizers

Ten plants of a certain type, grown under standard conditions and treated with a new 
brand of fertiliser, attain the following heights (in cm).
<pre><code>
25 28 24 23 27 30 24 21 28 30
</code></pre>

* The sample mean is 26.0 and the sample variance is 9.33.

* From extensive previous experiments, it is known that plants of the same type, grown under similar conditions but treated with a standard fertiliser, attain a mean height of 25.0 cm.

---

### Inputing Data into R environment

```r
Heights <- c(25,28,24,23,27,30,24,21,28,30)
```

```r
mean(Heights)
```

```
## [1] 26
```

```r
var(Heights)
```

```
## [1] 9.333333
```

```r
sd(Heights)
```

```
## [1] 3.05505
```

---

### Exercises

***Questions on the RSS Exam Paper***

(i) Write down a statistical model for the distribution of plant heights in the 
population from which the sample is assumed to be drawn, including the 
definition of any unknown parameters.

(ii) Obtain a 99% confidence interval for the population mean height of plants 
treated with the new fertiliser.

(iii) Carefully specifying your hypotheses, conduct a test at the 10% significance 
level to decide whether there is evidence to suggest that the new fertiliser 
produces plants of greater mean height than the standard fertiliser. State your 
conclusions.

(iv) Provide a 95% confidence interval for the population variance of the heights of 
plants treated with the new brand of fertiliser.

---

### 99% Confidence Interval

We will use `\(\mu=25\)` here

```r
t.test(Heights,mu=25,alpha=0.01)
```

```
## 
## 	One Sample t-test
## 
## data:  Heights
## t = 1.0351, df = 9, p-value = 0.3276
## alternative hypothesis: true mean is not equal to 25
## 95 percent confidence interval:
##  23.81455 28.18545
## sample estimates:
## mean of x 
##        26
```
The Confidence interval is (23.8,28.2)

---

### Using <tt> tidy() </tt>

The <tt>tidy()</tt> function from the {broom} R package

```r
library(broom)
t.test(Heights,mu=25,alpha=0.01) %>% tidy()
```

```
## # A tibble: 1 x 8
##   estimate statistic p.value parameter conf.low conf.high method     alternative
##      <dbl>     <dbl>   <dbl>     <dbl>    <dbl>     <dbl> <chr>      <chr>      
## 1       26      1.04   0.328         9     23.8      28.2 One Sampl~ two.sided
```

---

### Hypothesis Test

```r
t.test(Heights,mu=25,
       alpha=0.10,
       alternative="two.sided")
```

---

```r
t.test(Heights,mu=25,
       alpha=0.10,
       alternative="greater")
```

```
## 
## 	One Sample t-test
## 
## data:  Heights
## t = 1.0351, df = 9, p-value = 0.1638
## alternative hypothesis: true mean is greater than 25
## 95 percent confidence interval:
##  24.22904      Inf
## sample estimates:
## mean of x 
##        26
```

---

```r
library(broom)
t.test(Heights,mu=25,alpha=0.10,alterative="greater") %>% tidy()
```

---

### <tt>Desctools::VarCI()</tt>

#### Confidence Intervals for the Variance

Calculates confidence intervals for the variance.

Available approachs are the classical one using the ChiSquare distribution, a more robust version proposed by Bonett and the bootstrap options available in the package boot.

#### Usage

<pre><code>VarCI(x, 
      method = c("classic", "bonett", "norm", 
                  "basic", "stud", "perc", "bca"),
      conf.level = 0.95, 
      sides = c("two.sided", "left", "right"),
      na.rm = FALSE, R = 999)
</code></pre>

---

#### 95% Confidence Interval for Variance using standard method

```r
library(DescTools)
VarCI(Heights)
```

```
##       var    lwr.ci    upr.ci 
##  9.333333  4.415761 31.106624
```

#### 99% Confidence Interval for Variance using "bonett" method

```r
library(DescTools)
VarCI(Heights,
      method="bonett",
      conf.level = 0.95)
```

```
##       var    lwr.ci    upr.ci 
##  9.333333  4.913021 27.428935
```

---