2018-04-18

To understand the concept of p value is very important. To teach the the distribution of common statistic( \(\chi^2\) for chisq.test() , t for Student’s t-test , F for F-test) and concept of the p-value, plot.htest() function can be used.

https://cardiomoon.github.io/webr/index.html

You can install this package form the github. Currently, package
`webr`

is under construction and consists of only one
function - plot.htest().

```
#install.packages("devtools")
::install_github("cardiomoon/webr") devtools
```

The plot.htest() function is a S3 method for class “htest”. Currently, this function covers Welch Two Sample t-test, Pearson’s Chi-squared test, Two Sample t-test, One Sample t-test, Paired t-test and F test to compare two variances.

You can show the distribution of chi-squre statistic and p-value.

```
require(moonBook)
require(webr)
# chi-squared test
=chisq.test(table(acs$sex,acs$DM))
x x
```

```
Pearson's Chi-squared test with Yates' continuity correction
data: table(acs$sex, acs$DM)
X-squared = 3.1296, df = 1, p-value = 0.07688
```

`plot(x) `

You can show the distribution of t-statistic and p-value in one sample t-test.

`t.test(acs$age,mu=63)`

```
One Sample t-test
data: acs$age
t = 0.77978, df = 856, p-value = 0.4357
alternative hypothesis: true mean is not equal to 63
95 percent confidence interval:
62.52736 64.09574
sample estimates:
mean of x
63.31155
```

`plot(t.test(acs$age,mu=63))`

Before performing a t-test, you have to compare two variances.

```
=var.test(age~DM,data=acs)
x x
```

```
F test to compare two variances
data: age by DM
F = 1.2383, num df = 552, denom df = 303, p-value = 0.0376
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
1.012416 1.505776
sample estimates:
ratio of variances
1.238288
```

`plot(x)`

Based on the result of var.test(), you can perform t.test with default option(var.equal=FALSE).

```
=t.test(age~DM,data=acs)
x x
```

```
Welch Two Sample t-test
data: age by DM
t = 0.58982, df = 682.36, p-value = 0.5555
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.112568 2.068014
sample estimates:
mean in group No mean in group Yes
63.48101 63.00329
```

`plot(x)`

To compare means of body-mass index between male and female patients, perform F test first.

`var.test(BMI~sex,data=acs)`

```
F test to compare two variances
data: BMI by sex
F = 1.2078, num df = 254, denom df = 508, p-value = 0.07756
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.9794315 1.5008098
sample estimates:
ratio of variances
1.207759
```

`plot(var.test(BMI~sex,data=acs))`

Based on the result of F test, you can perform t-test using pooled variance.

```
=t.test(BMI~sex,data=acs,var.equal=TRUE)
x x
```

```
Two Sample t-test
data: BMI by sex
t = -0.50823, df = 762, p-value = 0.6114
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.6348532 0.3737344
sample estimates:
mean in group Female mean in group Male
24.19492 24.32548
```

`plot(x)`

You can show the distribution of t-statistic and p-value in paired t-test.

```
=t.test(iris$Sepal.Width,iris$Petal.Width,paired=TRUE)
xplot(x)
```

You can change the options of t.test.

```
=t.test(BMI~sex, data=acs,conf.level=0.99,alternative="greater",var.equal=TRUE)
xplot(x)
```

http://rpubs.com/cardiomoon/398623