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.

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

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

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.

```
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
```

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

```
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
```

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

```
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
```

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

```
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
```

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

```
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
```

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

```
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
```

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

You can change the options of t.test.