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 = 0.07688You 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.