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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.
These binaries (installable software) and packages are in development.
They may not be fully stable and should be used with caution. We make no claims about them.
Health stats visible at Monitor.