

ORIGINAL ARTICLE 

Year : 2022  Volume
: 9
 Issue : 1  Page : 5462 

How to conduct inferential statistics online: A brief handson guide for biomedical researchers
Shaikat Mondal^{1}, Swarup Saha^{2}, Himel Mondal^{3}, Rajesh De^{4}, Rabindranath Majumder^{5}, Koushik Saha^{2}
^{1} Department of Physiology, Raiganj Government Medical College and Hospital, Raiganj, West Bengal, India ^{2} Department of Anatomy, Rampurhat Government Medical College and Hospital, Rampurhat, West Bengal, India ^{3} Department of Physiology, Fakir Mohan Medical College and Hospital, Balasore, Odisha, India ^{4} Department of Community Medicine, RG Kar Medical College, Kolkata, India ^{5} Centre of Healthcare Science and Technology, Indian Institute of Engineering Science and Technology, Howrah, West Bengal, India
Date of Submission  29Oct2021 
Date of Acceptance  10Nov2021 
Date of Web Publication  23Mar2022 
Correspondence Address: Himel Mondal Department of Physiology, Fakir Mohan Medical College and Hospital, Balasore, Odisha India
Source of Support: None, Conflict of Interest: None  Check 
DOI: 10.4103/ijves.ijves_116_21
Introduction: Research data are first organized and visualized with the help of descriptive statistics. The next step is the inferential statistics. Result of the inferential statistics helps to conclude the finding. Many researchers and medical students may not have access to dedicated software for biostatistics. Aim: This study aimed to provide a guide on the conduct of common inferential statistics that can be done online. Methods: Common inferential statistical tests for both numerical data and categorical data were described in this study. All the tests were conducted online and the process is described step by step with example data. Results: The following tests were describedonesample ttest, onesample median test, unpaired ttest, Mann–Whitney Utest, paired ttest, Wilcoxon signedrank test, oneway analysis of variance (ANOVA), Kruskal–Wallis test, repeatedmeasure ANOVA, Friedman Test, Pearson correlation test, Spearman correlation test, Binomial test, Chisquare test, Fisher's exact test, and MacNemar test. All these tests could be conducted online from a computer connected to the internet. Conclusion: We could conduct common inferential statistical tests online without any installed software. Anyone without prior data analysis knowledge may conduct the tests with example data on any internet browser. We presume that these would help the medical undergraduate and postgraduate students.
Keywords: Analysis of variance, Chisquare test, free statistics, online statistics, P value, Pearson correlation, ttest
How to cite this article: Mondal S, Saha S, Mondal H, De R, Majumder R, Saha K. How to conduct inferential statistics online: A brief handson guide for biomedical researchers. Indian J Vasc Endovasc Surg 2022;9:5462 
How to cite this URL: Mondal S, Saha S, Mondal H, De R, Majumder R, Saha K. How to conduct inferential statistics online: A brief handson guide for biomedical researchers. Indian J Vasc Endovasc Surg [serial online] 2022 [cited 2022 May 28];9:5462. Available from: https://www.indjvascsurg.org/text.asp?2022/9/1/54/340497 
Introduction   
Descriptive statistics is the first step to get insight about research data. When researchers have knowledge about the central tendency and the distribution of the data, they decide on the inferential statistical analysis plan. In a previous article, we have described how one can conduct descriptive statistical test online without any dedicated statistical software package.^{[1]} Like descriptive statistical tests, inferential tests can also be conducted online. There are several websites that provide their service on public domain. Some of the websites and the tests they offer can be found in another article published elsewhere.^{[2]}
In this article, we would describe the steps of some common inferential statistical tests with detailed steps. We presume that the researchers and students in any resourcelimited settings would be benefitted from this brief guideline.
Methods   
Ethics
As this study does not involve any human or animal subjects, clearance from the institutional ethics committee was not necessary as per local guidelines. The online free services provided by different websites are on public domains. The data used in various statistical tests were fabricated for instructional purposes only.
Choice of statistical test
As shown in [Table 1] and [Table 2], we present the choice of statistical test for numerical and categorical data according to normality.^{[2],[3]} Hence, we need to get the normality test done before choosing a inferential test. We have described the steps to carry out a normality test online before starting the other tests.  Table 1: Some common inferential statistical tests for numerical or quantitative variables
Click here to view 
 Table 2: Some common inferential statistical tests for categorical or qualitative variables
Click here to view 
Websites
As shown in [Table 3], we listed the website used in this manuscript. The list is diverse but not a comprehensive one. Many of the websites are there that provide multiple tests; however, we tried to limit listing them not more than for two tests to keep the diversity. We presume this would help get maximum possible number of website for inferential statistical tests.  Table 3: Common inferential statistical test and websites offering free conduct of the tests
Click here to view 
Descriptive statistics
Example
You had a data set of weight (kg) of 16 research participants and you would like to get the central tendency (mean, median, and mode) and frequency distribution of the data.^{[4]}
Steps
 Go to https://www.socscistatistics.com/descriptive/frequencydistribution/default.aspx
 Paste the copied data in the box. Click on the “Generate” button
 You would get the mean, median, standard deviation (SD), skewness, minimum and maximum score, and other details with frequency distribution table.
Result
The mean, median, SD, and range was 67.29, 66.8, 13.72, and 47.3–96.5, respectively [Figure 1]a. Frequency distribution table showed that 31.2% was in 47–60.9, 43.8% was in 61–74.9, 18.8% was in 75–88.9, 6.2% was in 89–102.9 class. The class was automatically calculated according to the data. However, you can change the setting as per our necessity.  Figure 1: Descriptive and inferential statistical test result screenshot – (a) central tendency, (b) normality test, (c) onesample ttest,(d) onesample median test, (e) unpaired ttest, (f) MannWhitney Utest, (g) paired ttest, (h) Wilcoxon signed rank test, (i) oneway analysis of variance. High resolution figures are available in supplementary file (https://doi.org/10.6084/m9.figshare.16903144.v1)
Click here to view 
Normality test
Example
You had a data set of weight (kg) of 16 research participants and you would like to check if the data are normally distributed (Gaussian distribution) or not.^{[5]} For that, you fixed the null hypothesis (with α = 0.05) as “the data are normally distributed” and alternative hypothesis as “the data are not normally distributed.” If the P > 0.05, then you would fail to reject the null hypothesis. Simply, P > 0.05 = data normally distributed, P ≤ 0.05 = data are not normally distributed.
Steps
 Go to https://www.statskingdom.com/320ShapiroWilk.html
 Paste the copied data in in the “Data:” box; keep the name as “Population;” significance level as “0.05;” and outlier as “Included.” Click on the “Calculate” button
 You would get the result of Shapiro–Wilk normality test P value (if P > 0.05, then the data have normal distribution) along with mean, SD, median, skewness, QuantileQuantile (QQ) plot,^{[6]} and histogram.
Result
The P = 0.75. You could not reject the null hypothesis [Figure 1]b. Hence, it is assumed that the data were normally distributed.
Inferential statistical tests
Onesample ttest
Example
You had a data set of postprandial blood glucose level (in mg/dL) of 30 research participants. The reference value is 140 mg/dL. The data were “normally distributed” (Shapiro–Wilk test). You calculated the mean and it was found to be 144.87 mg/dL. From the mean, you could see it is greater than the reference. You would like to check if the difference was statistically significant or not.^{[7]}
Steps
 Go to https://www.graphpad.com/quickcalcs/OneSampleT1.cfm?Format = SD
 “Choose data entry format” as “Enter or paste up to 10,000 rows;” “Specify the hypothetical mean value” as “140;” paste the copied data in the “Enter data” box. Click on the “Calculate now” button
 You would get P value and its interpretation along with mean, standard deviation (SD), standard error of mean (SEM), t value, degree of freedom (df), and 95% confidence interval (CI).
Result
The blood glucose level in the sample (n = 30) (mean = 144.87, SD = 28.74) was similar to the reference value, t (29) = 0.93, P = 0.36 [Figure 1]c.
Onesample median test
Example
You had a data set of postprandial blood glucose level (in mg/dL) of 30 research participants. Your reference value is 140 mg/dL. The data were not normally distributed. Hence, you selected onesample median test (nonparametric test).^{[8]} You calculated the median and it was found to be 172 mg/dL. From the median, you could see it is greater than the reference. You would like to check if the difference was statistically significant or not.
Steps
 Go to https://atozmath.com/CONM/NonParaTest.aspx?q = mt
 Select “Method” as “5. Median test: 1 sample, 2 sample;” paste the copied data in the box; click on “Generate” button; put “140” in the “Median” box; keep “Significance Level” as “0.05” and “hypotheis” as “Onetailed.” Click on the “Find” button. You may need to “Click here to display next solution steps (part2/2)” to reveal the full result
 You would get the result at step 7 with χ^{2}, df, and P value.
Result
The median blood glucose level in the sample (median = 172, Q1Q3 = 110.75–189.25) was significantly different from the reference value of 140, χ^{2} (1) = 6.533, P < 0.05 [Figure 1]d.
Unpaired ttest
Example
You had a data set of age (years) of 30 males and 26 females. The data were normally distributed. You would like to check if the mean age between males and females are significantly different or not. The variance was calculated from https://www.statskingdom.com/220VarF2.html and the SD of the groups was not significantly different. Hence, you carried out unpaired ttest assuming equal variance.^{[9]} If there is unequal variance, you may follow the same steps below as the result would be shown both for equal variance and unequal variance.
Steps
 Go to https://www.usablestats.com/calcs/2 sampletandsummary = 1
 Click on the “Enter Raw Data” under the “Data” box; Paste the male data in “Sample 1” and female data in “Sample 2;” Click on the “Submit” button
 You would get the result under “Descriptive Statistics” box. Along with P value and t value, you would also get the mean, SD, SEM.
Result
There was no significant difference between the age of male (mean = 15.4, SD = 2.95) and female (mean = 14.5, SD = 2.67), t (54) = 1.18, P = 0.24 [Figure 1]e.
Mann–Whitney Utest
Example
You had a data set of body fat percentage of 19 males and females. You would like to check if there is any significant difference between body fat of males and females. The data were not normally distributed; hence, Mann–Whitney Utest was selected (nonparametric test). This test is also known as WilcoxonMann–Whitney test, Mann–WhitneyWilcoxon test, and Wilcoxon ranksum test.^{[10]} In cases where a column of data is having normal distribution and another column is not normally distributed, Mann–Whitney Utest is preferred.
Steps
 Go to https://www.socscistatistics.com/tests/mannwhitney/default2.aspx
 Paste copied data in Sample 1 and sample 2. Keep the “Significance Level:” as “0.05” and hypothesis as “Twotailed.” Click on “Calculate U”
 The U, z score, and P value would be shown.
Result
Body fat of females (median = 30.2, Q1Q3 = 27.5–32.5) was significantly higher than males (median = 19.4, Q1Q3 = 18–20.3), U = 73.5, P = 0.0019 [Figure 1]f
Quartile can be calculated from: https://www.calculatorsoup.com/calculators/statistics/quartilecalculator.php
Paired ttest
Example
You measured the systolic blood pressure (BP) (mmHg) before and after 3 min of exercise. From mean and SD, it was evident that BP was increased after exercise. You tested the data for normality and found a normal distribution. The differences between the measurements were also normally distributed and you had no outlier (calculated from https://statscalculator.com/outlier). Hence, you selected the paired ttest.^{[11]}
Steps
 Go to https://www.aatbio.com/tools/onetwosampleindependentpairedstudentttestcalculator
 Copy both the column of data and paste it in “Data Entry” box; click on the “Process data” button; in the “Calculation Options,” set “Type” as “Two Sample (Paired).” Click on the “Calculate ttest” button
 You would get the t value, P value. Besides, you would get the mean, median, SD, variance, range, 95% CI, histogram, box plot, QQ plot, etc.
Result
Systolic BP significantly increases after the exercise (mean = 135.8, SD = 2.75) compared to BP before the exercise (mean = 122.4, SD = 2.9), t (29) = 16.45, P < 0.0001 [Figure 1]g.
Wilcoxon signed rank test
Example
You had data of resting heart rate (HR) (in bpm) before and after a 6week Yoga practice. You would like to check if there is any change in HR after completing the Yoga practice session. The data were not normally distributed. Hence, according to guidelines, as shown in [Table 1], you selected Wilcoxon signedrank rest (nonparametric test).^{[12]} Quartile was calculated from http://www.alcula.com/calculators/statistics/quartiles/#gsc.tab = 0.
Steps
 Go to https://epitools.ausvet.com.au/paired
 Select the “Type of statistical test to be done” as “Wilcoxon signedrank test (nonparametric data);” keep the “confidence level” as “0.95;” make the “Digits after decimal point” “2;” copy two columns of data with column heading and paste in the box. Click on the “Submit” button
 The result with P value with Shapiro–Wilk test for normality, minimum, maximum, mean, SD, 95% CI, quartile, histogram, box plot, CI plot, etc., will be shown.
Result
The HR was significantly reduced after completion of yoga program (median = 68.5, Q1Q3: 62–75.5) compared to preyoga HR (median = 88, Q1Q3: 83.5–95.25), P < 0.0001 [Figure 1]h.
Oneway analysis of variance
Example
You had a data set of body weight (kg) of males, females, and intersex research participants. You would like test if the mean weight in this three groups were significantly different or not. The data were normally distributed. Hence, you selected the oneway analysis of variance (ANOVA).^{[13]}
Steps
 Go to https://goodcalculators.com/onewayanovacalculator
 Copy the data and paste it in the “OneWay ANOVA Calculator” section in “Group 1, Group 2, and Group 3.” You may add more group if necessary by clicking “+Add Group.” Click on “Calculate” to get the result
 You would get the P value, F value, df between groups and within groups along with mean, SD, SEM, etc., and a graphical comparison of the groups.
Result
The oneway ANOVA showed that there was a significant difference in the body weight of males (mean = 79.81, SD = 7.9), females (mean = 48.64, SD = 4.34), and intersex (mean = 58.5, SD = 4.5), F (2, 60) = 157.4, P < 0.0001 [Figure 1]i.
From ANOVA result, you cannot check which pair had significantly different weight. To know this, you need to run a post hoc (means “after the event”) test. You used the “Tukey's Honest Significant Difference (Tukey's HSD) from https://www.icalcu.com/stat/anovatukeyhsdcalculator.html and found that all group differences were significant (Group 1Group 2, Group 1Group 3, Group 2Group 3).^{[14]} If you want ANOVA and post hoc simultaneously, you may check the website https://astatsa.com/OneWay_Anova_with_TukeyHSD/. For the example data, select k = 3; uncheck box under K numbers; proceed to enter your treatment data; paste data; click on “Calculate ANOVA.” If the ANOVA result comes insignificant (P > 0.05), then, there is no question of running a post hoc test.
Tukey's HSD test for multiple comparisons showed that the mean body weight of male and females (P <.0001), male and intersex (P <.0001), and female and intersex (P <.0001) was significantly different.
Kruskal–Wallis test
Example
You had a data set of body fat percentage of three groups – sedentary, active, and athletes. You would like to check if the body fat percentage significantly differs among these groups. You checked the normality of the data and found that they were not normally distributed. Hence, you decided to carry out KruskalWallis test (nonparametric test).^{[15]}
Steps
 Go to https://mathcracker.com/kruskalwallis
 Copy each column of data and paste it in “Sample 1, Sample 2, and Sample 3” column; set the “Significance level (a) = “ as “0.05.” Click on “Solve”
 You would get the P value at the bottom of the page along with detailed calculation steps in the page.
Result
There was a significant difference among the median body fat of sedentary (median = 36, Q1Q3: 33.15–38.2), active (median = 22.35, Q1Q3: 18.5–23.4), and athletes (median = 17.35, Q1Q3: 14.5–18.5), χ^{2} (2) = 44.71, P < 0.0001 [Figure 2]a.  Figure 2: Inferential statistical test result screenshot – (a) Kruskal–Wallis test, (b) repeatedmeasure analysis of variance, (c) Friedman test, (d) Pearson correlation test, (e) Spearman correlation test, (f) Binomial test, (g) Chisquare test, (h) Fisher's exact test, (i) MacNemar test. High resolution figures are available in supplementary file
Click here to view 
To check which pair significantly showed different median, you need the Dunn's test (a post hoc test). However, while we wrote this article, there was no suitable calculator found online. Hence, you decided to do a pairwise Mann–Whitney Utests with Bonferroni correction (α = 0.05 will be divided by the number of groups to get α for the multiple Mann–Whitney Utest).^{[16]} Hence, α = 0.05/3 = 0.0166. In this test, a P < 0.0166 would be considered statistically significant. The test could be done from: https://www.statskingdom.com/170median_mann_whitney.html by setting the α = 0.0166. This website provides result with box plot and histogram. Alternatively, https://www.statskingdom.com/kruskalwalliscalculator.html may be used to conduct Kruskal–Wallis test along with pairwise comparison shown under “Multiple Comparison.”
A post hoc multiple pairwise Mann–Whitney Utest with Bonferroni correction showed that sedentary and active (P < 0.0001), sedentary and athletes (P < 0.0001), and active and sedentary (P < 0.0001) groups had significantly different body fat percentage.
Repeatedmeasure analysis of variance
Example
You had a data set of HR (bpm) of 20 athletes after mildintensity exercise, moderateintensity exercise, and vigorousintensity exercise. You would like to check if the HR differs in various grades of exercise. The data were normally distributed. Hence, repeatedmeasure ANOVA was selected appropriate for the data.^{[17]}
Steps
 Go to http://www.statisticslectures.com/calculators/anovawithin/index.php
 Click on the bullet button after “[3]” to “Select Number of Groups:” and click on “Change;” as you had 20 entries, keep “Select Number of entries:” as “50;” keep the significance level as “0.05;” manually enter subject wise data in “Input” section. Click on the “Submit” button
 You would get the “Output:” with df, F, and P value.
Result
Repeatedmeasure ANOVA showed that there was an significant difference in HR when the intensity of the exercise is increased (mild exercise mean = 68.85 SD = 2.5, moderate exercise mean = 79.05, SD = 2.46, and vigorous exercise mean = 109.45, SD = 3.78), F (2, 38) = 1329.881, P < 0.05 [Figure 2]b.
For repeatedmeasure ANOVA, if manual data entry seems time taking, you may try an alternative service provider https://www.socscistatistics.com/tests/anovarepeated/default.aspx).
You need to check which pair of data was significantly different. Hence, a post hoc test was necessary. When we wrote this article, we could not find any online calculator for a post hoc test for repeated measure ANOVA. Hence, pairwise test with Bonferroni correction was carried out.^{[18]} As you had three groups of repeated data, corrected α=0.05/3=0.0167 was used for the test. Paired ttests were conducted with moderatemild, vigorousmoderate, and vigorousmild pairs from https://www.statskingdom.com/160MeanT2pair.html.
A post hoc multiple paired ttest with Bonferroni correction showed that HR after moderateintensity significantly increased (P < 0.0001) compared to mild exercise, significantly increased in vigorousintensity exercise (P < 0.0001) compared to moderate exercise, and significantly increased (P < 0.0001) in vigorousintensity exercise compared to mild exercise.
Friedman test
Example
You had a data set of formative assessment marks of 23 students before and after the application of an innovative method of recording roll call where the students are engaged in academic activity during the roll call. You would like to test if the method improved the marks in formative assessment. The data were not normally distributed. Hence, Friedman's test (nonparametric test for repeated measure) was chosen instead of repeatedmeasure ANOVA.^{[19]}
Steps
 Go to https://www.socscistatistics.com/tests/friedman/default.aspx
 Paste the data in Treatment 1, Treatment 2, and Treatment 3; keep “Significance Level” as “0.05.” Click on “Calculate”
 You would get the P value.
Result
There was a statistically significant difference in the formative assessment score before the intervention (median = 5, Q1Q3 = 4–6) and in 1^{st} week (median = 7, Q1Q3 = 7–7), 2^{nd} week (median = 8, Q1Q3 = 7–8), and 3rd week (median = 8, Q1Q3 = 8–8) postintervention, χ^{2}_{r} (3, n = 23) = 49.99, P < 0.00001 [Figure 2]c.
As there was statistically significant difference, a post hoc test is needed to know which pair of observation differs significantly. This can be done from http://www.statstodo. com/FriedmanTWAV.php (click on “Javascript Program; “paste the data of four columns at a time; click on “Calculate Friedman's two way ANOVA;” scroll down to get the result of post hoc [Wilcoxon's Test, with Bonferroni adjustment] analysis).
A post hoc multiple Wilcoxon signedrank test with Bonferroni correction showed that there was a significant increase in academic performance from baseline score to post intervention week 1 (P < 0.01), week 2 (P < 0.01), and week 3 (P < 0.01). Although there was a significant difference between the scores of week 1 and week 3 (P < 0.05), the difference between the scores of week 1 and week 2, and week 2 and week 3 was not significantly different.
Pearson correlation test
Example
You had a data set of level of body fat (%) and lowdensity lipoprotein (LDL) cholesterol of 41 research participants. You would like to test if the body fat and LDLcholesterol is correlated. The data were normally distributed. Hence, you selected the Pearson correlation coefficient test.^{[20]}
Steps
 Go to http://www.wessa.net/rwasp_correlation.wasp
 Paste body fat data in “Data X:” and LDLcholesterol data in “Data Y:” (you can do the opposite also); keep chart option as default setting. Click on “Compute”
 You would get the r and P (onetailed and twotailed), coefficient of determination (r^{2}),^{[21]} along with 95% CI, mean, scatterplot, QQ plot of Data X and Data Y.
Result
The body fat and LDLCholesterol was found to be positively correlated, r (39) = 0.88, P < 0.0001 [Figure 2]d. The coefficient of “determination” was 0.78. It indicates that 78% of the differences in body fat (or LDLcholesterol) can be explained by a difference in LDLcholesterol (or body fat).
Spearman correlation test
Example
You had a data set of body weight and height of 22 research participants. You would like to test if the weight and height are correlated. The data were not normally distributed. Hence, you selected the Spearman correlation coefficient (nonparametric correlation test).^{[22]}
Steps
 Go to https://geographyfieldwork.com/Spearmans RankCalculator.html
 Select “Enter number (n) of pairs of measurements (minimum 5 maximum 75)” as “22” (22 pair data in the example); manually enter weight in “Data Set A” and height data in “Data Set B” Click. “Calculate.” You can click on “Scatter Graph” for a scatterplot
 You would get the Rho or R_{s} or r_{s} and P value and interpretation of the result.
Result
There was no correlation between the weight and height, r_{s} = 0.12, P >.05 [Figure 2]e. If manual data entry seems time taking, you may try an alternative service provider for Spearman correlation test from https://www.socscistatistics.com/tests/spearman/default2.aspx or http://www.wessa.net/rwasp_spearman.wasp.
Binomial test
Example
You made a customized Pasteur pipette. When 112 Wintrobe's tubes were filled with sample blood with the help of the Pasteur pipette, there were 78 successes without any air bubble in blood column and 34 failures with bubble in blood column. Hence, you had a categorical binary (two possible outcome) variable. Hence, the Binomial test was selected.^{[23]}
Steps
 Go to https://stattrek.com/onlinecalculator/binomial.aspx
 Put “0.5” in “Probability of success on a single trial;” “112” in “Number of trials;” “78” in “Number of successes (x).” Click on “Calculate” button
 You would get the binomial and cumulative probability P value.
Result
There is a probability of more than 50% success in filling tube with the modified Pasteur pipette, P = 0.000011. Furthermore, the probability of exactly, or more than 69.64% (78 successes in 112 trials) success with the new Pasteur pipette is there, P = 0.000019 [Figure 2]f.
You can also calculate the 95% CI from https://epitools.ausvet.com.au/ciproportion with sample size, positive result, and confidence level (0.95). It also provides the CI plot. The confidence limit in Wilson method was 0.61–0.77.
Chisquare test
Example
You conducted a preseminar survey on a sample of students (n = 92) and found that 13, 34, and 45 students had correct knowledge, no knowledge, and wrong knowledge on COVID19, respectively. A postseminar survey showed that 75, 7, and 10 showed correct knowledge, no knowledge, and wrong knowledge, respectively. From the percentage of correct response, you could see an increase in correct knowledge. However, you needed to test if the distribution of participants according to knowledge before and after the seminar statistically significant or not.^{[24]}
Steps
 Go to https://www.socscistatistics.com/tests/chisquare2/default2.aspx
 Write the “Group and Category Names;” click on “Next;” “Enter Your data Below;” click on “Next” button; keep “Significance Level:” as “0.05.” Click on “Calculate Chisquare”
 You would get the Chisquare value and P value.
Result
The relationship among correct knowledge, no knowledge, and wrong knowledge was significant, χ^{2} (2, n = 92) = 83.74, P < 0.00001 [Figure 2]g. Hence, the seminar on COVID19 had some impact on the knowledge among the students.
To know if correct knowledge was increased, you can observe the table. However, you need to test it statistically with a post hoc test. First, the table was divided into 2 × 2 with the following pair – correct versus noknowledge, noknowledge versus wrong knowledge, and correct versus wrong knowledge and Chisquare test were performed (from https://www.statskingdom.com/310GoodnessChi.html) with Bonferroni correction (α = 0.05/3 = 0.0166). It was found that correct versus noknowledge χ^{2} (1 n = 92) = 56.1, P < 0.00001 (significant); noknowledge versus wrong knowledge χ^{2} (1, n = 92) = 0.02, P = 0.89 (not significant); and correct versus wrong knowledge χ^{2} (1, n = 92) = 63.11, P < 0.00001 (significant). Fisher's exact test may also be conducted as a post hoc test if the frequency comes below 5.^{[25]}
If you need a simple 10 × 10 Chisquare test calculator, you may check the following website http://quantpsy.org/chisq/chisq.htm also.
If you need to calculate odd ratio (OR) and risk ratio, you may check https://www.socscistatistics.com/biostatistics/default2.aspx website.
Fisher's exact test
Example
You had a data set of smoking and constipation. Among the smoker participants, 3 of them had and 7 of them did not have constipation. Among the nonsmoker, 4 of them had and 6 of them did not have constipation. You prepared a 2 × 2 contingency table with the data of smoker and nonsmoker and constipation and noconstipation. You wanted to know if there was any association between smoking and constipation. As the frequency was <5, Fisher's exact test was selected.^{[26]}
Steps
 Go to https://www.langsrud.com/fisher.htm
 Place the values in the 2 × 2 box. Click on “COMPUTE”
 You would get the P values.
Results
There was no association between smoking and constipation, P = 0.99 [Figure 2]h.
If you need to analyze 2 × 3 contingency table, you may try https://www.danielsoper.com/statcalc/calculator.aspx?id = 58 website.
MacNemar test
Example
You had provided the research participants a new drug. You would like to test if there was reduction in body weight and body fat after taking the drug.^{[27]} You had 50 participants and after application of the drug, weight was reduced in 28 subjects (not reduced in 22 subjects) and body fat was reduced in 21 (not reduced in 29 subjects) subjects. The 2 × 2 contingency table had the following data: Left upper as 17 (weight reduced + fat reduced), right upper as 11 (weight not reduced + fat reduced), left lower as 4 (fatreduced + weight not reduced), and right lower as 18 (fat not reduced + weight not reduced).
Steps
 Go to http://vassarstats.net/propcorr.html
 Put the data in the empty 2 × 2 contingency table. Click on “Calculate”
 You would get onetailed and twotailed P values as “McNemar Test Result,” ORs, and 95% CI of OR.
Result
There was no difference in the proportion of participants with decreased body weight and body fat after the treatment, P = 0.12 [Figure 2]i. The OR was 2.75 and 95% CI was 0.875–8.637.
Results   
A total of 16 statistical testsonesample ttest, onesample median test, unpaired ttest, Mann– Whitney Utest, paired ttest, Wilcoxon signedrank test, oneway ANOVA, Kruskal–Wallis test, repeatedmeasure ANOVA, Friedman test, Pearson correlation test, Spearman correlation test, Binomial test, Chisquare test, Fisher's exact test, and MacNemar test was conducted online on various websites. These websites are on public domain. The list of the websites is shown in [Table 3].
Screenshots of the result page are presented in [Figure 1] and [Figure 2]. Full image can be accessed from the supplementary file found in Figshare (https://doi.org/10.6084/m9.figshare. 16903144.v1).
Discussion   
With an aim to provide a brief guide to some of the common inferential statistical tests, a total of 16 tests were described. However, this list is not at all comprehensive for inferential statistics used in biomedical researches. Many of the tests, especially used in some clinical studies and diagnostic testrelated studies (e.g., sensitivity, specificity, Receiver Operating curve, Kaplan– Meier estimate) are not discussed. Those would be discussed in the next article.
Although we were able to conduct the tests on a specific website while we wrote the manuscript, its existence in future is beyond our control. If some websites cease to operate, the readers are suggested to try new avenues. Hence, we listed a diverse websites. Among these websites, some (e.g., https://www.socscistatistics.com, https://www.socscistatistics.com) provide multiple tests.
Conclusion   
We discussed the conduct of some common inferential statistical tests that can be done online without any installed software. The steps for conducting the tests in the websites were described with the help of fabricated data. The method of reporting the results was also provided. Researchers in any resourcelimited settings with access to computer and internet may carry out the tests. We presume that the guide would be a helpful resource for the medical undergraduate and postgraduate students.
Supplementary file: https://doi.org/10.6084/m9.figshare. 16903144.v1
Acknowledgement
The corresponding author would like to thank Ahana Aarshi for allowing long hours working on this manuscript. Her repeated appearance near the laptop computer and demand to watch “Hasi song” (Hasi Ban Gaye  Hamari Adhuri Kahani) gave him a delightful break from the work. Above all, it was possible to finish the work with the continuous support from Sarika Mondal.
Financial support and sponsorship
Nil.
Conflicts of interest
There are no conflicts of interest.
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[Figure 1], [Figure 2]
[Table 1], [Table 2], [Table 3]
