|Year : 2022 | Volume
| Issue : 1 | Page : 54-62
How to conduct inferential statistics online: A brief hands-on guide for biomedical researchers
Shaikat Mondal1, Swarup Saha2, Himel Mondal3, Rajesh De4, Rabindranath Majumder5, Koushik Saha2
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||29-Oct-2021|
|Date of Acceptance||10-Nov-2021|
|Date of Web Publication||23-Mar-2022|
Department of Physiology, Fakir Mohan Medical College and Hospital, Balasore, Odisha
Source of Support: None, Conflict of Interest: None
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 described-one-sample t-test, one-sample median test, unpaired t-test, Mann–Whitney U-test, paired t-test, Wilcoxon signed-rank test, one-way analysis of variance (ANOVA), Kruskal–Wallis test, repeated-measure ANOVA, Friedman Test, Pearson correlation test, Spearman correlation test, Binomial test, Chi-square 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, Chi-square test, free statistics, online statistics, P value, Pearson correlation, t-test
|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 hands-on guide for biomedical researchers. Indian J Vasc Endovasc Surg 2022;9:54-62
|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 hands-on guide for biomedical researchers. Indian J Vasc Endovasc Surg [serial online] 2022 [cited 2022 May 28];9:54-62. 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. 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.
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 resource-limited settings would be benefitted from this brief guideline.
| Methods|| |
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., 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|
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|Table 2: Some common inferential statistical tests for categorical or qualitative variables|
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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|
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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.
- 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.
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) one-sample t-test,(d) one-sample median test, (e) unpaired t-test, (f) MannWhitney U-test, (g) paired t-test, (h) Wilcoxon signed rank test, (i) one-way analysis of variance. High resolution figures are available in supplementary file (https://doi.org/10.6084/m9.figshare.16903144.v1)|
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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. 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.
- 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, Quantile-Quantile (Q-Q) plot, and histogram.
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
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.
- 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).
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.
One-sample median test
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 one-sample median test (non-parametric test). 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.
- 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 “One-tailed.” Click on the “Find” button. You may need to “Click here to display next solution steps (part-2/2)” to reveal the full result
- You would get the result at step 7 with χ2, df, and P value.
The median blood glucose level in the sample (median = 172, Q1-Q3 = 110.75–189.25) was significantly different from the reference value of 140, χ2 (1) = 6.533, P < 0.05 [Figure 1]d.
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 t-test assuming equal variance. 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.
- 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.
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.
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 U-test was selected (nonparametric test). This test is also known as WilcoxonMann–Whitney test, Mann–WhitneyWilcoxon test, and Wilcoxon rank-sum test. In cases where a column of data is having normal distribution and another column is not normally distributed, Mann–Whitney U-test is preferred.
- 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 “Two-tailed.” Click on “Calculate U”
- The U, z score, and P value would be shown.
Body fat of females (median = 30.2, Q1-Q3 = 27.5–32.5) was significantly higher than males (median = 19.4, Q1-Q3 = 18–20.3), U = 73.5, P = 0.0019 [Figure 1]f
Quartile can be calculated from: https://www.calculatorsoup.com/calculators/statistics/quartile-calculator.php
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 t-test.
- Go to https://www.aatbio.com/tools/one-two-sample-independent-paired-student-t-test-calculator
- 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 t-test” button
- You would get the t value, P value. Besides, you would get the mean, median, SD, variance, range, 95% CI, histogram, box plot, Q-Q plot, etc.
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
You had data of resting heart rate (HR) (in bpm) before and after a 6-week 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 signed-rank rest (nonparametric test). Quartile was calculated from http://www.alcula.com/calculators/statistics/quartiles/#gsc.tab = 0.
- Go to https://epitools.ausvet.com.au/paired
- Select the “Type of statistical test to be done” as “Wilcoxon signed-rank 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.
The HR was significantly reduced after completion of yoga program (median = 68.5, Q1-Q3: 62–75.5) compared to preyoga HR (median = 88, Q1-Q3: 83.5–95.25), P < 0.0001 [Figure 1]h.
One-way analysis of variance
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 one-way analysis of variance (ANOVA).
- Go to https://goodcalculators.com/one-way-anova-calculator
- Copy the data and paste it in the “One-Way 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.
The one-way 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/anova-tukey-hsd-calculator.html and found that all group differences were significant (Group 1-Group 2, Group 1-Group 3, Group 2-Group 3). 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.
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 Kruskal-Wallis test (nonparametric test).
- Go to https://mathcracker.com/kruskal-wallis
- 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.
There was a significant difference among the median body fat of sedentary (median = 36, Q1-Q3: 33.15–38.2), active (median = 22.35, Q1-Q3: 18.5–23.4), and athletes (median = 17.35, Q1-Q3: 14.5–18.5), χ2 (2) = 44.71, P < 0.0001 [Figure 2]a.
|Figure 2: Inferential statistical test result screenshot – (a) Kruskal–Wallis test, (b) repeated-measure analysis of variance, (c) Friedman test, (d) Pearson correlation test, (e) Spearman correlation test, (f) Binomial test, (g) Chi-square test, (h) Fisher's exact test, (i) MacNemar test. High resolution figures are available in supplementary file|
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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 U-tests with Bonferroni correction (α = 0.05 will be divided by the number of groups to get α for the multiple Mann–Whitney U-test). 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/kruskal-wallis-calculator.html may be used to conduct Kruskal–Wallis test along with pair-wise comparison shown under “Multiple Comparison.”
A post hoc multiple pairwise Mann–Whitney U-test 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.
Repeated-measure analysis of variance
You had a data set of HR (bpm) of 20 athletes after mild-intensity exercise, moderate-intensity exercise, and vigorous-intensity exercise. You would like to check if the HR differs in various grades of exercise. The data were normally distributed. Hence, repeated-measure ANOVA was selected appropriate for the data.
- Go to http://www.statisticslectures.com/calculators/anovawithin/index.php
- Click on the bullet button after “” 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.
Repeated-measure 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 repeated-measure 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. As you had three groups of repeated data, corrected α=0.05/3=0.0167 was used for the test. Paired t-tests were conducted with moderate-mild, vigorous-moderate, and vigorous-mild pairs from https://www.statskingdom.com/160MeanT2pair.html.
A post hoc multiple paired t-test with Bonferroni correction showed that HR after moderate-intensity significantly increased (P < 0.0001) compared to mild exercise, significantly increased in vigorous-intensity exercise (P < 0.0001) compared to moderate exercise, and significantly increased (P < 0.0001) in vigorous-intensity exercise compared to mild exercise.
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 repeated-measure ANOVA.
- 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.
There was a statistically significant difference in the formative assessment score before the intervention (median = 5, Q1-Q3 = 4–6) and in 1st week (median = 7, Q1-Q3 = 7–7), 2nd week (median = 8, Q1-Q3 = 7–8), and 3rd week (median = 8, Q1-Q3 = 8–8) postintervention, χ2r (3, n = 23) = 49.99, P < 0.00001 [Figure 2]c.
A post hoc multiple Wilcoxon signed-rank 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
You had a data set of level of body fat (%) and low-density lipoprotein (LDL) cholesterol of 41 research participants. You would like to test if the body fat and LDL-cholesterol is correlated. The data were normally distributed. Hence, you selected the Pearson correlation coefficient test.
- Go to http://www.wessa.net/rwasp_correlation.wasp
- Paste body fat data in “Data X:” and LDL-cholesterol 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 (one-tailed and two-tailed), coefficient of determination (r2), along with 95% CI, mean, scatterplot, Q-Q plot of Data X and Data Y.
The body fat and LDL-Cholesterol 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 LDL-cholesterol) can be explained by a difference in LDL-cholesterol (or body fat).
Spearman correlation test
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).
- 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 Rs or rs and P value and interpretation of the result.
There was no correlation between the weight and height, rs = 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.
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.
- Go to https://stattrek.com/online-calculator/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.
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.
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 COVID-19, 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.
- 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 Chi-square”
- You would get the Chi-square value and P value.
The relationship among correct knowledge, no knowledge, and wrong knowledge was significant, χ2 (2, n = 92) = 83.74, P < 0.00001 [Figure 2]g. Hence, the seminar on COVID-19 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 no-knowledge, no-knowledge versus wrong knowledge, and correct versus wrong knowledge and Chi-square 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 no-knowledge χ2 (1 n = 92) = 56.1, P < 0.00001 (significant); no-knowledge versus wrong knowledge χ2 (1, n = 92) = 0.02, P = 0.89 (not significant); and correct versus wrong knowledge χ2 (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.
If you need a simple 10 × 10 Chi-square 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
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 non-smoker, 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 non-smoker and constipation and no-constipation. You wanted to know if there was any association between smoking and constipation. As the frequency was <5, Fisher's exact test was selected.
- 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.
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.
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. 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 (fat-reduced + weight not reduced), and right lower as 18 (fat not reduced + weight not reduced).
- Go to http://vassarstats.net/propcorr.html
- Put the data in the empty 2 × 2 contingency table. Click on “Calculate”
- You would get one-tailed and two-tailed P values as “McNemar Test Result,” ORs, and 95% CI of OR.
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 tests-one-sample t-test, one-sample median test, unpaired t-test, Mann– Whitney U-test, paired t-test, Wilcoxon signed-rank test, one-way ANOVA, Kruskal–Wallis test, repeated-measure ANOVA, Friedman test, Pearson correlation test, Spearman correlation test, Binomial test, Chi-square 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 test-related 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 resource-limited 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
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
Conflicts of interest
There are no conflicts of interest.
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[Figure 1], [Figure 2]
[Table 1], [Table 2], [Table 3]