REVIEW ARTICLE 
https://doi.org/10.5005/jpjournals100281623 
Superiority Trial: Sample Size Calculation and its Interpretation in Health Research
^{1,35}Department of National Institute of Nursing Education, Postgraduate Institute of Medical Education and Research (PGIMER), Chandigarh, India
^{2}Department of Endocrinology, Postgraduate Institute of Medical Education and Research (PGIMER), Chandigarh, India
Corresponding Author: Ashok Kumar, Department of National Institute of Nursing Education, Postgraduate Institute of Medical Education and Research (PGIMER), Chandigarh, India, Phone: +91 9855012233, emails:
ajangir_27@yahoo.in; ajangir5@gmail.com
Received on: 05 August 2022; Accepted on: 12 April 2023; Published on: 26 July 2023
ABSTRACT
Sample size calculation in any randomized clinical trial is an essential step to avoid over and underestimation of the outcomes. Health researchers should prior estimate the sample size which helps in the validity and reliability of the superiority trial. This paper highlights the stepbystep calculation of sample size in a superiority trial using appropriate and simple formulas and its interpretation with suitable examples in different conditions so that health researchers can better understand this important part of a clinical trial.
How to cite this article: Kumar A, Pal R, Nagi M, et al. Superiority Trial: Sample Size Calculation and its Interpretation in Health Research. J Postgrad Med Edu Res 2023;57(3):131–136.
Source of support: Nil
Conflict of interest: Dr Sukhpal Kaur is associated as the National Editorial Board member of this journal and this manuscript was subjected to this journal’s standard review procedures, with this peer review handled independently of this editorial board member and her research group.
Keywords: Health research, Randomized clinical trial, Sample size calculation, Superiority trial
WHAT IS ALREADY KNOWN?

Sample size calculation in health research.

Importance of sample size calculation in the clinical trial.

Sample size calculation is considered a difficult part of the research.
WHAT THIS PAPER ADDS?

Indepth calculation with suitable examples along with simple explanations.

Enhance confidence to calculate sample size in superiority trials in health research.

Also, add the interpretation of sample size calculation results.
INTRODUCTION
In healthcare research, there are mainly three types of randomized controlled trials (RTCs); superiority trial, equivalence trial, and noninferior trial.^{1} In a superiority trial, a researcher is interested to research out that new treatment (drug A) is superior to standard (old) treatment (drug B).^{2} In this article, we will discuss sample size calculation in the superiority trial and its interpretation in detail which is a vital part of the trial.^{3} Prior calculation of sample size is very important in any clinical trial or study to avoid over and underestimation of the study results.^{4}
On the basis of the nature of the outcome of the study, there are mainly two conditions in which the superiority trial can be conducted.
Condition 1: When decrement (reduction) in dependent (outcome) variable is considered better (superior):
For quantitative outcome—decrement (reduction) in mean systolic blood pressure (SBP) in hypertensive patients or decrement (reduction) in mean anxiety score in patients.
For qualitative outcome—a decrement (reduction) in the proportion of bedsores in bedridden patients or a decrement (reduction) in the proportion of mortality in patients.
Condition 2: When increment (improvement) in dependent (outcome) variable is considered better (superior):
For quantitative outcome—increment (improvement) in quality of life (QOL) score in breast cancer patients.
For qualitative outcome: increment (improvement) in the success rate of any procedure or increment in satisfaction rate.
CONDITION 1
Superiority Margin (θ_{0})
Suppose we want to see a mean difference of −10 or a proportion difference of −10% (−0.10) between the new treatment group (drug A) and the old treatment group (drug B) at a certain time point.
Here, θ = −10 or −10%
But on the contrary, we also want that this difference should not come equal to or greater than = θ_{0} = −5 (superiority margin), if this difference would come equal to or greater than −5, for example, suppose −4.5. So, this violates the condition of superiority trial and we have to conclude that drug A is not superior to drug B.
As we want to achieve the difference between the two treatments as −10; but not equal to or greater −5 (θ_{0} = −5) to fulfill the superiority trial condition. We have to judicially (logically) decide about the superiority margin in the superiority trial. Generally, it is taken as 0.
Hypothesis of Superiority Trial^{5}^{,}^{6}
Condition 1A—when the dependent variable is quantitative in nature (mean) and decrement (reduction) in the dependent (outcome) variable is considered superior.
For example 1A—in an RTC, we want to see the effect of a new therapy in the reduction of mean SBP of patients by 10 mm Hg at 1month followup compared to the control group taking the standard therapy.
Here;
µ_{n} = mean of SBP in new therapy group at 1month followup = 130 mm Hg
µ_{s} = mean of SBP in standard (old) therapy group at 1month followup = 140 mm Hg
θ = difference = µ_{n}–µ_{s} = 130–140 = −10 mm Hg
θ_{0} = superiority margin (as assumed by the researcher)= −5 mm Hg
H_{0}: null hypothesis = µ_{n}–µ_{s} ≥ θ_{0} (here θ_{0} = −5)
H_{1}: alternate (research) hypothesis = µ_{n}–µ_{s}< θ_{0} (here θ_{0} = −5)
Formula for Sample Size Calculation for Superiority Trial^{1}^{,}^{3}^{,}^{7}^{,}^{9}
or
Here,
r = ratio of sample size of group II (n2 or standard (old) treatment group) and sample size of group I (n1 or new treatment group) (usually r = 1; because n1 = n2; the number of subjects is same in each group).
Z = Standard Normal deviate function value (Table 1).
α (type I error) or β (type II error) value  Power of study = 1– β  α value at 95% confidence interval  Normal deviate function  Corresponding Z value 

0.2  1–0.2 = 0.8 = 80%  Z_{1–β} = Z_{1–0.2}  0.842  
0.15  1.036  
0.1  1–0.1= 0.9 = 90%  Z_{1–β} = Z_{1–0.1}  1.282  
0.05  Onesided = α = 5% = 0.05  Z_{1–α}  1.645  
0.025  Twosided = α/2 = 2.5% = 0.025  Z_{1–α/2}  1.960  
0.01  2.326  
0.001  3.090 
Note: Usually, the researchers take β = 0.2 for 80% power of study and β = 0.1 for 90% power of study and α = 0.05 for onesided test, and α/2 = 0.05/2 = 0.025 for twosided test. A onesided test is used for superiority and noninferiority trials whereas a twosided test is used for the equivalence trial; Researchers usually calculate sample size on bold parameters (corresponding Z value) in health research. So these values are given in bold (highlighted).
α = type I error rate; usually it is taken as 5% = 0.05.
β = type II error rate = 20% = 0.2; so, power of study = 1–β = 1–0.2 = 0.8 or 80%; if β = 0.10 then power of study will be 1–0.10 = 0.90 = 90%).
µ_{n} = mean of SBP in the new therapy group at 1month followup = 130 mm Hg.
µ_{s} = mean of SBP in standard (old) therapy group at 1month followup = 140 mm Hg.
θ_{0} = superiority margin (as assumed by the researcher) = −5 mm Hg.
θ = difference = µ_{n}–µ_{s} = 130–140 = −10 mm Hg.
σ = population/sample standard deviation (generally it is taken from the previous research studies or expert’s assumptions and in this case, it is assumed as σ =10).
Now, using formula (i) or (ii) to calculate the sample size for the superiority trial:
Put the values in the above formula and calculate the desired sample size:
How to Write Down “Sample Size Calculation Paragraph”
On the basis of the above calculation—“the researcher has to recruit 50 subjects in each group (N_{T} = total 100 subjects) in the study to achieve the reduction of mean SBP by at least 10 mm Hg at 1 month of followup in the new treatment group (drug A) as compared to standard treatment group (drug B) assuming standard deviation of 10 and superiority margin of −5 mm Hg at 80% power of the study and 95% confidence interval.”
Adjustment for Dropout Rate/Attrition Rate
In followup/longitudinal studies, there may be chances of loss to followup or dropout of the study subjects. So, it is kept in mind before the study plan and sample size calculation and proper adjustment are done, and the final sample size calculation is done after adjustment of the anticipated dropout/attrition rate.
Formula for the Adjustment for Dropout Rate/Attrition Rate
Here,
N_{A} = sample size after adjustment of the dropout/attrition rate.
N_{T} = total sample size before adjustment.
D = dropout/attrition rate, that is, 10% = 0.10; 20% = 0.20
At a dropout rate of 10%, the final sample size would be:
So, with a 10% dropout rate, the researcher has to recruit a total of 112 subjects instead of 100.
And the sample size calculation paragraph would be like this—“the researcher has to recruit 50 subjects in each group (total 100 subjects) in the study to achieve the reduction of mean SBP by at least 10 mm Hg at 1 month of followup in the new treatment group (drug A) as compared to standard treatment group (drug B) assuming standard deviation of 10 and superiority margin of −5 mm Hg at 80% power of the study and 95% confidence interval. Anticipating a 10% dropout rate, a total of 112 subjects (56 subjects in each group) would be recruited in the current study.”
Condition 1B—when the dependent variable is qualitative in nature (frequency/proportion) and decrement (reduction) in the dependent (outcome) variable is considered superior.
For example 1B—in an RTC, we want to see the effect of a new therapy in the reduction of the prevalence of bedsores among the patients by 25% (0.25) at 3 months followup compared to a control group who is taking standard therapy.
Here;
p_{n} = proportion of bedsore in new therapy group = 35% = 0.35
p_{s} = proportion of bedsores in standard (old) therapy group = 60% = 0.60
θ = difference = p_{n}–p_{s}= 0.35–0.60 = −0.25
θ_{0} = superiority margin (as assumed by the researcher) = −10% = −0.10
H_{0}: null hypothesis = p_{n}–p_{s} ≥ θ_{0} (here θ_{0} = −10%)
H_{1}: alternate (research) hypothesis = p_{n}–p_{s}<−θ_{0} (here θ_{0} = −10%)
Formula for Sample Size Calculation for Superiority Trial^{1}^{,}^{3}^{,}^{7}^{}^{9}
or
Here,
= ratio of sample size of group II (n2 or standard (old) treatment group) and sample size of group I (n1 or new treatment group) (usually r = 1; because n1 = n2; the number of subjects is same in each group).
Z = standard normal deviation function value (see Table 1).
α = type I error rate = 5% = 0.05
β = type II error rate = 20% = 0.2; so, power of study = 1–β = 1–0.2 = 0.8 or 80%; if β = 0.10 then power of study will be 1–0.10 = 0.90 = 90%).
p_{n} = proportion of bedsore in new therapy group = 35% = 0.35
p_{s}= proportion of bedsores in standard (old) therapy group = 60% = 0.60
θ = difference = p_{n}–p_{s}= 0.35–0.60 = −0.25
θ_{0} = superiority margin (as assumed by the researcher) = −10% = −0.10
Now, using formula (iii) or (iv) to calculate the sample size for the superiority trial:
Put the values in the above formula and calculate the desired sample size:
How to Write Down “Sample Size Calculation Paragraph”
On the basis of the above calculation—the researcher has to recruit 130 subjects in each group (N_{T} = total 260 subjects) in the study to achieve the reduction of bedsore proportion/prevalence by at least 20% at 3 months of followup in the new treatment group (drug A) as compared to standard treatment group (drug B) assuming superiority margin of −10% at 80% power of the study and 95% confidence interval.”
Formula for the Adjustment for Dropout Rate/Attrition Rate
Here,
N_{A} = sample size after adjustment of the dropout/attrition rate.
N_{T} = total sample size before adjustment.
D = dropout/attrition rate, that is, 10% = 0.10; 20% = 0.20
At a dropout rate of 10%, the final sample size would be:
So, with a 20% dropout rate, the researcher has to recruit a total of 325 subjects instead of 260.
And the sample size calculation paragraph would be like this— “the researcher has to recruit 130 subjects in each group (total 260 subjects) in the study to achieve the reduction of bedsore prevalence/proportion by at least 20% at 3 months of followup in the new treatment group (drug A) as compared to standard treatment group (drug B) assuming superiority margin of −10% at 80% power of the study and 95% confidence interval. Anticipating a 20% dropout rate, a total of 326 subjects (163 subjects in each group) would be recruited in the current study.”
CONDITION 2
Superiority Margin (θ_{0})
Suppose we want to see a mean difference of 20 or a proportion difference of 20% between the new treatment group (drug A) and the old treatment group (drug B) at a certain time point.
Here, θ = 20
But on the contrary, we also want that this difference should not come equal to or below = θ_{0} = 10 or 10% (superiority margin), if this difference would come equal to or below 10 for example suppose 9.25. So, this violet the condition of the superiority trial and we have to conclude that drug A is not superior to drug B.
As we want to achieve the difference between the two treatments as 20 or 20%; but not equal to or below 10 (θ_{0} = 10 or 10%). It means this difference must be above at least 10 or 10% (θ_{0}) to fulfill the superiority trial condition. We have to judicially (logically) decide about the superiority margin in the superiority trial. Generally, it is taken as 0.
Hypothesis of Superiority Trial^{5}^{,}^{6}
Condition 2A—when the dependent variable is quantitative in nature (mean) and increment or improvement in the dependent (outcome) variable is considered better (superior).
For example 2A—in an RTC, we want to see the effect of a new therapy in the improvement in the mean QOL score of the patients by 20 at 3 months followup compared to a control group who is taking standard therapy.
Here;
µ_{n} = mean of QOL score in new therapy group = 70
µ_{s} = mean of QOL score in standard (old) therapy group = 50
θ = difference = µ_{n}–µ_{s} = 70–50 = 20
θ_{0} = superiority margin (as assumed by the researcher) = 10
Hence, H_{0}: null hypothesis = µ_{n}–µ_{s} ≤ θ_{0} (here, θ_{0} = 10)
H_{1}: Alternate (research) Hypothesis: µ_{n}–µ_{s} > θ_{0}
Formula for Sample Size Calculation^{1}^{,}^{3}^{,}^{7}^{}^{9}
or
Here,
= ratio of sample size of group II (n2 or standard (old) treatment group) and sample size of group I (n1 or new treatment group) (usually r = 1; because n1= n2; the number of subjects is same in each group.)
Z = standard normal deviation function value (see Table 1).
α = type I error rate = 5% =0.05
β = type II error rate = 20% = 0.2; so, power of study = 1–β = 1–0.2 = 0.8 or 80%; if β = 0.10 then power of study will be 1–0.10 = 0.90 = 90%).
µ_{n} = mean of QOL score in new therapy group = 70.
µ_{s} = mean of QOL score in standard (old) therapy group = 50
θ = difference = µ_{n}–µ_{s} = 70–50 = 20
θ_{0} = superiority margin (as assumed by the researcher) = 10
σ = population/sample standard deviation (generally it is taken from the previous research studies or expert assumptions and in this case, it is assumed as σ =15).
Now, using formula (i) or (ii) to calculate the sample size for superiority trial:
Put the values in the above formula and calculate the desired sample size:
Condition 2B—when the dependent variable is qualitative in nature (frequency/proportion) and increment or improvement in the dependent (outcome) variable is considered better (superior).
For example—in an RTC, we want to see the effect of a new therapy in increment in success rate among the patients by 20% (0.20) at 3 months followup compared to a control group who is taking standard therapy.
Here;
µ_{n} = p_{n} = proportion of success in new therapy group = 45% = 0.45
µ_{s} = p_{s}= proportion of success in standard (old) therapy group = 25% = 0.25
θ = actual difference = p_{n}–p_{s}= 0.45–0.25 = 0.20
θ_{0} = superiority margin (as assumed by the researcher) = 0.10
H_{0}: null hypothesis = p_{n}—p_{s} ≤ θ_{0} (here, θ_{0} = 10)
H_{1}: Alternate (research) hypothesis—p_{n}–p_{s} > θ_{0}
Formula to Calculate Sample Size^{1}^{,}^{3}^{,}^{7}^{}^{9}
or
Here,
= ratio of sample size of group II (n2 or standard (old) treatment group) and sample size of group I (n1 or new treatment group) (usually r = 1; because n1= n2; the number of subjects are same in each group.)
Z = standard normal deviation function value (see Table 1).
α = type I error rate = 5% =0.05
β = type II error rate = 20% = 0.2; so, power of study = 1–β = 1–0.2 = 0.8 or 80%; if β = 0.10 then power of study will be 1–0.10 = 0.90 = 90%).
p_{n} = proportion of success in new therapy group = 45% = 0.45
p_{s} = proportion of success in standard (old) therapy group = 25% = 0.25
θ = difference = p_{n}–p_{s}= 0.45–0.25 = 0.20
θ_{0} = superiority margin (as assumed by the researcher) = 0.10
Now, using formula iii (or iv) to calculate sample size for the superiority trial:
Put the values in the above formula and calculate the desired sample size:
EXAMPLES
Condition 1A—when the dependent variable is quantitative in nature (mean) and decrement (reduction) in the dependent (outcome) variable is considered superior.
Example 1 A—in a superior RTC, the researcher wants to assess the effect of a new exercise program on the mean cholesterol level of patients at 6 weeks compared to the control group. All the parameters are as follows:
µ_{n} = mean cholesterol level of patients in a new exercise program at 6 weeks = 150 mg/dL
µ_{s} = mean cholesterol level of patients in a standard (old) therapy group at 6 weeks = 220 mg/dL
θ = difference = µ_{n}–µ_{s} = 150–220 = −70 mg/dL
θ_{0} = superiority margin (as assumed by the researcher)= −25 mg/dL
σ = population/sample standard deviation (generally it is taken from the previous research studies or expert’s assumptions and in this case, it is assumed as σ = 80) (Table 2).
Requirement  Details 

Type I error (α)  0.05 {At 95% confidence interval (onesided)} 
Type II error (β) Power of study = 1–β  At β = 0.1 so, power of study = 1–β = 1–0.1 = 0.9 = 90% At β = 0.2 so, power of study = 1–β = 1–0.2 = 0.8 = 80% 
Primary outcome/endpoint  Quantitative (mean difference) or qualitative/categorical (proportion/percentage difference) 
Difference of primary outcome/endpoint between two groups  θ = μn–μs [for quantitative outcome/endpoint (mean difference)] θ = ρn–ρs [for qualitative/categorical outcome/endpoint (proportion/percentage difference)] 
Superiority margin  θ0 = Superiority margin (as assumed by the researcher) 
Population/sample standard deviation  σ/s (generally it is taken from previous research studies or expert assumptions) 
0.05, Researchers usually calculate sample size on Type I error at 0.05 and corresponding Z value (bold numbers) in health research. So these values are given in bold (highlighted).
Formula for Sample Size Calculation for Superiority Trial
(note: Z1–α = 1.645 at 5% level of significance; Z_{1–β} = 0.842 at 80% power of study.)
Put the values in the above formula and calculate the desired sample size:
So, the researcher has to enroll at least 32 patients in each group (a total of 64) for this trial.
Condition 1B—when the dependent variable is qualitative/dichotomous in nature (frequency/proportion) and decrement (reduction) in the dependent (outcome) variable is considered superior.
Example 1B—in a superiority RTC, the researcher wants to assess the effect of mobile app strategy on the reduction of composite poststroke complications among bedridden stroke survivors compared to standard care at 3 months followup. All the parameters are as follows:
p_{n} = proportion of composite poststroke complications in mobile app group = 33.8% = 0.338.
p_{s}= proportion of composite poststroke complications in standard (old) therapy group = 56.4% = 0.564.
θ = difference = p_{n}–p_{s}= 0.338–0.564 = −0.226
θ_{0} = superiority margin (as assumed by the researcher) = −5% = −0.05
Formula for Sample Size Calculation for Superiority Trial
(note: Z1–α = 1.645 at 5% level of significance; Z_{1–β} = 0.842 at 80% power of study.)
Put the values in the above formula and calculate the desired sample size:
So, the researcher has to enroll at least 91 patients in each group (a total of 182) for this trial.
Condition 2A—when the dependent variable is quantitative in nature (mean) and increment or improvement in the dependent (outcome) variable is considered better (superior).
Example 2A—in a superiority RTC, the researcher wants to assess the effect of a new diet regimen in the improvement in the mean hemoglobin (Hb) among anemic patients at 1month followup compared to the control group. All the parameters are as follows:
µ_{n} = mean Hb in the new diet regimen group at 1 month = 14.0 mg/dL.
µ_{s} = mean Hb in control group = 10 mg/dL.
θ = difference = µ_{n}–µ_{s} = 14–10 = 4 mg/dL.
θ_{0} = superiority margin (as assumed by the researcher) = 2.
σ = population/sample standard deviation (generally it is taken from the previous research studies or expert assumptions and in this case, it is assumed as σ =3).
Formula for Sample Size Calculation for Superiority Trial
(note: Z1–α = 1.645 at 5% level of significance; Z_{1–β} = 0.842 at 80% power of study.)
Put the values in the above formula and calculate the desired sample size:
So, the researcher has to enroll at least 28 patients in each group (a total of 56) for this trial.
Condition 2B—when the dependent variable is qualitative/dichotomous in nature (frequency/proportion) and increment or improvement in the dependent (outcome) variable is considered better (superior).
Example 2B—In a superior RTC, the researcher wants to assess the effect of a new therapy on the satisfaction rate among the breast cancer survivors at 3 months followup compared to the control group.
µ_{n} = p_{n} = proportion of satisfaction in new therapy group = 80% = 0.80.
µ_{s} = p_{s}= proportion of satisfaction in control group = 50% = 0.50.
θ = actual difference = p_{n}–p_{s}= 0.80–0.50 = 0.30.
θ_{0} = superiority margin (as assumed by the researcher) = 15% = 0.15.
Formula for Sample Size Calculation for Superiority Trial
(note: Z1–α = 1.645 at 5% level of significance; Z_{1–β} = 0.842 at 80% power of study.)
Put the values in the above formula and calculate the desired sample size:
So, the researcher has to enroll at least 113 patients in each group (a total of 226) for this clinical superiority trial.
(note: dropout/nonresponse rate adjustment—we have to adjust sample size for dropout rate or nonresponse rate).
CONCLUSION
Sample size calculation in clinical superiority trials is essential. Health researchers should be acquainted with calculated sample sizes in clinical trials. This article provides thorough and stepbystep calculation and interpretation of sample size calculation in superiority trials in health research.
ORCID
Ashok Kumar https://orcid.org/0000000216914953
Rimesh Pal https://orcid.org/0000000348599393
Manisha Nagi https://orcid.org/0000000182845979
Maninderdeep Kaur https://orcid.org/0000000334468851
Sukhpal Kaur https://orcid.org/0000000337241310
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