Journal of Chinese Integrative Medicine: Volume 10, 2012 Issue 3

Estimation of sample size and testing power (Part 6)

1.

Liang-ping Hu (Consulting Center of Biomedical Statistics, Academy of Military Medical Sciences, Beijing 100850, China E-mail: lphu812@sina.com)

2.

Xiao-lei Bao (Consulting Center of Biomedical Statistics, Academy of Military Medical Sciences, Beijing 100850, China )

3.

Xue Guan (Consulting Center of Biomedical Statistics, Academy of Military Medical Sciences, Beijing 100850, China )

4.

Shi-guo Zhou (Consulting Center of Biomedical Statistics, Academy of Military Medical Sciences, Beijing 100850, China )

ABSTRACT: The design of one factor with k levels (k≥3) refers to the research that only involves one experimental factor with
k levels (k≥3), and there is no arrangement for other important non-experimental factors. This paper introduces the estimation of sample size and testing power for quantitative data and qualitative data having a binary response variable with the design of one factor with
k levels (k≥3).

Received December 29, 2011; accepted December 31, 2011; published online March 15, 2012. Full-text LinkOut at PubMed. Journal title in PubMed:
Zhong Xi Yi Jie He Xue Bao.

The design of one factor with k levels (k≥3) refers to the research that only involves one experimental factor with k levels (k≥3) and there is no arrangement for other important non-experimental factors because the researchers want to balance the influence of the important non-experimental factors by randomized grouping^{[1]}. If the experimental factor is independent among research subjects, then the subjects can be divided into k groups by complete randomization; otherwise, the research subjects should be selected from k subpopulations. In this paper, we will introduce the estimation of sample size and testing power for quantitative data and qualitative data having a binary response variable with the design of one factor with k levels (k≥3).

1 Quantitative data with the design of one factor with k levels (k≥3)

1.1 Estimation of sample size 1.1.1 Formula The formula for sample size estimation for quantitative data with the design of one factor with k levels (k≥3) is as follows^{[1-3]}:

In formula (1), n stands for the sample size of each group, which is required to be equal; refer to the estimated population mean and the estimated population standard deviation of the group i; , in which k stands for the number of the groups; , in which δ is the non-central parameter of the non-central F distribution under certain conditions; and v1=k-1 is the degree of freedom of the experimental factor with the design of one factor with k levels (k≥3). The value of ψ can be obtained by SAS function, or by the ψ table in reference books. The iterative algorithm is adopted for estimation. First, compute the ψ value by using the known information which includes α, β, v_{1}=k-1 and v_{2}=∞ or by checking the ψ table, and apply it to formula (1) in order to get n(1); then compute the ψ value by using the known information including v_{1}=k-1 and v_{2}=k(n_{(1)}-1) or by checking the ψ table and apply it to formula (1) in order to get n_{(2)}, and so on... The computation stops when the nearest two results are stable, and this is the required sample size. 1.1.2 Example 1 Three methods were adopted in treatment of patients suffering from poststroke depression. The neurological rehabilitation status was observed. The means of the Scandinavian Stroke Scale (SSS) score after treatment by the three methods were estimated to be 11.0, 10.0 and 9.0, respectively, and the standard deviations were 3.0, 3.0 and 2.0, respectively. If the difference of the three methods was required to be statistically significant, how many patients were needed for each group (α and β were set to be 0.05 and 0.10, respectively)? Analysis: Example 1 deals with the sample size estimation of the population mean comparison for quantitative data with the design of one factor with three levels. Based on the given information, we can calculate that =(11.0+10.0+9.0)/3=10.0, )^{2}=(11.0-10.0)^{2}+(10.0-10.0)^{2}+(9.0-10.0)^{2}=2.0. The given information also includes v_{1}=3－1=2, v_{2}=∞, α=0.05 and β=0.10, thus we know that ψ_{0.05,0.10,2,∞}=2.52 by checking the ψ table, and n_{(1)}=(2.52)^{2}×(22.0/3)/［2.0/(3-1)］≈46.6. Let n_{(1)}=47, based on the known information v_{1}=3－1=2, v_{2}=3×(47-1)=138, α=0.05 and β=0.10, we get ψ_{0.05,0.10,2,138}≈ψ_{0.05,0.10,2,120}＝2.55 and n_{(2)}=(2.55)^{2}×(22.0/3)/［2/(3-1)］≈47.7. Let n_{(2)}=48, based on the known information v_{1}=3-1=2, v_{2}=3×(48-1)=141, α=0.05 and β=0.10, we get ψ_{0.05,0.10,2,141}≈ψ_{0.05,0.10,2,120}=2.55 and n_{(3)}=(2.55)^{2}×(22.0/3)/［2/(3-1)］≈47.7. Let n_{(3)}=48, and n=48 is the finial sample size; that is, each group needs 48 patients, and the total sample size would be 144. Besides, the following two SAS programs can be applied to estimate the sample size. Bellow is the first SAS program:

data example_1; input method$ score CellWgt; datalines; A 11.0 1 B 10.0 1 C9.0 1 ; run; ods html; PROC GLMPOWER data=example_1; class method; model score=method; weight CellWgt; contrast "A vs. B" method1 -10; contrast "A vs. C" method10 -1; contrast "B vs. C" method01 -1; POWER stddev = 2.7 alpha= 0.05 ntotal = . power= 0.90; run; ods html close; quit;

Program explanation: In the beginning of the program, the estimated population means and the sample size ratio of the three groups are specified. The option “stddev = 2.7” in the glmpower procedure refers to the mean of the estimated population standard deviations of the three groups. We can also input the different estimated population standard deviations by changing the option into “stddev = 2.0 2.5 3.0” or “stddev = 2.0 to 3.0 by 0.5”. The option “alpha = 0.05” specifies the value of α; “ntotal = .” requires to estimate the population sample size; “power = 0.90” specifies the testing power 1-β to be 0.90. The parameter values in the program can be altered in similar situations. Output and explanation:

Fixed scenario elements

Dependent variable

Score

Weight variable

CellWgt

Alpha

0.05

Error standard deviation

2.7

Nominal power

0.9

Computed N Total

Index

Type

Source

Test degree of freedom

Error degree of freedom

Actual power

N total

1

Effect

method

2

141

0.906

144

2

Contrast

A vs. B

1

459

0.900

462

3

Contrast

A vs. C

1

114

0.900

117

4

Contrast

B vs. C

1

459

0.900

462

The above results show that if the testing power to find out that the population means of the SSS score of the three groups are unequal and required to be 90%, 48 patients are needed for each group, namely, a total of 144 patients are needed. Furthermore, if the testing power to find out that the population means of the SSS score of each two groups are unequal and required to be 90%, 154 patients are needed for each group, that is, a total of 462 patients are needed. The other SAS program is as follows:

Program explanation: The option “groupmeans =11.0|10.0|9.0” specifies the estimated population means of the three groups; “stddev = 2.7” specifies the estimated population standard deviations of the three groups; “groupweights = (1 1 1)” specifies the sample size ratio of the three groups. The parameter values can be modified in similar situations. Output and explanation:

Overall F test for one-way ANOVA

Fixed Scenario Elements

Method

Exact

Alpha

0.05

Group means

11 10 9

Standard deviation

2.7

Group weights

1 1 1

Nominal power

0.9

Computed N Total

Actual power

N total

0.906

144

The results show that if the testing power to find out that the population means of the SSS score of the three groups are unequal and required to be 90%, 48 patients are needed for each group, that is, a total of 144 patients are needed. 1.2 Estimation of testing power The following example is used to demonstrate the estimation of testing power for quantitative data with the design of one factor with k levels (k≥3). Example 2 A researcher expected to examine the difference of the serum soluble CD8 antigen level (U/mL) between patients suffering from mild and severe aplastic anemia and healthy people in order to reflect the extent of hematopoietic dysfunction caused by the disorder of the immune status. The researcher selected 10 subjects from the three types of population respectively (patients suffering from mild aplastic anemia, patients suffering from severe aplastic anemia and healthy people), and examined the CD8 antigen level. Below was the result (x±s): the healthy group 290±174, the mild group 658±155 and the severe group 763±127. The researcher analyzed the data by adopting the analysis of variance of quantitative data with the design of one factor with three levels. Estimate the testing power. Analysis: Example 2 involves the testing power estimation for quantitative data with the design of one factor with three levels. The following SAS program can be applied.

Program explanation: The option “groupmeans =290| 658 | 763” specifies the estimated population mean of each group; “stddev = 174 155 127”specifies the estimated population standard deviation of each group; “groupweights = (1 1 1)” specifies the sample size ratio of each group. Output and explanation:

Overall F test for one-way ANOVA

Fixed scenario elements

Method

Exact

Alpha

0.05

Group means

290 658 763

Group weights

1 1 1

Total sample size

30

Computed power

Index

Standard deviation

Power

1

174

>0.999

2

155

>0.999

3

127

>0.999

The result shows that the testing power was >0.999.

2 Qualitative data with the design of one factor with k levels (k≥3) having a binary response variable

2.1 Sample size estimation of multirate comparison 2.1.1 Formula

In formula (2), n refers to the sample size of each group, which is required to be equal; p_{max} and p_{min} stand for the maximum rate and the minimum rate. When the difference of p_{max} and p_{min} (p_{d}) is known, p_{max}=0.5+p_{d}/2, p_{min}=0.5-p_{d}/2; λ is the non-central parameter δχ^{2} of the non-central χ^{2} distribution under certain conditions (df=k-1), which can be computed by SAS function or by checking the λ table in reference books. k stands for the number of the groups. 2.1.2 Example 3 Three drugs used to expel the intestinal worm parasite were applied to a pilot test. The stool examination showed that the egg negative conversion rates of drug A, drug B and drug C were 80%, 85% and 95%, respectively. How many patients were needed for each group in the clinical trial? Analysis: Example 3 deals with the sample size estimation of multirate comparison. The needed SAS program is as follows: Program explanation: In the first line, “alpha=0.05” specifies the probability of making type Ⅰ error; “beta=0.10” specifies the probability of making type Ⅱ error; “k=3” specifies the number of the groups. In the second line, “p=0.80,0.85,0.95” specifies the egg negative conversion rate of each group. The value of λ can be obtained by checking the λ table or by invoking the SAS function. The values of the parameters can be altered in similar situations.

%let alpha=0.05; %let beta=0.10; %let k=3; %let p=0.80,0.85,0.95; data example_3; p1=max(&p); p2=min(&p); lamda=CNONCT(CINV(1-&alpha,&k-1),&k-1,&beta); n=ceil(lamda/(2*(arsin(sqrt(p1))-arsin(sqrt(p2)))**2)); file print; PUT #3 @10' ' n 'patients were needed for each group. '; run;

The result: 112 patients are needed for each group. 2.2 Testing power estimation of multirate comparison 2.2.1 Formula First, compute the value of λ by formula (2), then invoke the SAS function to solve the value of β in function λ=CNONCT(CINV(1-α, k-1), k-1, β) and at last compute the testing power (power=1-β). 2.2.2 Example 4 In example 3, suppose there are overall 300 patients in the clinical trial and each group has 100 patients. Estimate the testing power. Analysis: Example 4 deals with the testing power estimation of multirate comparison. The needed SAS program is as follows:

%let alpha=0.05; %let k=3; %let n=100; %let p=0.80,0.85,0.95; data example_4; p1=max(&p); p2=min(&p); lamda=&n*2*(arsin(sqrt(p1))-arsin(sqrt(p2)))**2; do beta=0.001 to 0.999 by 0.001; lamda_1=CNONCT(CINV(1-&alpha,&k-1),&k-1,beta); output; if abs(lamda_1-lamda)/lamda<=0.01 then goto ok; end; ok:power=1-beta; file print; PUT #3 @10'The testing power reaches ' power '。'; run;

Program explanation: The value of beta in the statement “do…, …end” can not be 0 or 1, otherwise, it will be unable to calculate the value of lamda_1, which may lead to a wrong result. The statement “if abs(lamda_1-lamda)/lamda<=0.01 then goto ok;” requires to set a reasonable error (for instance, the maximum error is set to be 1%), otherwise the result may also turn out to be wrong. Output: The testing power reaches 0.867.

References

1.

Hu LP. The application of statistical triple-type theory in experimental designs[M]. Beijing: People’s Military Medical press, 2006. 57, 222-224, 318-319. Chinese.

2.

Hu LP. SAS research design and statistical analysis[M]. Beijing: People’s Medical Publishing House, 2010. 200-202, 208-209. Chinese.

3.

Hu LP, Gao H. Guidance of statistical analysis by SAS[M]. Beijing: Publishing House of Electronics Industry, 2010. 700-701. Chinese.

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