Research Article 
Open Access 

Oyeka ICA^{1} and Okeh UM^{2*} 
^{1}Department of Applied Statistics, Nnamdi Azikiwe University, Awka, Nigeria 
^{2}Department of Industrial Mathematics and Applied Statistics, Ebonyi State University, Abakaliki, Nigeria 
^{*}Corresponding author: 
Okeh UM
Department of Industrial Mathematics and Applied Statistics
Ebonyi State University Abakaliki, Nigeria
Email: [email protected] 


Received February 05, 2013; Published March 22, 2013 

Citation: Oyeka ICA, Okeh UM (2013) Statistical Analysis of Response from One Period Cross Over Design in Clinical Trial. 2:643 doi:10.4172/scientificreports.643 

Copyright: © 2013 Oyeka ICA, et al. This is an openaccess article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited. 

Abstract 

This paper proposes and presents a chisquare statistical method for the analysis of response from one period cross over design for two sample data in which the sampled populations may be measurements that are numeric (assuming real values) and nonnumeric assuming only values on the nominal scale. Test statistics are developed for testing the null hypothesis that subjects who receive each of the treatments first do not differ in their response as well as the null hypothesis that subjects exposed to one of the treatment or experimental conditions first do not on the average differ in their responses with those exposed to the other treatment or experimental condition first. Estimates of the proportions responding positive; experiencing no change in response or responding negative are provided for subjects exposed to each treatment first as well as for the two treatments together. The proposed method which is illustrated with some sample data can be used with either numeric or nonnumeric data and is shown to be at least as powerful as the traditional two sample (small) ttest. 

Keywords 

Cross over; Treatment; Chisquare; Design; Patients 

Introduction 

Suppose subjects for a clinical trial are first matched on characteristics associated with the outcome understudy such as a disease and randomly assigned the treatments T1 and T2. In particular, suppose as in a cross over design, each subject serves as his own control, that is, each patient receives each treatment. One half of the sample of 2n patients or subjects is randomly selected to be given the two treatments in one order and the other half to be given the treatments in the reversed order. That is n of the random sample of the 2n patients or subjects is given treatment, T1 first and treatment T2 later and the remaining n subjects is given treatment T2 first and treatment T1 later. A number of factors must be guarded against in analyzing the data from such studies. However, the order in which the treatments are given may affect the response [1]. A test that is valid when order effects are present has been described [2]. Another factor to be guarded against is the possibility that a treatment’s effectiveness may be long lasting and hence may affect the response to the treatment given after it. When this socalled carry over effect operates and when it is unequal for the two treatments, then for comparing their effectiveness, only the data from the first period may be used [3]. Specifically, the responses by the subjects given one of the treatments first must be compared with the responses by the subjects given the other treatment first. In this paper we present a method for analyzing data from a crossover design in which each subjects serves as his own control and analysis is based on responses by patients given one of the treatments first and responses by patients given the other treatment first. Here allowance is made for the possibility that patients or subjects may die or drop out of the study. 

The Proposed Method 

In general, let n_{j} subjects or patients be randomly assigned for treatment with T_{j} first: for j=1,2 when n_{1} and n_{2} are not necessarily equal. Let y_{ij} the beresponse by the i^{th} subject administered treatment T _{j} first for i=1,2,…, n_{j}, j=1,2. 

Two possibilities present themselves here namely: y_{ij} may be numeric assuming real values or it may be nonnumeric assuming only values on the nominal scale of measurement. If the test score y_{ij} is the numeric, assuming responses or values in the range (c_{1},c_{2}) where c_{1} and c_{2} are real numbers (c_{1}<c_{2}) that indicate that the subject test normal, condition of interest absent, response is negative, etc. Values of y_{ij} that are less than or equal to c_{1} and values that are greater than or equal to c_{2} indicate the opposite conclusion; i.e., the patient tests are positive, the condition is present, response is abnormal, there is no improvement, etc. If the response y_{ij} are on the nominal scale of measurement then y _{ij} may assume values such are positive, nondefinitive or negative: present, non definitive or absent; yes, nondefinitive or no, etc. 

If y_{ij} is numeric, let 

(1) 

for i=1,2,...,n_{j}; j=1,2 

If y_{ij} is nonnumeric but assumes values on a nominal scale of measurement, let 

(2) 

for i=1,2,...,n_{j}; j=1,2 

Note that by specification allowance has been made for the possibility that patients or subjects may drop out that is, they are lost to the study. If patients do not drop out of the study then n_{j}=n for j=1,2. 

For both equations 1 and 2, let 

(3) 

Where 

(4) 

(5) 

Now 

(6) 

and 

(7) 

That is 

(8) 

Also, 

That is 

(9) 



Which when simplified and evaluated using equations (8) and (9) yields 

(10) 

Note that the sample estimates of are respectively given as 

(11) 

Where are respectively the numbers of in the frequency distribution of these values in U_{ij}. The sample estimate of the difference between and namely
is 

(12) 

Where i=1,2,...,n_{j}; j=1,2 

or equivalently using equation (12), we have that the estimated variance of W_{j} is from equations 10 and 11 

(13) 

(14) 

The null hypothesis that the subjects or patients who take treatment T_{j} first are as likely to test positive (abnormal, yes) do not differ in their response which is as negative (normal, no) that is equivalent to testing of null hypothesis 



vs 

(15) 

(16) 

for j=1,2 has approximately the chisquare distribution with 1 degree of freedom for sufficiently large sample size n_{j}. H_{0j} is rejected at a specified α level of significance otherwise H_{0} is accepted where is obtained from an appropriate chisquare table with 1 degree of freedom at α level of significance. 

Of greater interest however is testing the null hypothesis H_{0} that patients or subjects who take treatment T_{1} first have the same positive response rate as patients who take treatment T_{2} first. This is equivalent to testing the null hypothesis 



vs 

(17) 

The null hypothesis may be tested using the test statistics 

(18) 

Which under H_{0} has a chisquare distribution with 1 degree of freedom for sufficiently large values of n_{1} and n_{2} where 

(19) 

Now 



Now U_{r1}U_{s2} can only assume the values 1,0 and 1. It assumes the value 1 if U_{r1} and U_{s2} both assume the value 1 or both assume the value 1 or both assume the value 1 with probability It assumes the value 0 if U_{r1} and U_{s2} both assume the value 0 or U_{r1} assume the value 0, no matter the value assumed by U_{s2} or U_{s2} assumes the value 0 no matter the value assumed by U_{r1} with probability. 

U_{r1}U_{s2} assumes the value 1 if U_{r1} assumes the value 1 and U_{s2} assumes the value 1 or vice versa with probability 

Hence using these values, evaluating and simplifying we have 



Using these values in Equation 19 with Equation 10, we have that 



(20) 

Or equivalently 

(21) 

Therefore, the test statistic of equation 18 may be written as 

(22) 

which has a chisquare distribution with 1 degree of freedom for sufficiently large n_{1} and n_{2}. The null hypothesis of equation 17 is rejected at the α level of significance if 

Otherwise the null hypothesis is accepted. Also the test statistic of equation 22 may equivalently be expressed in terms of sample proportions as: 

(23) 

If the null hypothesis H_{0} of equation 17 is rejected, then each H_{0j}; for j=1,2 is tested to determine which of the groups treated first with treatment T_{j}; j=1,2 may have led to the rejection of the overall null hypothesis of equation 17. 

Illustrative Example 

A clinician is interested in determining whether or not a certain condition is present in a population. He collected a random sample of ‘n’=34 subjects from this population and exposed each of them to two types of diagnostic procedures T_{1} and T_{2} at two different points in time. A subsample of n_{1}=14 subjects are screened with procedure T_{1} first and the remaining subsample of n_{2}=20 subjects are at the same time screened with procedure T_{2} first. This process is repeated with the same subjects in the reverse order a little while later. The results for the tests administered first on the subjects are as follows where a plus sign (+) indicates conditions present or positive response; a minus sign () indicates condition absent or negative response; and a zero (0) indicates condition indeterminate or nonspecific: 

Test T_{1}: ; +; 0; +; ; 0; 0; +; 0; 0; +; ; +; +; 

Test T_{2}: 0; 0; ; +; +; ; +; 0; +; ; +0; ; 0; +; ; +; +; +; +; 


We here use these data to illustrate the proposed method. 

Results 

Now using equation 1 with the data we have that 

Hence from equation 11, we have that 



W_{1}=63=3; W_{2}=105=5. 

Now from equation 13, we have that the variances of W_{1} and W_{2} are respectively 

Var (W_{1})=14(0.429+0.214(0.4290.214)^{2})=14(0.597)=8.358 

and 

Var (W_{2})=20(0.500+0.250(0.5000.250)^{2})=20(0.687)=13.740. 

The difference between the sample proportions of subjects responding positive and negative when screened with test T_{1} first is with estimated variance 



Similarly, the difference between the proportion of sample subject responding positive and negative when screened with test T_{2} first is 

with estimated variance 



Hence, the difference in the proportions of sample subjects responding positive when screened with test T_{1} first compared with when screened with test T_{2} first is 

with estimated variance 



Now notice that the estimated value of seem to indicate that test T_{2} may have greater tendency of revealing positive responses by subjects more than test T_{1}. To ascertain whether this tendency is statistically significant, we have from equation 23 that 

which with 1 degree of freedom is not 

statistically significant, leading to a nonrejection of the null hypothesis of equation 17. It would be instructive to compare the results obtained using the proposed method with what would have been obtained if the traditional twosample method of analysis had been used with the data. To do this, we would compare the sample proportion of subjects who test positive when screened with test T_{1} first namely with the proportion of subjects who test positive when screened with test T_{2} first namely 

The corresponding Chisquare test statistic is
which with 1 degree of freedom is also not statistically significant again leading to a nonrejection of the null hypothesis. 

Discussions 

However, although the proposed method and the traditional method here both lead to a nonrejection of the null hypothesis, the relative sizes of the corresponding chisquare values nonetheless suggest that the traditional method is likely to lead to an acceptance of the null hypothesis (Type II Error) more frequently and hence is likely to be less powerful than the proposed method. Furthermore, the proposed method unlike the traditional method enables the statistical comparisons of subjects’ responses under each treatment in the event that the overall or initial null hypothesis is rejected. It also enables the simultaneous estimation of the proportions of subjects under each treatment and overall, whose response in the tests is either positive, indeterminate or negative which provide additional useful information for policy purposes. 

Conclusion 

We have here proposed and developed a method for the analysis of data generated from a crossover type study design in which analysis is based only on the sample subjects exposed to the two experimental or treatment conditions first. Test statistics are developed for testing the null hypothesis that subjects who receive each of the treatments first do not differ in their response as well as the null hypothesis that subjects exposed to one of the treatment or experimental conditions first do not on the average differ in their responses with those exposed to the other treatment or experimental condition first. 

Estimates of the proportions responding positive; experiencing no change in response or responding negative are provided for subjects exposed to each treatment first as well as for the two treatments together. 

The proposed method which is illustrated with some sample data can be used with either numeric or nonnumeric data and is shown to be at least as powerful as the traditional two sample small ttest. 


References 

 Meiser P, Free SM, Jackson GL (1958) Remoderation of methodology in studies of pains relief. Biometrics 14: 330342.
 Gart JJ (1969) An exact test for comparing matched proportions in crossover designs. Biometrika 56: 7580.
 Grizzle JE (1965) The twoperiod changeover design and its use in clinical trials. Biometrics 21: 467480.


