Biostatistics. Advantages of a randomized controlled trial over a trial with systematic allocation - Essay Example

?Biostatistics Question Advantages of a randomized controlled trial over a trial with systematic allocation Random controlled trials are experimental set ups with treatment and control groups, and elements of a study’s sample space have an equal probability of selection and allocation to either the treatment of control group. Systematic sampling however follows a defined approach with unequal probability for each sample space elements. One of the advantages of randomized control trials over systematic allocation is its ability to eliminate bias. Researchers and participants in randomized controlled trials lack influence of sample selection and allocation to either of the experimental groups and this eliminates chances of sampling bias. Eliminated bias in the randomized trials has significant effects on research processes such as enhanced reliability and validity which systematic allocation into biased sample may compromise (Miller, Strang and Miller 2010, p. 38). Randomized controlled trials also offer higher probability of homogeneity of background information on research participants within control and treatment groups as compared to systematic allocation that may have a set of background characteristics in one group and generate another group with different background characteristics. Such variations may result from bias, challenge comparability of observations from a study’s group, and identifies systematic allocations with validity and reliability challenges. Randomized controlled trials also have the advantage of probability sampling basis, which offers validity to data analysis, over the non-probability based systematic sampling (Friedman, Furberg and DeMets 2010, p. 71-72). Lack of a standard approach to systematic allocation also identifies benefit of randomized trials because variation in sampling criteria may be formulated to achieve bias and periodic sampling is an example (Miller, Strang and Miller 2010, p. 38). Variable data type and justification Number of adverse events Number of adverse events is a quantitative variable because it assumes numeric values. It can further be classified as discrete quantitative variable because it can only assume positive whole numbers (Weiers 2010, p. 8). Air blast sensitivity Air blast sensitivity, based on applied scale in the study, is a quantitative variable because of its numeric value on an ordinal scale that further classifies it as a discrete quantitative variable (Weiers 2010, p. 8; Neuhaus et al. 2013, p. 352). Tactile hypersensitivity Tactile hypersensitivity, based on the research, is a qualitative variable because it lacks numeric values. Its values are either yes or no and are further subjective, subject to study participants’ judgment (Weiers 2010, p. 8; Neuhaus et al. 2013, p. 352). Age Age is a quantitative variable because it assumes a numeric values and can further be categorized as a continuous quantitative variable because of its scope that can assume unlimited range of non-negative real numbers (Weiers 2010, p. 8). Gender Gender is a qualitative variable because it lacks numerical value (Weiers 2010, p. 8). Question 2 Graphical summary of the mean tactile sensitivity and variability The table bellow summarizes the mean and standard, for variability, for the tactile sensitivity values across the three groups at the end of 28 days. Table 1: Mean and standard deviation after 28 days Treatment Mean Standard deviation test A 21.48 11.86 Test B 20.58 11.32 Control 11.77 5.95 This data can be presented in graphical form as follows. Graph 1: Graph of mean and standard deviation after 28 days Interpretation of the sample mean and standard deviation of each group Mean of test group A indicates that each of the group subjects is expected to have an average score of 21.48. The group’s standard deviation, 11.48, defines the mean deviation of each of the group’s scores from the group’s mean. Mean for the test group B shows that members reported an average score of 20.58 and the standard deviation shows the average degree of variation of individual scores from the group mean, 11.32. Mean of the control group indicates an average score of 11.77 for group members with an average variance of 5.95 from the group mean. Difference in mean tactile sensitivity Given that that: Mean for test group A= 21.48 Standard deviation for test group 1= 11.86 NA= 49 Mean for test group B= 20.58 Standard deviation B= 11.32 NB= 52 Mean for control group= 11.77 Standard deviation for control group= 5.95 NC= 48 Difference between test group A and control group means Difference between means= 21.48-11.77 =9.17 Confidence interval= (MA-MC) ± t(95,0.05)*SD, Where SD is the standard error for the mean difference =9.17±1.9855*(1.93) =9.17± 3.835 (5.335, 13.003) Between test groups B and control group Difference between means= 20.58-11.77 =8.81 Confidence interval= (Mb-MC) ± t(98,0.05)*SD, Where SD is the standard error for the mean difference =8.81±1.999*(1.849) =8.81± 3.696 =(5.114, 12.506) Interpretation of findings Ninety-five percent confidence intervals between test groups A and control group and between test group B and control group do not include zero. This means that the treatment group means are significantly different from control group mean. Alternative analysis of the data The data could also be analyzed using analysis of variance tool instead of the student t distribution. Test of hypothesis on the new set of data Analysis of variance for the data accepts a hypothesis that all the means are equal, p>0.05. No significant difference exists between the group means. (Comrey and Lee 2006, p. 106, 107) Question 3 Given a normal age distribution for enrolling students with a true population mean of 22.5 and standard deviation of 4.5 years Difference between population and sample mean Population mean is a parameter that conveys true values of population mean while sample mean is a statistic for a section of the population and is only an estimator of the true population mean (Rosenthal 2012, p. 164). Most likely age of an enrolling student selected at random The most likely age is 22.5 because mean of a normally distributed data coincides with the mode. Age range for 95 percent of the population 95 percent confidence interval= µ±S*z0.05 =22.5±1.96*4.5 =22.5±8.775 =(13.725, 31.275) Histogram sketch Question 4 Comment on the number of children attending different schools The data suggest that more students attend state schools than independent schools and this contributed to the high probability of state school students. Appropriate test for analyzing the data Analysis of variance’s F test is the appropriate analysis tool (Huitema 2011, p. 180). Analysis The following table summarizes analysis results. ANOVA Source of Variation SS df MS F P-value F crit Between Groups 1472.667 1 1472.667 3.051105 0.155607 7.708647 Within Groups 1930.667 4 482.6667 Total 3403.333 5         Interpretation The high p value, p=0.156>0.05 means that there is to correlation between oral health and social deprivation based on type of school. Works cited Comrey, A and Lee, H 2006, Elementary statistics: A problem solving approach, Lulu.com, Morrisville. Friedman, L Furberg, C and DeMets, D 2010, Fundamentals of clinical trials, Springer, New York. Huitema, B 2011, The analysis of covariance and alternatives: Statistical methods for experiments, quasi-experiments, and single case studies, John Wiley & Sons, Hoboken. Miller, P Strang, J and Miller, P 2010, Addiction to research methods, John Wiley & Sons, Hoboken. Neuhaus, K et al. 2013, ‘Effectiveness of calcium sodium phosposilicate containing prophylaxis paste in reducing dentine hypersensitivity immediately and 4 weeks after a single application: A double-blind randomized controlled trial,’ Journal of Clinical Periodontal, Vol. 2013, Issue. 40, pp. 349-357. Rosenthal, J 2012, Statistics and data interpretation for social work, Springer Publishing Company, New York. Weiers, R 2010, Introduction to business statistics, Cengage Learning, Mason. Read More