Hypothesis tests single population means

One-Way Analysis of Variance n A hypothesis test used to test the null hypothesis that the means of two or more samples are equal. The samples must be independent, the populations from which the samples were obtained are assumed to be normally distributed with equal variances, and there is a single factor with M levels of classification or treatment used for classification or treatment used for classification of each sample. There are thus M samples, Yi, Y2,. ..,Ym, each with Nj elements, yji, yj2, ->yjNj- In n simple linear model, yji would be given by ... 

During an experiment one has a set of data which seems to be formed from two groups. Is it wise to calculate a single mean of all data because difference between means of subpopulations results from simple fluctuations, or have we to form two subpopulations with significantly different means The hypothesis that all data are Null hypothesis parts of a sole population is the so-called null hypothesis. The rejection of the null hypothesis leads to establishing a significant difference between both groups of data. The t-test is used to decide this question. 

Assuming a normal multivariate distribution, with the same covariance matrices, in each of the populations, (X, X2,..., Xp) V(7t , 5), the multivariate analysis of variance MANOVA) for a single factor with k levels (extension of the single factor ANOVA to the case of p variables), permits the equality of the k mean vectors in p variables to be tested Hq = jl = 7 2 = = where ft. = fl, fif,..., fVp) is the mean vector of p variables in population Wi. The statistic used in the comparison is the A of Wilks, the value of which can be estimated by another statistic with F-distribution. If the calculated value is greater than the tabulated value, the null hypothesis for equality of the k mean vectors must be rejected. To establish whether the variables can distinguish each pair of groups a statistic is used with the F-distribution with p and n — p — k + i df, based on the square of Mahalanobis distance between the centroids, that permits the equality of the pairs of mean vectors to be compared Hq = jti = ft j) (Aflfl and Azen 1979 Marti n-Alvarez 2000). 

The assessment of bioequivalence is based on 90% confidence intervals for the ratio of the population geometric means (test/reference) for the parameters under consideration. This method is equivalent to two one-sided tests with the null hypothesis of bio-inequivalence at the 5% significance level. Two products are declared bioequivalent if upper and lower limits of the confidence interval of the mean (median) of log-transformed AUC and Cmax each fall within the a priori bioequivalence intervals 0.80-1.25. It is then assumed that both rate (represented by Cmax) and extent (represented by AUC) of absorption are essentially similar. Cmax is less robust than AUC, as it is a single-point estimate. Moreover, Cmax is determined by the elimination as well as the absorption rate (Table 2.1). Because the variability (inter- and intra-animal) of Cmax is commonly greater than that of AUC, some authorities have allowed wider confidence intervals (e.g., 0.70-1.43) for log-transformed Cmax, provided this is specified and justified in the study protocol. 

---