Chi-squared values

In the case of the 4x3 contingency Table 32.4 we obtain a chi-square value of... 

If the normal approximation to the binomial distribution is valid (that is, not more than 20% of expected cell counts are less than 5) for drug therapy and symptom of headache, then you can use the Pearson chi-square test to test for a difference in proportions. To get the Pearson chi-square / -value for the preceding 2x2 table, you run SAS code like the following ... 

Bartlett s is based on the calculation of the corrected y2 (chi-square) value by the formula... 

Bartlett s is very sensitive to departures from normality. As a result, a finding of a significant chi square value in Bartlett s may indicate nonnormality rather than heteroscedasticity. Such a finding can be brought about by outliers, and the sensitivity to such erroneous findings is extreme with small sample sizes. 

When k = 2, the Kruskal-Wallis chi-square value has 1 df This test is identical to the normal approximation used for the Wilcoxon Rank-Sum Test. As noted in previous sections, a chi square with 1 df can be represented by the square of a standardized normal random variable. In the case oik = 2, the //-statistic is the square of the Wilcoxon Rank-Sum Z-test (without the continuity correction). 

The commercially available software (Maximum Entropy Data Consultant Ltd, Cambridge, UK) allows reconstruction of the distribution a.(z) (or f(z)) which has the maximal entropy S subject to the constraint of the chi-squared value. The quantified version of this software has a full Bayesian approach and includes a precise statement of the accuracy of quantities of interest, i.e. position, surface and broadness of peaks in the distribution. The distributions are recovered by using an automatic stopping criterion for successive iterates, which is based on a Gaussian approximation of the likelihood. 

Since the error measure is greater than the chi square value, the measurements are not acceptable. According to the Almoey indicators, the variable X2 is most likely tD tie corrupted by gross error. 

The log-likelihood function at the maximum likelihood estimates is -28.993171. For the model with only a constant term, the value is -31.19884. The t statistic for testing the hypothesis that (3 equals zero is 5.16577/2.51307 = 2.056. This is a bit larger than the critical value of 1.96, though our use of the asymptotic distribution for a sample of 10 observations might be a bit optimistic. The chi squared value for the likelihood ratio test is 4.411, which is larger than the 95% critical value of 3.84, so the hypothesis that 3 equals zero is rejected on the basis of these two tests. 

The mathematical basis of the test includes an assumption that the y2 values are continuous . In other words, they could take any value. The reality, however, is that, when we count canisters (or any other set of discrete items), the results are discontinuous - we may observe 1, 2 or 3 canisters, etc., but not a fractional value. The subsequent chi-square values are therefore also discontinuous - some values of X2 could never arise because they do not match any outcome based upon whole numbers of canisters. This mis-match between the assumptions made by the test and the reality of the data introduces a bias that may inflate the y2 value and make the data look a little more significant that it really is. 

With four possible categories of numbers and with the total fixed, there are three degrees of freedom This corresponds to a probability level of approximately 0.25 and indicates that, if all the pumps were operating under similar conditions, a chi-squared value as large as 5.4 would be expected one time in four by chance alone. Thus there is a 25 percent chance that the supervisor could be wrong in the analysis of the operation of pump 1. 

This chi-squared value, at three degrees of freedom, is between the 0.02 and 0.01 level and there is approximately only 1 chance in 75 of obtaining this distribution if the pumps were in fact operating the same. 

The judgment may be based on the reduced chi-square values l for the fits with the two respective models. However, even if fitting with the two models involves the same number of degrees of freedom v, it should not be said that model 2 is significantly preferable to model 1 just because xl for model 2 is smaller than l for model 1. One must answer the question Is the difference significant Here we need to look at the probability distribution for the ratio of reduced chi squares. Such a ratio, should conform to a... 

Obtain the reduced chi-square value xl given by Eq. (31b) and carry out the appropriate statistical tests for goodness of fit, including inspection of the weighted residuals for systematic trends (see Goodness of Fit). 

DF Chi-square values for one-sided test t Values for one-sided test ... 

Fluorescence decay profiles are analysed as a sum of up to three exponentials by an iterative reconvolution procedure which has been described elsewhere (17-19) and is based on the Marquardt algoritlim. Goodness of fit is judged by inspection of the weighted residuals, autocorrelation function of the weighted residuals, reduced chi-square value and the Durbin-Watson parameter. 

Step 1. Using n — 1 as the number of degrees of freedom, look up, in the chi-square (x2) tables, that chi-square value which is (1) so small that anything less than it would occur by chance less than 2.5% of the time (xi2), or (2) so large that anything greater than it would occur by chance less than 2.5% of the time (X22). These would be values for P = 0.025 and P - 0.975, respectively. The values of chi-square between these two limits would occur by chance 95% of the time. (Note Since in the chi-square tables the nearest columns to the desired P = 0.975 and P = 0.025 may be P = 0.98 and P = 0.02, it may be more convenient to use 96 % confidence limits to avoid the necessity for extrapolation.)... 

Graphs of s vs. number of revolutions are plotted. Also, the use of the usual chi-square test is illustrated for comparing distributions to determine whether a batch is randomly mixed. In addition, a table is given for classifying mixtures according to the relative frequency of occurrence of chi-square values. 

The authors mention the fact that probability can be taken as a measure of the quality of blending. This is somewhat analogous to the Weiden-baum and Bonilla method for classifying the quality of a mixture in accordance with the probability of occurrence by chance of the chi-square value that is obtained when the mixture spot sample distribution is compared with the theoretical normal distribution (W2). 

Select the computed chi-squared value, SRR Dim myRange2 As Range... 

This chi-square value does not lie in the critical region. Therefore, at a confidence level of 95%, the variance does exceed 0.8 cSF. Therefore, the null hypothesis is rejected and the alternative hypothesis is accepted. The measurements should be repeated in order to reduce the measurement variability. 

Ln(X) is greater than or equal to the chi-squared value associated f-r degrees of freedom and critical value a (Bain and Engelhardt, 1987). If LLf and LLr are the log-likelihoods for the full and reduced models, respectively, then an alternative form of the likelihood ratio test is given by... 

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