![]() ![]() in these results, the degrees of freedom (df) is 4. minitab uses the degrees of freedom to determine the p value associated with the test statistic. ![]() thus, the engineer concludes that the variables are associated and that the performance of the presses varies depending on the shift. both p values are less than the significance level of 0.05. ![]() For this data, the pearson chi square statistic is 11.788 (p value = 0.019) and the likelihood ratio chi square statistic is 11.816 (p value = 0.019). therefore, at a significance level of 0.05, you can conclude that the association between the variables. the likelihood chi square statistic is 11.816 and the p value = 0.019. in these results, the pearson chi square statistic is 11.788 and the p value = 0.019. Key results: p value for pearson chi square, p value for likelihood ratio chi square. ![]()
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