![]() P-value > α : Cannot conclude the data do not follow a normal distribution (Fail to reject H 0) If the p-value is larger than the significance level, the decision is to fail to reject the null hypothesis because there is not enough evidence to conclude that your data do not follow a normal distribution. The data considerations topics might identify a different analysis that you can use. The analysis does not work well with nonnormal data.The analysis works well with nonnormal data that were transformed into normal data.The analysis works well with nonnormal data.The data considerations topics indicate one the following: If your data do not follow the normal distribution, read the data considerations topic for any other analyses that you want to perform. P-value ≤ α: The data do not follow a normal distribution (Reject H 0) If the p-value is less than or equal to the significance level, the decision is to reject the null hypothesis and conclude that your data do not follow a normal distribution. A significance level of 0.05 indicates a 5% risk of concluding that the data do not follow a normal distribution when they actually do follow a normal distribution. Usually, a significance level (denoted as α or alpha) of 0.05 works well. For example, the p-value/2=0.0141/2=0.007 < 0.05, hence the concluse for the one side test is to reject the hypothesis and therefore the new method improve the grading score.To determine whether the data do not follow a normal distribution, compare the p-value to the significance level. If one wants to test whether one group is greater(smaller) than the other, p-value can be divided by 2. Note that SAS perform a two-sided test, meaning the hypothesis is to compare a significant difference between two groups. The conclusion is to reject the null hypothesis and that the the reading grade of two methods are significantly different. For the hypothesis of comparing control and treatment, t-value=-3.83, and the p-value is 0.0141.For confidence interval for control-treatment = (-21.9317 -4.0683).In this example, the p-value = 0.0318 < 0.05, so we should read the "Satterthwaite" section. When the p-value (shown under "Pr>F") is no more than 0.05, then the variances are unqueal then read the "Satterthwaite" section of the result.When the p-value (shown under "Pr>F") is greater than 0.05, then the variances are equal then read the "Pooled" section of the result.The test on Equality of Variances is given at the end, and is repeated below, Note that the results show both "Pooled" and "Satterthwaite" sections, which is based on sample variances check. Proc ttest data=read sides=2 alpha=0.05 h0=0 Note that the test is two-sided (sides=2), the significance level is 0.05, and the test is to compare the difference between two means (mu1 - mu2) against 0 (h0=0). In this demo example, two samples (control and treatment) are independent, and pass the Normality check. Two dependent samples and does not follow Normal distribution, suggest Signed Rank test.Two independent samples and does not follow Normal distribution, suggest WMW test.Two dependent samples and follow Normal distribution, suggest Paired T-test.When the assumptions are not met, other methods are possible based on the two samples: The two samples follow normal distributions, and can be done with Normality check.There is one continuous dependent variable and one categorical independent variable (with 2 levels).Infile "H:\sas\data\reading.csv" dlm=',' firstobs=2 Grade (continuous) ~ method (categorical: 2 levels) The problem is to test whether the two methods make a difference? The model you can set up for this problem is After they are trained with the method, their performance is measured as grades. Users will be randomly assigned either one method. There is a new method (treament, or t) and a standard method (control, or c). Suppose there is a study to compare two study methods and see how they improve the grades differently. One variable to be measured and compared between two conditions (samples). ![]() A common experiment design is to have a test and control conditions and then randomly assign a subject into either one. The idea of two sample t-test is to compare two population averages by comparing two independent samples. Compare two independent samples with t-test. ![]()
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