![]() TitleString = sprintf('Condition %i\n p-value of %0.2f',k,PValues(k)) The p-value is given together with h, which tells you whether the null hypothesis is rejected (value of 0) or not (value of 1). % Group data for easy referencing in plots It's quite easy to compute: Without much information about your data I re-arranged them into single row vectors for comparisons.Ĭond2 = Ĭond3 = It looks like you want to perform 2 sample (paired) t-test, in which case you want to use the ttest2 function. My question is how to do T-test for the fMRI data? H1: Condition1 ≠ Condition2Īnd should I compute based on these:1.Difference between the mean intensities of each conditionĢ -1 3 -1 -1 -1 -2 1 2 -3 -> under class 1 stimulus I want to test difference in signal between two conditions(class 1 stimulus vs rest condition), (class 2 stimulus vs rest condition) and (class 3 stimulus vs rest condition). The first two rows are under class 1 stimulus the next two rows are under class 2 stimulus, the next next two rows are under class 3 stimulus, the last three rows are under no stimulus(rest condition). The significance is 0.0017, which means that by chance we would have observed values of t more extreme than the one in this example in only 17 of 10,000 similar experiments! A 95% confidence interval on the mean is, which includes the theoretical (and hypothesized) difference of -0.5.I have a fMRI data matrix, the size of which is 9*10 (I randomly put the value in it). The result h = 1 means that we can reject the null hypothesis. Notice that the true difference is only one half of the standard deviation of the individual observations, so we are trying to detect a signal that is only one half the size of the inherent noise in the process. We test the hypothesis that there is no true difference between the two means. The observed means and standard deviations are different from their theoretical values, of course. We then generate 100 more normal random numbers with theoretical mean 1/2 and standard deviation 1. This example generates 100 normal random numbers with theoretical mean 0 and standard deviation 1. tail = 0 specifies the alternative (default).The tstat element is the value of the t statistic, and df is its degree of freedom.Īllows specification of one- or two-tailed tests, where tail is a flag that specifies one of three alternative hypotheses: ![]() Returns a structure stats with two elments, tstat and df. ci in this case is a 100(1-alpha)% confidence interval for the true difference in means. For example if alpha = 0.01, and the result, h, is 1, you can reject the null hypothesis at the significance level 0.01. Gives control of the significance level alpha. significance is the probability that the observed value of T could be as large or larger by chance under the null hypothesis that the mean of x is equal to the mean of y.Ĭi is a 95% confidence interval for the true difference in means. Where s is the pooled sample standard deviation and n and m are the numbers of observations in the x and y samples. ![]() The significance is the p-value associated with the T-statistic ![]() The result, h, is 1 if you can reject the null hypothesis at the 0.05 significance level alpha and 0 otherwise. Performs a t-test to determine whether two samples from a normal distribution (in x and y) could have the same mean when the standard deviations are unknown but assumed equal. Hypothesis testing for the difference in means of two samples Ttest2 (Statistics Toolbox) Statistics Toolbox
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