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➡ Click here: Chi square vs t test
The above presents the data in an unambigous manner. If your table is larger, try the free demos of basic statistics only and statistics, nonlinear regression and scientific graphics. The test for whether the slope of regression line is in significant difference from zero.
In the 2 x 2 case of the chi-square test of independence, expected frequencies less than 5 are between considered acceptable if Yates' correction is employed. That is, one must specify that Group1 - the experimental group in this case - is coded as 1, and Group2 - the control group in this case - is coded as 2. Test statistics that follow a chi-squared distribution arise from an pan of independent normally distributed data, which is valid in many cases due to the. In some of these cases McNemar's test for significance of change would have been more appropriate. Null: The means of the two groups are not significantly different. It tests the frequencies of one prime for different values of another nominal variable.
As a result an alternative critical ratio test was devised that gives identical results to the confidence interval. So for the proportion, for example person 1 had. In a chi-squared test, you draw a table of your observed frequencies and your predicted frequencies and calculate the chi-squared value.
Different statistical tests - If there is a theoretical reason for doing so, the following table will allow you to enter your own E ij's.
The type of data you are dealing with will determine the best statistical test to use Chi-squared test The chi-squared test is used with categorical data to see whether any difference in frequencies between your sets of results is due to chance. For example, a ladybird lays a clutch of eggs. You expect that all of the clutch will hatch, but only three-quarters of them do. Is the failure of some of the clutch to hatch statistically significant, and if it is, what could be the reason for it? In a chi-squared test, you draw a table of your observed frequencies and your predicted frequencies and calculate the chi-squared value. You compare this to the critical value to see whether the difference between them is likely to have occurred by chance. If your calculated value is bigger than the critical value, you reject your null hypothesis. T-test The t-test enables you to see whether two samples are different when you have data that are continuous and normally distributed. The test allows you to compare the means and standard deviations of the two groups to see whether there is a statistically significant difference between them. For example, you could test the heights of the members of two different biology classes. Mann—Whitney U-test The Mann—Whitney U-test is similar to the t-test. It is used when comparing ordinal data ie data that can be ranked or has some sort of rating scale that are not normally distributed. Measurements must be categorical — for instance, yes or no — and independent of each other eg a single person cannot be represented twice. For example, the Mann—Whitney U-test could be used to test the effectiveness of an antihistamine tablet compared to a spray in a group of people with hay fever. To do this, you would split the group in half, then give each half a different treatment and ask each person how effective they thought it was. The test could be used to see whether there is a difference in the perceived efficacy of the two treatments. Standard error and 95 per cent confidence limits The standard error and 95 per cent confidence limits allow us to gauge how representative of the real world population the data are. If there is a statistically significant relationship, you can reject the null hypothesis, which may be that there is no link between the two variables. Wilcoxon matched pairs test Like the Mann—Whitney U-test, this test is used for discontinuous data that are not normally distributed but do have a link between the two datasets. For example, when asking people to rank how hungry they feel before a meal and doing so again after they have eaten — because the same person is providing both answers, the datasets are not independent.