Our data supports the claim that the number of clubs a student is a member of has a negative effect on his or her hours of free time. Therefore, we REJECT the null hypothesis. I calculated a p-value of 0.006, which is less than the standard assumed significance level of 0.05. In the tcdf function of your calculator, input your lower limit (-999), your upper limit (-2.72), and the degrees of freedom (24). This means that you have to divide the given p-value by 2 → \(0.012 \div 2 = 0.006\) What are these karate moves T-tests, ANOVAs, Factorial ANOVAs, regression, multiple regression, factor analysis, structural equation modeling, mixed models. However, we want to conduct a ONE-sided test. Remember that the p-value displayed on the table is for a TWO-sided test.
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