normal 8 normal 8 normal 8 normal 8 normal 8 normal 8 normal 8 normal 8 4.5 7.2 3.4 9.1 1.2 1.33 0.98 1.03 0.78 0.56 2 The table shows the means and variances from 5 experimental conditions. Compute variance of the means. 9.717 0.01 showCorrectAnswer 2 Each mean is based on n=5 observations. Compute the MSB based on the variance of the means (These are the same values as previously shown). 48.585 0.01 Multiply the variance of the means by n=5. showCorrectAnswer 2 Find the MSE by computing the mean of the variances. .936 0.01 showCorrectAnswer Which best describes the assumption of homogeneity of variance? Homogeneity of variance is the assumption that the variances in the populatons are equal. false The populations are both normally distributed to the same degree. false The between and within population variances are approximately the same. true the variances in the populatons are equal. When performing a one factor ANOVA (between subjects) it is important that each subject only provide a single value. If a subject were to provide more than one value the independence of each value would be lost and the test provided by an ANOVA not be valid. True. When a subjects provides more than one data point the values are not independent therefore violating one of the assumptions of between-subjects ANOVA. True True false False If the MSE and MSB are approximately the same, it is highly likely that population means are different. False If the null hypothesis that all the population means are equal is true then both MSB and MSE estimate the same quantity. false True true False You want to make a strong case that the different groups you have tested come from populations with different means. Your case is strongest: When the population means differ, MSB estimates a quantity larger than does MSE. A high ratio of MSB to MSE is evidence that the population means are different. false MSE/MSB is high. false MSE/MSB = 1. false MSB/MSE is low. true MSB/MSE is high Why can't an F ratio be below 0? F is defined as MSB/MSE. Since both MSB and MSE are variances and negative variance is impossible, an F score can never be negative. true Neither MSB nor MSE can ever be a negative value. false MSB is never less than 1 false MSE is never less than 1 Consider an experiment in which there are 7 groups and within each group there are 15 participants. What are the degrees of freedom for the numerator (between)? k-1 = 7-1 = 6 6 0.0001 Consider an experiment in which there are 7 groups and within each group there are 15 participants. What are the degrees of freedom for the denominator (within)? N-k = 105-7 = 98 98 0.0001 The F distribution has a: true positive skew false no skew false negaative skew The F distribution has a long tail to the right which means it has a positive skew. An independent groups t test with 12 degrees of freedom was conducted and the value of t was 2.5. What would the F be in a one-factor ANOVA? 6.25 0.005 F equals t squared. These values are from three independent groups. What is the F in a one-way ANOVA? (If you are using a program, make sure to reformat the data as described.) 0 3 0.005 showCorrectAnswer If the sum of squares total were 100 and the sum of squares condition were 80, what would the sum of squares error be? 20 0.001 Sum of squares total equals sum of squares condition + sum of squares error. If the sum of squares total were 100, the sum of squares condition were 80 in an experiment with 3 groups and 8 subjects per group, what would the F ratio be? 420.005 Divide sums of squares by degrees of freedom to get mean squares. Then divide MSB by MSE to get F which equals 42.