I have sample size of 51, number of levels is 71. I used One-way Repeated Measures Anova in SPSS. But, Mauchly's test of sphericity fails to give me a significant value (it gives "." result). Why? Can I still use the result of Greenhouse-Geisser significance test?
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I'm doing a two-way between-within ANOVA in SPSS. I have two groups with 9 subjects each (so total = 18), and 24 levels of one repeated measure.  I understand why Mauchly's test of Sphericity has no meaning when there are are only 2 levels of a repeated measures factor, but I notice (using General Linear Model.....repeated measures in SPSS) that Mauchly's test of Sphericity also appears to be undefined (or gives the useless output of Mauchly's W = '.0' , p = '.') when the number of levels of a repeated measure is equal to or greater than the number of cases (subjects). In these instances, even though Mauchly's statistic is not calculated, Greenhouse-Geisser, Huynh-Feldt, and Lower-Bound Epsilon values are calculated.  I would be really happy if someone could provide some insight on why Mauchly's statistic is not calculated in these cases and what should be done to assess sphericity in the absence of Mauchly's statistic.
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I am new to statistics and I currently deal with ANOVA. I carry out an ANOVA test in R using  aov(dependendVar ~ IndependendVar)   I get – among others – an F-value and a p-value.   My null hypothesis ($H_0$) is that all group means are equal.   There is a lot of information available on , but I don't know how to read an F-statistic and how F and p are connected.   So, my questions are:   How do I determine the critical F-value for rejecting $H_0$? Does each F have a corresponding p-value, so they both mean basically the same? (e.g., if $p&lt;0.05$, then $H_0$ is rejected)
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I am using a MANOVA test to compare nine different dependent variables (from neuropsychological and neuropsychiatric assessment) between three groups. The output shows a significant influence from GROUP on my variables ($p &lt; .001$).   Ofcourse, I am interested in how the three groups influence every dependent variable. I have studied Field's "Discovering Statistics Using IBM SPSS Statistics" chapter 16, and he states that the preferred post-hoc analysis is a discriminant analysis, because of the linear combination in which the dependent variables are related to group membership in a MANOVA. Discriminant analysis could account for this linear combination, so Field states.  Otherwise, I read some literature, on basic statistical sites, where is stated that I can use multiple univariate ANOVA's with Bonferroni correction, and use a post-hoc on these univariate ANOVA's when they are significant.   Which of these method's is better? That is, which will make my chance of a Type I or II error the least?
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It looks like you can use coding for one categorical variable, but I have two categorical and one continuous predictor variable. Can i use multiple regression for this in SPSS and if so how? thanks!
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I'm probably being a bit simplistic here, but from what I understand the Bonferroni post hoc test is simply when we adjust the alpha to account for inflated error when several post hoc tests are conducted. You divide the alpha by number of tests. So what exactly does SPSS do when we click the button for Bonferroni?   Shouldn't the dividing of alpha be done by us in interpreting the result, e.g. if we do 10 post hoc tests our alpha criterion should be .005 so if the p value is .012 then it is not significant, but SPSS hasn't done anything there, we have just changed our interpretation. So what exactly is it adjusting?
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This is my question: I have to perform a mini-meta-analysis on two studies(independent, balanced sample) with a Repeated measures design and with two within factors. I have the F statistic and the p value of two significant interactions in study 1 and study 2. I want to present a unique p value, so I have to combine the p values. Is it correct to transform the p value in z score and after do I have to average the z scores and find the correspondent p value? Can you suggest a paper with a similar procedure?
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A woman I was working for asked me to do a one-way ANOVA on some data. I replied that the data were repeated measures (time series) data, and that I thought the assumption of independence was violated. She replied that I should not worry about the assumptions, just do the test and she would take into account that the assumptions might not have been met.  That did not seem right to me. I did some research, and found this wonderful blog post by David Robinson, , which exposed me to the No Free Lunch Theorem. I have looked at the original paper and some follow on stuff, and frankly the math is a bit over my head.  The gist of it -- according to David Robinson -- seems to be that the power of a statistical test comes from its assumptions. And he gives two great examples. As I wade through the other articles and blog posts about it, it seems to always be referenced in terms of either supervised learning or search.  So my question is, does this theorem apply to statistical tests in general? In other words, can one say that the power of a t-test or ANOVA comes from its adherence to the assumptions, and cite the No Free Lunch Theorem?   I owe my former boss a final document regarding the work I did, and I would like to know if I can reference the No Free Lunch Theorem in stating that you cannot just ignore the assumptions of a statistical test and say you'll take that into account when evaluating the results.
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I am trying to analyze data based on a sample size of around 50 and would appreciate advice on which statistical test would be most appropriate. Participants, on a single questionnaire, were asked to complete two different sections. Variable 1: rating Likert items on a scale from 1-5 (strongly agree to disagree). Variable 2: rating Likert items of a validated tool on a scale of 1-9, which is averaged to obtain a single number.  My research question is: Do those who 'agree' with the statement in the first section significantly differ in terms of their scores on the variable 2, compared to those who 'disagree'? (ex. Do participants tend to 'agree' with "I like to read books for pleasure." have higher IQs (variable 2) than those who 'disagree'?)  I was advised by a colleague to run a one-way ANOVA with variable 1 (categorical) as IV and variable 2 (continuous) as DV.   Question 1: My independent variable is the three groups: 'agree', 'neutral', and 'disagree.' (I combined strongly agree with agree because of low frequencies and likewise for strongly disagree and disagree.) However, there are many more participants in the 'agree' category than in the 'disagree' category. The distribution is, on average, around 44 agrees, 3 neutrals, and 3 disagrees. Do the (very) unequal sample sizes prevent me from performing ANOVA, or any other tests? Which test would be most appropriate for this analysis?  Question 2: I would like to run analyses on all of the Likert items, which is more than 10 statements. (ex. Seeing how IQs correlate with agreement to many statements including “I like to read books for pleasure,” “I watch a lot of TV,” “I exercise more than 30 minutes each week,” etc.) I was planning to do a Bonferroni correction to adjust for the multiple tests. Would this be correct, and help make my analyses more robust? Note: The continuous data are not normally distributed (there is negative skewness), but I seem to be able to correct the skewness using logarithmic transformation.  Any advice or suggestions for papers to read would be appreciated. Thank you!
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I did a normality test for my data and as usual, SPSS will give both Shapiro-Wilk and Kolmogorov-Smirnov numbers. Is it okay if I choose to use only the number from Shapiro-Wilk? because the number from SW is the ones that showed significant level (0.183)  while KS is not (0.046). is there any references that I can use to support my decision as to using only SW when both SW and KS were given? thanks.
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1) I had a dataset with missing data for baseline variables and outcome variables. Through multiple imputation in SPSS (10 imputations, 50 iterations, PMM for scale variables) I imputed the missing data for the baseline variables. When I analyse the data (i.e. only the 10 imputation data sets) and use the independent sample t-test for a continuous variable results are pooled. This includes mean and p-value, but not standard deviation. Is it correct that I can take the average of the 10 standard deviations calculated for the imputed data sets to calculate the pooled standard deviation?  2) The pooled p-value from the independent t-test on 10 imputation data sets in SPSS is not the average of the 10 p-values that are calculated: am I correct? As I believe a correction is made for the fact that the p-value is based on imputed data?  3) I want to impute the missing data for the outcome variables as well. I did not include those variables in the first imputation process since one of the outcome variables is affected by the value of one of the baseline variables for which data was missing. So I figured that I would first impute the baseline variable, create a 'complete' data set for it and than impute the outcome variables using that data and the other baseline variables. Is it actually possible to impute a second time? And how would I do that? Now (after the first imputation process) I have a dataset which is 11 times as large as originally (original data set + 10 x imputed data set). Do I use all 10 imputation data sets to run the imputation process again which would result in another tenfold of that data set (so 100 fold of my original data set size)? And is it true this would lead to a larger standard deviation for scale variables?
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I have some non-Gaussian distributed variable and I need to check if there are significant differences between the values of this variable in 5 different groups.  I have performed Kruskal-Wallis one-way analysis of variance (which came up significant) and after that I had to check which groups were significantly different. Since the groups are kind of sorted (the values of the variable in the first group are supposed to be lower than the values of the variable in the second group which are supposed to be lower than the values of the variable in the third group and so on) I only performed 4 tests:  Group 1 vs Group 2 Group 2 vs Group 3 Group 3 vs Group 4 Group 4 vs Group 5   I have performed this analysis with two different methods. I started by using Dunn's Multiple Comparison Test but nothing came up significant. On the other hand if I use Mann-Whitney test and correct for the number of tests (4) using Bonferroni, 3 tests come up significant.  What does it mean? Which results should I trust?
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I am struggling with using the correct test for count data in R. The dataset I have is the number of nymphs produced by three aphid species on wheat.  So this is count data with one response variable (number of nymphs) and one explanatory variable with three levels (i.e. three different species).  I want to test whether there are any differences between the three species with the number of nymphs they produce.  I think that I should not use an ANOVA with count data. I have tried a GLM with QuasiPoisson, but I am struggling to interpret the output and be sure that I have used the correct test. Essentially I want to equivalent for a one-way anova but for count data.
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Is it possible for one-way (with $N&gt;2$ groups, or "levels") ANOVA to report a significant difference when none of the $N(N-1)/2$ pairwise t-tests does?   In  @whuber wrote:      It is well known that a global ANOVA F test can detect a difference of means even in cases where no individual [unadjusted pairwise] t-test of any of the pairs of means will yield a significant result.   so apparently it is possible, but I do not understand how. When does it happen and what the intuition behind such a case would be? Maybe somebody can provide a simple toy example of such a situation?  Some further remarks:   The opposite is clearly possible: overall ANOVA can be non-significant while some of the pairwise t-tests erroneously report significant differences (i.e. those would be false positives). My question is about standard, non-adjusted for multiple comparisons t-tests. If adjusted tests are used (like e.g. Tukey's HSD procedure), then it is possible that none of them turns out to be significant even though the overall ANOVA is. This is covered here in several questions, e.g.  and . Update. My question originally referred to the usual two-sample pairwise t-tests. However, as @whuber pointed out in the comments, in the ANOVA context, t-tests are usually understood as post hoc contrasts using the ANOVA estimate of the within-group variance, pooled across all groups (which is not what happens in a two-sample t-test). So there are actually two different versions of my question, and the answer to both of them turns out to be positive. See below.
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I am currently arguing with someoe on how to correctly treat data with multiple measurements for each subject. In this case data was gathered for each subject within a short time for different conditions within each subject. All measurements gather exactely the same variable, just multiple.  One option now is to just group the data by conditions and not care that multiple data points come from one subject. However the data points from each subject are probably not completely independent.  The other alternative is, to first take the mean of all measurements for each condition from each subject and then compare the means. However this will probably impact the significance, since in the final analysis it is not taken into account, that the means have less error.  How can you correctly analyse such data? Is this somehow taken care of in SPSS? In principle it should be possible to calculate the error margin when calculating a mean and than considering this in the final analysis, but I do not guess that SPSS is somehow doing this calculation behind my back.
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I have two dependent variables, one is positively skewed (significantly), the other is negatively skewed (not significant). I can apply a log10 transformation to improve the skewness on the first one.   Questions:   Can I then statistically compare the two variables in a mixed ANOVA test (the two variables are steps of a factor "emotion" and I compare between factor "group"), given that one is transformed and the other isn't?  If not, what options do I have?
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I'm analyzing reaction time data from a repeated measures ANOVA with the following design: Factor 1 (between-group): GROUP (controls, clinical) Factor 2 (within-group): TASK TYPE (social, non-social) Factor 3 (within-group): DECISION DIFFICULTY (easy, hard)  ANOVA results indicate a significant 2-way interaction (GROUP&times;DIFFICULTY). No other main effects, 2-way interactions, nor the 3-way interaction are significant (all $p &gt; .4)$.  Since there was no effect of task, I pooled data across the 2 task types. I'm not sure what is the right way to go from here in terms of reporting and interpreting the results. My specific questions are:   Should I always perform follow-up tests with a significant 2-way interaction, or should I just stop there and interpret based on the figures (and without further stats)? Between- or within-group post-hoc: Should post-hoc tests be restricted to an independent t-test between groups (i.e. comparing "pooled easy" data between groups, and "pooled hard" data between groups)? Can I also perform and report a within-group paired t-test comparing "pooled easy" to "pooled hard" within each group? Interpretation: If neither independent t-test from (2) is significant, can I still interpret the 2-way interaction based on a significant paired t-test from (2)? (The clinical group's reaction times appear to increase with task difficulty, whereas reaction times don't change with task difficulty in controls.)   In the absence of significant between-group t-tests, can I still report these results based on the 2-way interaction and a within-group t-test as "task difficulty affects reaction times in the clinical group and not the control group"?  Thank you. I appreciate any help you can give!
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I'm pretty new to statistics and I need your help. I have a small sample, as follows:    H4U   0.269   0.357   0.2   0.221   0.275   0.277   0.253   0.127   0.246   I ran the Shapiro-Wilk test using R:    shapiro.test(precisionH4U$H4U)   and I got the following result:   W = 0.9502, p-value = 0.6921   Now, if I assume the significance level at 0.05 than the p-value is larger then alpha (0.6921 > 0.05) and I cannot reject the null hypothesis about the normal distribution, but does it allow me to say that the sample has a normal distribution?  Thanks!