HOW-TO #5 A Cornucopia of Statistical Goodies: Understanding t-tests, ANOVA, and Calculating Percent Change Analysis of variance, or ANOVA, is a statistical procedure commonly used by researchers from a wide array of disciplines, including psychology and neuroscience. ANOVAs are often used in the FMRI community to make comparisons of signal activation across subjects. For this reason, we provide a brief introduction to ANOVA in this how-to. Be warned, however, that thick volumes have been written on analysis of variance, and the information provided in this how-to can in no way duplicate those efforts. This how-to will simply provide a very general overview of ANOVA (no ugly equations, I promise!). In addition, there will be a brief discussion regarding independent- versus paired-samples t-tests. Finally, we will conclude with a discussion on calculating percent change and why this normalization method is important when running ANOVA. TABLE OF CONTENTS ----------------- I. Independent Samples t-test II. Paired Samples t-test III. One-way Analysis of variance (ANOVA) IV. Two-way ANOVA V. Parameter Normalization - Calculating Percent ChangeI. t-test for INDEPENDENT SAMPLES ------------------------------Instead of going directly into an explanation of ANOVA, it is perhaps more beneficial to begin with ANOVA's younger sibling, the t-test. Since ANOVA is based on the principles of t-tests, a review of the t-test will hopefully make ANOVA more comprehensible. When exactly should a t-test be used? When making simple, straightforward comparisons between two independent samples, the independent samples t-test is usually the statistic of choice. For clarity, we begin with an example that implements the independent samples t-test. EXAMPLE of Independent Samples t-test: ------------------------------------- Suppose a standardized vocabulary test is given to a group of students. his sample consists of 120 high school seniors, with an equal number of males and females. In other words, our study consists of two independent samples: 60 males and 60 females who are high school seniors. We are curious to know if there are any significant sex differences in performance on the standardized vocabulary test. What statistical procedure can be used to best address this question? The Independent samples t-test. An independent samples t-test is used to compare the means of two different samples. In this example, our two samples are MALES and FEMALES, which make up our single independent variable, "sex." An independent samples t-test is appropriate when there is one independent variable (IV) with two levels. For instance, a comparison of young and older adults' performance on a memory test could be examined using an independent samples t-test. This type of t-test could also be used to examine if there is significantly less sneezing in patients administered Allergy Drug A versus other patients who are given a placebo. In general, if a study consists of two samples that share a variable of interest, and group membership does not overlap, then a comparison on some dependent variable can be made between these two groups. The first thing to do is clearly label our independent and dependent variables (also known as "factors"). Recall that the independent variable (IV) is the factor being manipulated by the experimenter, whereas the dependent variable (DV) is the factor being measured (i.e., the result of interest). In this case, the IV is 'sex' with two levels: male and female (Note: 'sex' or 'gender' is sometimes called a quasi-independent variable because the experimenter cannot assign a sex to each subject. The subject already comes to the experiment as either male or female). The dependent variable in this example is 'vocabulary test score'. A data spreadsheet for this example may look like this: SEX ID # VOCAB. SCORE (max=100) --- ---- ---------------------- F 001 85 F 002 96 F 003 72 . . . . . . F 060 64 M 001 56 M 002 100 M 003 82 . . . . . . M 060 45 The question of interest is whether the mean test score for the females in our sample will differ significantly from the mean test score of the males in our sample. To answer this question, an independent samples t-test must be performed. The output from the t-test can be presented in a table and a figure, as shown below: Descriptive Statistics: DV IV N Mean SD SE Mean -- -- --- ---- -- ------- Vocab Males 60 75.6 10.6 1.37 Females 60 76.4 13.9 1.79 Independent Samples t-test: t-value df p-value ------- -- ------- 1.22 118 .224 The results suggest that our DV "vocabulary test score" did not differ significantly between the two levels of our IV "sex." That is, males and females did not differ significantly in their performance on the vocabulary test (t=1.22, p>.05). The results are often made clearer if a figure is drawn. In fact, it is probably good practice to plot out the results to gain a better understanding of the results, especially when examining main effects and interactions in ANOVA. As my old stats teacher used to say, "If in doubt, plot it out." With a basic understanding of the Independent samples t-test, we can quickly cover the next topic: Paired Samples t-tests.II. t-test for PAIRED SAMPLES ------------------------------The paired samples t-test is usually based on groups of individuals who experience both conditions of the variable of interest. For instance, one study might examine the effects of Drug A versus Drug B on a single sample of 100 diabetics. Subjects in this sample would receive Drug A one week, and Drug B the next. Another study might look at a sample of older adults and their mean reaction times to two types of stimulus conditions: incongruent color-word pairs (e.g., the word "blue" printed in red ink) and congruent color-word pairs (e.g., the word "blue" printed in blue ink). In each of these cases, participants receive both drug/stimulus conditions. EXAMPLE of Paired Samples t-test: -------------------------------- Suppose our group of 120 high-school seniors was given two different vocabulary quizzes. The question of interest in this case is whether test performance varies significantly, depending on the quiz given. In other words, do participants perform similarly or significantly different between the two quizzes? In this example, our independent variables or factors of interest are Quiz 1 and Quiz 2. Our dependent measure will be the score or performance for each quiz. All 120 subjects will be administered these two quizzes, and the paired samples t-test will be used to determine if participants' test scores differ significantly between the two quizzes. A data spreadsheet for this example may look like this: SUBJECT ID # QUIZ 1 SCORE QUIZ 2 SCORE ------------ ------------ ------------ 001 66 72 002 71 80 003 80 83 . . . . . . 118 64 70 119 56 66 120 32 30 The question of interest is whether the mean score for Quiz 1 differs significantly from the mean score for Quiz 2. To answer this question, a paired samples t-test must be performed. The output from the t-test may look something like this: Descriptive Statistics: Variables N Correlation p Mean SD SE --------- --- ----------- --- ----- --- --- Quiz 1 120 .673 .000 69.2 8.4 0.77 Quiz 2 79.0 6.2 0.57 Paired Samples t-test: t-value df p-value ------- -- ------- -2.87 119 .005 The paired-sample analysis indicates that for the 120 participants, the mean score on the second quiz (M=79.0) was significantly higher (p=.005) than the mean score on the first quiz (M=69.2). Figure 2 clearly shows that on average, subjects performed higher on Quiz 2 than on Quiz 1. These results also indicate that a significant positive correlation exists between these two variables (r=.673, p<.001), suggesting that students who score high on one of the quizzes tend to score high on the other. Likewise, students who score low on one quiz also tend to also score low on the other. With the basics of t-tests covered, it is now time to proceed to the ever- elusive Analysis of Variance (ANOVA).III. ONE-WAY ANALYSIS OF VARIANCE (ANOVA) ------------------------------------Analysis of variance (ANOVA) is a statistical procedure used for comparing sample means. The results of the ANOVA can be used to infer that the means of the corresponding population distributions also differ. Whereas t-tests compare only two sample distributions, ANOVA is capable of comparing many. As a study becomes increasingly complicated, more complex statistical measures must also be used. In the examples above, a comparison of two means was made, making the t-test a sufficient statistical tool. However, if a study involves one independent variable with more than just two levels or means, then a one-way ANOVA will be an appropriate alternative to the t-test. EXAMPLE of One-Way ANOVA: ------------------------ Suppose we were conducting a study on high school students' opinions regarding the cafeteria food, and asked a sample of 100 students to rate the quality of the cafeteria food on a scale of 1 to 10 (1 = "I'd rather eat dog food", 10 = "Pretty darn good"). Our main objective for this study is to determine if there is a difference in the rating of cafeteria food, depending on a student's grade level. As such, our independent variable or factor would be GRADE LEVEL, consisting of 4 levels: 9th, 10th, 11th, and 12th grades. Our dependent variable would be CAFETERIA FOOD RATING. One may ask, could multiple independent samples t-tests be done to examine the differences in cafeteria food ratings between class grades? That is, couldn't we just do several independent t-tests rather than a single one- way ANOVA? For example: Comparison of cafeteria food ratings for: 9th vs. 10th grade 9th vs. 11th grade 9th vs. 12th grade 10th vs. 11th grade 10th vs. 12th grade 11th vs. 12th grade The answer is both 'yes' and 'no'. In theory, it IS possible to do multiple t-tests instead of a single ANOVA, but in practice, it may not be the smartest strategy. ANOVA offers numerous advantages that t-tests cannot provide. Specifically, ANOVA is a very robust measure that allows the user to avoid the dreaded "inflated alpha" that may arise from conducting several t-tests. Side Note - Watch out for Inflated Alpha! ----------------------------------------- What exactly is inflated alpha? Suppose we run several t-tests and set our p-value to less than .05. A p<.05 tells us that the probability of getting statistically significant results simply by chance is less than 5%. However - and this is a big "however" - conducting multiple t-tests may lead to what is known as an "inflated alpha." This is when the null hypothesis is erroneously rejected. In this example, the null hypothesis is that no significant differences exist in food rating between class grades. Erroneous rejection of the null hypothesis is also known as a Type 1 error or a false positive. As more independent t-tests are done, the probability of getting a Type 1 error increases. In our example, there are only six possible t-tests that can be done, so alpha inflation is not a major problem. However, suppose one was testing the effectiveness of ten different types of allergy drugs. The number of t-test comparisons that can be done in this sample would be 45! See below: Drug 1 vs. Drug 2 Drug 2 vs. Drug 9 Drug 5 vs. Drug 6 Drug 1 vs. Drug 3 Drug 2 vs. Drug 10 Drug 5 vs. Drug 7 Drug 1 vs. Drug 4 Drug 3 vs. Drug 4 Drug 5 vs. Drug 8 Drug 1 vs. Drug 5 Drug 3 vs. Drug 5 Drug 5 vs. Drug 9 Drug 1 vs. Drug 6 Drug 3 vs. Drug 6 Drug 5 vs. Drug 10 Drug 1 vs. Drug 7 Drug 3 vs. Drug 7 Drug 6 vs. Drug 7 Drug 1 vs. Drug 8 Drug 3 vs. Drug 8 Drug 6 vs. Drug 8 Drug 1 vs. Drug 9 Drug 3 vs. Drug 9 Drug 6 vs. Drug 9 Drug 1 vs. Drug 10 Drug 3 vs. Drug 10 Drug 6 vs. Drug 10 Drug 2 vs. Drug 3 Drug 4 vs. Drug 5 Drug 7 vs. Drug 8 Drug 2 vs. Drug 4 Drug 4 vs. Drug 6 Drug 7 vs. Drug 9 Drug 2 vs. Drug 5 Drug 4 vs. Drug 7 Drug 7 vs. Drug 10 Drug 2 vs. Drug 6 Drug 4 vs. Drug 8 Drug 8 vs. Drug 9 Drug 2 vs. Drug 7 Drug 4 vs. Drug 9 Drug 8 vs. Drug 10 Drug 2 vs. Drug 8 Drug 4 vs. Drug 10 Drug 9 vs. Drug 10 The above example illustrates a study that might be in danger of alpha inflation if multiple t-tests were conducted rather than a single one-way ANOVA. Of course, if one insists on running multiple t-tests, then the best way to avoid alpha inflation is to adjust the p-value to a more conservative level (e.g., A p-value of .05 divided by 45 comparisons results in a p=.001). Even with this correction, running this many t-tests is a bit cumbersome (and will inevitably send any editor reading the Results section of your manuscript into a tizzy). It is probably better to run the ANOVA, determine if any differences exist in the first place, and then run specific t-tests to determine where those differences lie. Enough about Inflated Alpha. Back to our example... A spreadsheet for our study on high school students' rating of cafeteria food might look like this: GRADE ID # FOOD RATING (range 1-10) ----- ---- ----------- 9 001 10 9 002 5 . . . . . . 10 101 8 10 102 3 . . . . . . 11 201 5 11 202 9 . . . . . . 12 301 7 12 302 9 . . . . . . A one-way ANOVA can be conducted on this data, with "cafeteria food rating" as our dependent measure, and "grade" (with four levels) as our independent measure. The mean food ratings for each class grade are shown in the output and figure below: Descriptive Statistics: IV levels N Mean Food Rating SD SE Min Max --------- --- ---------------- -- -- --- --- 9th grade 20 8.35 1.53 0.34 6 10 10th grade 24 7.75 2.13 0.44 4 10 11th grade 45 8.04 2.26 0.34 2 10 12th grade 11 6.27 3.32 1.00 2 10 ----------------------------------------------------------------- TOTAL 100 7.84 2.29 0.23 2 10 One-way ANOVA: DV: Food Rating Sum Squares df Mean Square F p ----------- -- ----------- --- --- Between Groups 34.297 3 11.432 2.272 .085 Within Groups 483.143 96 5.033 --------------------------------- TOTAL 517.440 99 The F-score and p-value of a one-way ANOVA will indicate whether the main effect of the independent variable "class grade" was significant. In other words, a significant F-statistic would tell us that class grade had a significant effect on cafeteria food rating. In this example, our p-value of .085 suggests that a marginally significant difference exists within comparisons of food ratings among our four grade levels. Although the ANOVA results tell us that grade-related differences exist in cafeteria food ratings, it does not tell us 'where' those differences lie. Is it between 9th and 12th graders? 10th and 11th? Additional post-hoc tests must be done to address this issue. The results of the post-hoc tests are shown below: Post Hoc Tests (note: these are independent t-tests): Grade(I) Grade(J) Mean (I-J) SE p-value ------- ------- ---------- ---- ------- 9th vs. 10th .60 .679 .379 9th vs. 11th .31 .603 .613 9th vs. 12th 2.08* .842 .015 10th vs. 11th -.29 .567 .605 10th vs. 12th 1.48 .817 .074 11th vs. 12th 1.77* .755 .021 The asterisks (*) indicate there are two pairs of groups whose means differ significantly (p<.05) from each other. 12th graders seem to dislike the cafeteria food significantly more than both 9th and 11th graders (p=.015 and .021 respectively). Seniors also dislike the cafeteria food marginally more than 10th graders (p=.074) Why did the overall ANOVA show only a marginally significant difference (p=.085), while the pair-wise comparisons yielded two differences (i.e., 9th vs. 12th and 11th vs. 12th) that were strongly significant? This is because the overall ANOVA compares all values simultaneously, thus weakening statistical power. The post-hoc tests are simply a series of independent t-tests.IV. TWO-WAY ANALYSIS OF VARIANCE ----------------------------A two-way (or two-factor) ANOVA is a procedure that designates a single dependent variable and utilizes exactly two independent variables to gain an understanding of how the IV's influence the DV. WARNING: When conducting ANOVA, it is relatively easy to get certain software packages (like SPSS, SAS, SPM, AFNI, etc.) to conduct a two-way, three-way, even four- or five-way analysis of variance. The programs may do all the arithmetic, which results in impressive-looking output, but it is up to the user to properly interpret those results. The ease of calculating analysis of variance on the computer often masks the fact that a successful study requires many hours of careful planning. In addition, a one-way ANOVA is fairly straightforward and easy to interpret, but a two-way ANOVA requires some training and frequently involves a thorough examination of tables and figures before interpretation is clear. Understanding a three-way ANOVA usually requires an experienced researcher, and interpretation of a four-way ANOVA is often nightmarish in nature, even for the most skilled researcher. EXAMPLE of Two-Way ANOVA: ------------------------- Suppose we were interested in determining if performance on a math test would differ between males and females given chocolate, vanilla, or strawberry ice cream before taking the test (yes, this is a silly example but just go with it). In this example, we have two independent variables and one dependent variable (side note: more than one DV would require a MANOVA). As such, the design of our experiment is set up in the following manner: IV1: SEX (Females, Males) IV2: FLAVOR (Chocolate, Vanilla, Strawberry) DV: MATH TEST SCORE In this example, we have a 2 x 3 between-subjects design. The design is labeled "between-subjects" because participants fall under only one level of each independent variable. For instance, subjects are either male or female, and they are either given chocolate, vanilla, or strawberry ice cream. There is no overlap of our subject groups. As such, our data spreadsheet might look like this: SEX FLAVOR ID# MATH TEST SCORE --- ------ --- --------------- F chocolate 001 104 F chocolate 002 89 . . . . . . . . F vanilla 101 98 F vanilla 102 100 . . . . . . . . F strawberry 201 80 F strawberry 202 77 . . . . . . . . M chocolate 301 98 M chocolate 302 106 . . . . . . . . M vanilla 401 78 M vanilla 402 70 . . . . . . . . M strawberry 501 88 M strawberry 502 102 . . . . . . . . By running a two-way ANOVA on the data, we hope to answer the following questions: First, is there a main effect of SEX? That is, do males and females differ significantly on their performance on the math test? Second, is there a main effect of FLAVOR, where a significant difference in math performance can be found between chocolate, vanilla, and strawberry ice cream eaters? Finally, is there a significant SEX by FLAVOR interaction? That is, how do sex and flavor interact in their effect on math test performance? Can any sex differences in math performance be found between the three different ice cream flavors? The output of our two-way ANOVA is shown below: Descriptive Statistics: DV = Math Score Flavor Sex Mean Score Std. Dev N ------ --- ---------- -------- -- Chocolate Female 103.95 18.14 20 Male 106.85 13.01 13 TOTAL 105.09 16.15 33 Vanilla Female 100.00 12.31 26 Male 98.46 11.82 13 TOTAL 99.49 12.01 33 Strawberry Female 102.83 10.68 18 Male 90.73 21.23 15 TOTAL 97.33 17.18 33 ----------------------------------------------------------- TOTAL Female 102.03 13.90 64 Male 98.29 17.20 41 TOTAL 100.57 15.30 105 Two-way ANOVA: Source Sum Squares df Mean Square F p ------ ----------- --- ------------ --- --- Corrected 2350.41 5 470.08 2.12 .070 Model Intercept 996892.70 1 996892.70 4487.39 .000 FLAVOR 1262.73 2 631.36 2.84 .063 SEX 316.56 1 316.56 1.43 .235 FLAVOR * SEX 960.44 2 480.22 2.16 .121 Error 21993.31 99 222.16 Total 1086378.00 105 Corrected Err. 24343.71 104 Main Effect of FLAVOR: --------------------- The above results indicate that our main effect of FLAVOR was marginally significant (F(2,99)=2.84, p=.063), suggesting that math scores differed somewhat between the three flavor groups. To clarify this main effect further, it is always helpful to plot out the results. Figure 4 shows that collapsed across SEX, chocolate tasters scored marginally higher on the math test than strawberry and vanilla tasters. Vanilla and strawberry tasters did not seem to differ on their math test performance. Main Effect of SEX: ------------------ Our main effect of SEX was not statistically significant (F(1,99)=1.43, p=.235), indicating no sex-related differences in math test performance. Again,this result can be seen clearly by collapsing across FLAVOR, and plotting the mean math scores for males and females (see Figure 5): SEX by FLAVOR Interaction: ------------------------- Finally, there was no significant SEX by FLAVOR interaction (F(2,99)=2.16, p=.121). By plotting out the results (see Figure 6), we clearly see that male and female chocolate tasters performed virtually identically on the math test. The same result can be seen between male and female vanilla tasters. However, amongst the strawberry tasters, we see a pattern (although not statistically significant) of females performing slightly higher on the math test than males. As this example illustrates, it is always a good idea to plot out the results. A schematic view of the data makes it much easier to decipher where the significant main effects and interactions may lie. Even if the results are not statistically significant, a plot will show the pattern of results, which can be just as informative as the statistical findings.V. PARAMETER NORMALIZATION - Calculating Percent Change ----------------------------------------------------Although this section may seem to deviate a bit from the previous sections on t-tests and ANOVAs, the discussion of calculating percent change is actually quite relevant to statistical testing. Primarily, the calculation of percent signal change is a type of normalization of the data, which is necessary when comparing groups whose baselines differ. When making comparisons across subjects or groups of subjects, normalization of the data becomes an important issue. If certain parameters of interest are not normalized before group comparisons are made, then the results may be misleading, suggesting there are differences between groups when in fact there are not. To fully understand this issue, an example is in order: EXAMPLE of Percent Change: ------------------------- Suppose we tested a sample of young adults and a sample of older adults on a reaction time task. The reaction time task has two conditions: the first one involves color blocks presented on the computer screen. Each participant must identify the color the block is painted in as quickly as possible. The second condition consists of color words (e.g., "BLUE") printed in an incongruent ink color (e.g., red). The goal is to ignore the word and name the color ink. This second condition is often referred to as the "Stroop" task. The Stroop task typically slows down reaction time because participants must first suppress the urge to read the word, and instead, name the ink color. As such, the Stroop condition should result in significantly longer response times than the color blocks condition for both age groups. Our 2 x 2 mixed factorial design is composed of the following independent and dependent variables: IV1-AGE: 1. Young 2. Older IV2-CONDITION: 1. Color blocks 2. Stroop task DV - RESPONSE TIME (measured in msec) In this case, AGE is a between-subjects variable because subjects are either young or older, but not both. Alternatively, CONDITION is a within subjects variable because participants are exposed to both the color blocks and the Stroop task. As such, our design is identified as "mixed" because there are both between- and within-subject variables in our experiment. The following hypotheses can be formulated for this experiment: First, do we expect to find a main effect of AGE? If so, we might predict that with age, reaction time declines, which results in significantly slower response times for older adults than young adults. Second, do we expect to find a main effect of CONDITION? We predict that the Stroop condition will result in significantly slower response times for both age groups than the colored blocks condition. This is because reading is more automatic than color naming. As such, subjects must take the time to suppress the reading response in order to correctly name the ink color. Finally - and most importantly - do we expect to find a significant AGE x CONDITION interaction? In other words, do older adults show more slowing in the Stroop condition relative to the color blocks condition than do young adults? If so, we might conclude that with age, not only does response time increase, it also becomes more difficult to suppress irrelevant information, making the Stroop task more difficult for older than young adults. One reason for such a finding may be that the ability to block out irrelevant information becomes compromised with age. The output for this study may look something like this: Descriptive Statistics: DV = Mean Response time (in milliseconds) Young Adults Older Adults ------------ ------------- A) Color Blocks 500 ms 1000 ms B) Stroop Task 550 ms 1100 ms ------------------------------------------------------- Difference (B-A): 50 ms 100 ms By examining the absolute differences between the Stroop task and the Color Blocks, we find a difference of 50 ms for young adults and 100 ms for older adults. From this finding, one might conclude, "Wow, older adults are twice as slow as young adults when they are presented with the Stroop task. It is twice as hard for older adults to deal with the Stroop task, which involves the simultaneous presentation of two pieces of incongruent information. This must mean that with age, older adults develop inhibitory deficits that deter them from suppressing irrelevant information." However, the interpretation of these results may be inaccurate, because it is not accounting for the fact that the baselines for young and older adults are different. If the color blocks condition is our baseline/neutral condition, we see that on average, young adults respond at around 500 milliseconds. On the other hand, older adults' baseline is much slower, at 1000 milliseconds. This finding indicates that in general, older adults have slower response times than young adults. As such, we must account for these differences in baseline, by calculating the percent change in response time between the baseline condition and the Stroop condition. That is, how much slower is each group in the Stroop condition relative to their baseline? Is it 5 percent for young and 10 percent for older adults? Is it 10 percent for young and 12 percent for older? Converting our absolute response times to percent change will allow us to answer this question. One way to calculate the percent change is by using the following simple equation: A = Baseline Condition (e.g., color blocks response time) B = Stimulus Condition (e.g., Stroop task response time) ((B - A)/A) * 100 = percent change We can apply this formula to each of our age groups: Young Adults: ((550 - 500)/500) * 100 = 10% Older Adults: ((1100 - 1000)/1000) * 100 = 10% When the data are normalized, we realize that there really aren't any age- related differences in Stroop interference between young and older adults. For both age groups, the Stroop condition slowed participants by 10% compared to the baseline condition. Although older adults are slower in general than young adults, both age groups experience the same amount of interference from the Stroop condition. Therefore, these results do not suggest that inhibitory deficits increase with age. As such, the AGE x CONDITION interaction should not be statistically significant. A plot of the interaction (Figure 7) verifies this conclusion: FMRI and Calculating Percent Signal Change: ------------------------------------------ The above example showed the importance of normalization of data in a psychological study. This normalization process is just as important in FMRI research. When comparing parameters that quantify activation across subjects (i.e., an ANOVA will be run to examine activation levels across subjects in response to a stimulus), these parameters should be normalized by calculating the percent signal change. This is because FMRI signal amplitude varies for different subjects, runs, scanning sessions, regressors, image reconstruction software, and modeling strategies. Amplitude measures (i.e., regression coefficients) can be turned to percent signal change from the baseline. For an FMRI example of how the percent signal change is calculated using AFNI, see the AFNI_howto section of this tutorial.CONCLUSION ----------The goal of this review was to take some anguish and anxiety out of statistics. Since we cannot take the statistics out of research, we must conquer and embrace it. Hopefully, any questions you may have had regarding t-tests, ANOVAs, and parameter normalization have been adequately answered. Although the examples shown here come from behavioral research (I'm a psychologist, this is what I know), the statistics can be applied to FMRI research as well, on a voxel-by- voxel basis. Although the BOLD signal is a somewhat more complicated dependent variable than say, a reaction time or a test score, the statistics are the same.REFERENCES ----------If this review has whetted your appetite for more statistics, the following references selected by members of the Scientific and Statistical Computing Core at the NIMH (http://afni.nimh.nih.gov/sscc) may be of interest to you. Good luck and happy reading! Picks by Dr. Cox: ---------------- Bickel, P. J. & Doksum, K. A. (1977). Mathematical Statistics. Holden-Day. Casella, G. & Berger, R. L. (1990). Statistical Inference. Brooks/Cole. Picks by Dr. Chen: ----------------- Kutner, M.H., Nachtschiem, C.J., Wasserman, W., & Neter, J. (1996). Applied Linear Statistical Models, 4th edition. McGraw-Hill/Irwin. Neter, J., Wasserman, W., & Kutner, M.H. (1990). Applied Linear Statistical Models: Regression, Analysis of Variance, and Experimental Designs, 3rd edition. Richard d Irwin Publishing. Picks by Dr. Christidis: ----------------------- Howell, D. C. (2003). Fundamental Statistics for the Behavioral Sciences, 5th edition. International Thomson Publishing. Hays, W. L. (1994). Statistics, 5th edition. International Thomson Publishing. Helpful website: www.statsoftinc.com/textbook/stanman.html