Nearly all granting agencies require an estimate of an adequate sample size to detect the effects hypothesized in the study. But all studies are well served by estimates of sample size, as it can save a great deal on resources.
Why? Undersized studies can’t find real results, and oversized studies find even insubstantial ones. Both undersized and oversized studies waste time, energy, and money; the former by using resources without finding results, and the latter by using more resources than necessary. Both expose an unnecessary number of participants to experimental risks.
The trick is (more…)
One of the main assumptions of linear models such as linear regression and analysis of variance is that the residual errors follow a normal distribution. To meet this assumption when a continuous response variable is skewed, a transformation of the response variable can produce errors that are approximately normal. Often, however, the response variable of interest is categorical or discrete, not continuous. In this case, a simple transformation cannot produce normally distributed errors.
A common example is when the response variable is the counted number of occurrences of an event. The distribution of counts is discrete, not continuous, and is limited to non-negative values. There are two problems with applying an ordinary linear regression model to these data. First, many distributions of count data are positively skewed with many observations in the data set having a value of 0. The high number of 0’s in the data set prevents the transformation of a skewed distribution into a normal one. Second, it is quite likely that the regression model will produce negative predicted values, which are theoretically impossible.
An example of a regression model with a count response variable is the prediction of the number of times a person perpetrated domestic violence against his or her partner in the last year based on whether he or she had witnessed domestic violence as a child and who the perpetrator of that violence was. Because many individuals in the sample had not perpetrated violence at all, many observations had a value of 0, and any attempts to transform the data to a normal distribution failed.
An alternative is to use a Poisson regression model or one of its variants. These models have a number of advantages over an ordinary linear regression model, including a skew, discrete distribution, and the restriction of predicted values to non-negative numbers. A Poisson model is similar to an ordinary linear regression, with two exceptions. First, it assumes that the errors follow a Poisson, not a normal, distribution. Second, rather than modeling Y as a linear function of the regression coefficients, it models the natural log of the response variable, ln(Y), as a linear function of the coefficients.
The Poisson model assumes that the mean and variance of the errors are equal. But usually in practice the variance of the errors is larger than the mean (although it can also be smaller). When the variance is larger than the mean, there are two extensions of the Poisson model that work well. In the over-dispersed Poisson model, an extra parameter is included which estimates how much larger the variance is than the mean. This parameter estimate is then used to correct for the effects of the larger variance on the p-values. An alternative is a negative binomial model. The negative binomial distribution is a form of the Poisson distribution in which the distribution’s parameter is itself considered a random variable. The variation of this parameter can account for a variance of the data that is higher than the mean.
A negative binomial model proved to fit well for the domestic violence data described above. Because the majority of individuals in the data set perpetrated 0 times, but a few individuals perpetrated many times, the variance was over 6 times larger than the mean. Therefore, the negative binomial model was clearly more appropriate than the Poisson.
All three variations of the Poisson regression model are available in many general statistical packages, including SAS, Stata, and S-Plus.
References:
- Gardner, W., Mulvey, E.P., and Shaw, E.C (1995). “Regression Analyses of Counts and Rates: Poisson, Overdispersed Poisson, and Negative Binomial Models”, Psychological Bulletin, 118, 392-404.
- Long, J.S. (1997). Regression Models for Categorical and Limited Dependent Variables, Chapter 8. Thousand Oaks, CA: Sage Publications.
Researchers are often interested in setting up a model to analyze the relationship between some predictors (i.e., independent variables) and a response (i.e., dependent variable). Linear regression is commonly used when the response variable is continuous. One assumption of linear models is that the residual errors follow a normal distribution. This assumption fails when the response variable is categorical, so an ordinary linear model is not appropriate. This article presents a regression model for a response variable that is dichotomous–having two categories. Examples are common: whether a plant lives or dies, whether a survey respondent agrees or disagrees with a statement, or whether an at-risk child graduates or drops out from high school.
In ordinary linear regression, the response variable (Y) is a linear function of the coefficients (B0, B1, etc.) that correspond to the predictor variables (X1, X2, etc.). A typical model would look like:
Y = B0 + B1*X1 + B2*X2 + B3*X3 + … + E
For a dichotomous response variable, we could set up a similar linear model to predict individuals’ category memberships if numerical values are used to represent the two categories. Arbitrary values of 1 and 0 are chosen for mathematical convenience. Using the first example, we would assign Y = 1 if a plant lives and Y = 0 if a plant dies.
This linear model does not work well for a few reasons. First, the response values, 0 and 1, are arbitrary, so modeling the actual values of Y is not exactly of interest. Second, it is really the probability that each individual in the population responds with 0 or 1 that we are interested in modeling. For example, we may find that plants with a high level of a fungal infection (X1) fall into the category “the plant lives” (Y) less often than those plants with low level of infection. Thus, as the level of infection rises, the probability of a plant living decreases.
Thus, we might consider modeling P, the probability, as the response variable. Again, there are problems. Although the general decrease in probability is accompanied by a general increase in infection level, we know that P, like all probabilities, can only fall within the boundaries of 0 and 1. Consequently, it is better to assume that the relationship between X1 and P is sigmoidal (S-shaped), rather than a straight line.
It is possible, however, to find a linear relationship between X1 and a function of P. Although a number of functions work, one of the most useful is the logit function. It is the natural log of the odds that Y is equal to 1, which is simply the ratio of the probability that Y is 1 divided by the probability that Y is 0. The relationship between the logit of P and P itself is sigmoidal in shape. The regression equation that results is:
ln[P/(1-P)] = B0 + B1*X1 + B2*X2 + …
Although the left side of this equation looks intimidating, this way of expressing the probability results in the right side of the equation being linear and looking familiar to us. This helps us understand the meaning of the regression coefficients. The coefficients can easily be transformed so that their interpretation makes sense.
The logistic regression equation can be extended beyond the case of a dichotomous response variable to the cases of ordered categories and polytymous categories (more than two categories).
I recently had this question in consulting:
I’ve got 12 out of 645 cases with Mahalanobis’s Distances above the critical value, so I removed them and reran the analysis, only to find that another 10 cases were now outside the value. I removed these, and another 10 appeared, and so on until I have removed over 100 cases from my analysis! Surely this can’t be right!?! Do you know any way around this? It is really slowing down my analysis and I have no idea how to sort this out!!
And this was my response:
I wrote an article about dropping outliers. As you’ll see, you can’t just drop outliers without a REALLY good reason. Being influential is not in itself a good enough reason to drop data.
Should you drop outliers? Outliers are one of those statistical issues that everyone knows about, but most people aren’t sure how to deal with. Most parametric statistics, like means, standard deviations, and correlations, and every statistic based on these, are highly sensitive to outliers.
And since the assumptions of common statistical procedures, like linear regression and ANOVA, are also based on these statistics, outliers can really mess up your analysis.
Despite all this, as much as you’d like to, it is NOT acceptable to
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SPSS has a nice little feature for adding and averaging variables with missing data that many people don’t know about.
It allows you to add or average variables, while specifying how many are allowed to be missing.
For example, a very common situation is a researcher needs to average the values of the 5 variables on a scale, each of which is measured on the same Likert scale.
There are two ways to do this in SPSS syntax.
Newvar=(X1 + X2 + X3 + X4 + X5)/5 or
Newvar=MEAN(X1,X2, X3, X4, X5).
In the first method, if any of the variables are missing, due to SPSS’s default of listwise deletion, Newvar will also be missing.
In the second method, if any of the variables is missing, it will still calculate the mean. While this seems great at first, the researcher may wish to limit how many of the 5 variables need to be observed in order to calculate the mean. If only one or two variables are present, the mean may not be a reasonable estimate of the mean of all 5 variables.
SPSS has an option for dealing with this situation. Running it the following way will only calculate the mean if any 4 of the 5 variables is observed. If fewer than 4 of the variables are observed, Newvar will be system missing.
Newvar=MEAN.4(X1,X2, X3, X4, X5).
You can specify any number of variables that need to be observed.
(This same distinction holds for the SUM function in SPSS, but the scale changes based on how many are being averaged. A better approach is to calculate the mean, then multiply by 5).
This works the same way in the syntax or in the Transform–>Compute menu dialog.
First Published 12/1/2016;
Updated 7/20/21 to give more detail.
