hypothesis testinginteractioninterpretationregression
I have difficulties interpreting an interaction term in a linear regression with regards to my hypothesis.
Consider this basic example:
H0: Better school grades lead to higher income and this
relation is the same for men and for women.
DV: income
IV: school grade, sex
How do I interpret a non-significant, positive interaction term grade * sex?
Would it be correct to accept H0?
Rather than saying the relationship is stronger, I think it's more precise to say that weight increases significantly more quickly with height for males than for females. Strength of relationship would be measured by measures like $R^2$, and these are affected not only by the rate of increase of one variable with another, but by the amount of noise in the data. e.g. if the data were something like this:
maleheight <- rnorm(1000, 70, 3)
femaleheight <- rnorm(1000, 65, 2.5)
maleweight <- maleheight*2.2 + rnorm(1000, 0, 20)
femaleweight <- femaleheight*1.3 + rnorm(1000, 0, 10)
height <- c(maleheight, femaleheight)
weight <- c(maleweight, femaleweight)
male <- c(rep(1, 1000), rep(0, 1000))
data <- data.frame(cbind(height, weight, male))
and the model
m1 <- with(data, lm(weight~height + male + height*male))
summary(m1)
shows your pattern, but the relationship looks stronger for women
It looks like you're confused as to what an interaction is in general. In particular, from the bit "suggesting a positive interaction - as HR increases so does PD", it sounds like you believe that the interaction term for a pair of independent variables (IVs) tells you how the IVs are associated with each other. It doesn't. Rather, it tells you how the dependent variable (DV) is associated with the product of the two IVs.
To interpret an interaction, it's often less helpful to look at the coefficients themselves than to look at the effect of interaction in combination with the associated main effects. For example, we can plot the effects of HP, BD, and HP × BD like this (change HP and BP values as appropriate for your actual data):
d = expand.grid(hp = c(-1, 0, 1), bd = c(-1, 1))
d = transform(d, effect = -.0324*hp + 1.4095*bd + 1.1849*hp*bd)
library(ggplot2)
qplot(bd, color = ordered(hp), effect, data = d)
We see that as HP increases, the effect of BD on the DV also increases.
Best Answer
"Accepting H$_{0}$" is always a logical fallacy (i.e. lack of significance is always "failed to reject"). Interpretively, this means you did not find evidence of the interaction
grade*sex.The reason why you can only state that you did not find evidence of X with tests for difference is that these tests only provide evidence of how likely you are to see $\hat{\beta}_{\text{grade}\times\text{sex}}$ if H$_{0}$ is true, and your test only yields your desired power to reject for at least as large as one not all (smaller) possible values under H$_{\text{A}}$.
If you want to state that you found or did not find evidence of an absence of X, then you need to use, for example, tests for equivalence (say, using two one-sided tests) where H$_{0}$ no longer takes the form H$_{0}^{+}\text{: }\theta=0$, but rather takes the form H$_{0}^{-}\text{: }|\theta|=\Delta$, where $\Delta$ is a researcher-specified value meaning "too small a difference to care about". (the '$+$' and '$-$' in the superscript indicate null hypotheses for difference and for equivalence, respectively.)
To perform an equivalence test on
grade*sex(i.e. you want to provide evidence that there is no interaction), you will need a few things:grade*sex(i.e. the coefficient $\hat{\beta}_{\text{grade}\times\text{sex}}$)Given that, then:
H$_{0}^{-}\text{: }|\beta_{\text{grade}\times\text{sex}}| \ge \Delta$, which gives two one-sided null hypotheses:
H$_{01}\text{: }\beta_{\text{grade}\times\text{sex}} \ge \Delta$, and
H$_{02}\text{: }\beta_{\text{grade}\times\text{sex}} \le -\Delta$
The test statistics corresponding to both of these are:
$$t_{1} = \frac{\Delta - \hat{\beta}_{\text{grade}\times\text{sex}}}{s_{\hat{\beta}_{\text{grade}\times\text{sex}}}}$$ $$t_{2} = \frac{\hat{\beta}_{\text{grade}\times\text{sex}}+ \Delta}{s_{\hat{\beta}_{\text{grade}\times\text{sex}}}}$$
These are both right-side/upper tail tests, so you get the p-values:
$p_{1}=\text{P}\left(T_{df} \ge t_{1}\right)$, and
$p_{2}=\text{P}\left(T_{df} \ge t_{2}\right)$
If both H$_{01}^{-}$ and H$_{02}^{-}$ are rejected with $p\le\alpha$ (not $p \le \alpha/2$), then, taken together with the failure to reject H$_{0}^{+}$ you can conclude you found evidence that the
grade*sexis equivalent to zero, given $\alpha$ and $\Delta$.However, if you reject only one or reject neither of H$_{01}^{-}$ and H$_{02}^{-}$, then, taken together with the failure to reject H$_{0}^{+}$ you can't conclude anything: your results are indeterminate because your data are underpowered.