To simplify the wording let's just call the variables male and female.
The main question aside, this is not a typical test for interaction. By specifying:
$$wage = \beta_0 + \beta_1 male + \beta_2 male \times height + \epsilon$$
you are implicitly stating that height does not matter for female at all. Usually, a full interaction test should contain the variables that are used to compose the interaction:
$$wage = \beta_0 + \beta_1 male + \beta_2 height + \beta_3 male \times height + \epsilon$$
That way, the males have:
$$wage = \beta_0 + \beta_1 male + \beta_2 height + \beta_3 male \times height + \epsilon$$
And the females have:
$$wage = \beta_0 + \beta_2 height + \epsilon$$
In your version, the female will only have the constant (intercept), which could likely be a wrong specification.
Back to the question about:
wage = sex+ sex* height + reverse_sex * weight + constant + error
The actual interaction tests should then be:
$$wage = \beta_0 + \beta_1 male + \beta_2 height + \beta_3 male \times height + \beta_4 female + \beta_5 weight + \beta_6 female \times weight + \epsilon$$
A couple points here. First, male and female are completely collinear so one of them will be omitted:
$$wage = \beta_0 + \beta_1 male + \beta_2 height + \beta_3 male \times height + \beta_4 weight + \beta_5 female \times weight + \epsilon$$
For males, these terms remain:
$$wage = \beta_0 + \beta_1 male + \beta_2 height + \beta_3 male \times height + \beta_4 weight + \epsilon$$
For females, these terms remain:
$$wage = \beta_0 + \beta_2 height + \beta_4 weight + \beta_5 female \times weight + \epsilon$$
So, it's technically fine, the $\beta_5$ is still the extra "effect" of weight for female.
Second, this is unnecessarily complicating everything because your proposed model:
$$wage = \beta_0 + \beta_1 male + \beta_2 height + \beta_3 male \times height + \beta_4 weight + \beta_5 female \times weight + \epsilon$$
is essentially the same as:
$$wage = \beta_0 + \beta_1 male + \beta_2 height + \beta_3 male \times height + \beta_4 weight + \beta_5 male\times weight + \epsilon$$
The $\beta_5$ will likely flip sign, but the magnitude is the same. It's basically the difference in slopes between males and females. If males' slope is $a$ smaller than females'; females' slope is $a$ bigger than males'. You'll also find the t-statistic will also flip sign, but p-values are the same. There is no need to split hair here.
Let's say I only wanted to investigate how weight affects wage on
females, but not males. Would it be possible to incorporate this in
one equation? Or would I need a separate regression for each sex?
So, let's just actually show it:
set.seed(81226)
male <- sample(c(1,0), 100, replace=T)
female <- 1 - male
weight <- rnorm(100, 150, 35)
wage <- 25000 - 5 * weight + 1 * male + 2.5 * (male * weight) +
rnorm(100, 0, 100)
m01 <- lm(wage ~ male + weight + male*weight)
summary(m01)
m02<- lm(wage ~ female + weight + female*weight)
summary(m02)
plot(weight, wage, pch=16, col=(male+1))
lines(weight[female==1], m01$fitted[female==1])
lines(weight[male==1], m01$fitted[male==1], col="red")
The first regression using male is:
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 24995.6097 55.7790 448.118 < 2e-16 ***
male 83.2834 73.5968 1.132 0.261
weight -5.0967 0.3627 -14.053 < 2e-16 ***
male:weight 2.0805 0.4723 4.405 2.75e-05 ***
The second regression using female is:
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 25078.8931 48.0124 522.342 < 2e-16 ***
female -83.2834 73.5968 -1.132 0.261
weight -3.0162 0.3026 -9.969 < 2e-16 ***
female:weight -2.0805 0.4723 -4.405 2.75e-05 ***
Graphically, the relationship is:
The red is males, and the black is female. In the first model, female only got the coefficient -5.0967, that is the slope of the black line. The slope of the red line has an adjustment of 2.0805, which is (-5.0967 + 2.0805). The 2.0805 is then the "difference in slopes," aka, the interaction. If both lines are parallel, effect of weight on wage is the same for both sex.
Now, the second model uses female. The slope for males is -3.0162, which is actually just (-5.0967 + 2.0805) from above. The females' slope has a further adjustment of -2.0805 (notice the sign flip), ending up with -5.0967.
I hope this helps clarifying that your question "effect of weight on female" is the same as "absence of such effect of weight on male." Your proposed question sounds making sense, but to people who understand regression it is closer to a needless gesture: if males got a benefit, the females would of course suffer from the same magnitude of penalty.
