TOST and its two null hypotheses Background
I have a device which is used to size potatoes. I want to use statistics to assess how accurate this device is.
To that end, I've collected two data sets, $X$ and $Y$, where $X$ is the set of measurements collected by the device for a single one-ton box of potatoes, and $Y$ is the set of measurments taken by hand for the same box. Note that, while $X$ and $Y$ pertain to the same box, the two data sets are not paired in the sense that one cannot say, for each $x_i$, that it corresponds to a given $y_i$; indeed, there are around 10% more elements in $X$ than in $Y$.
In assessing the accuracy of the device, I am very concerned with how well it reports the size profile of a given box, and not particularly concerned with how well it measures any one potato. Both $X$ and $Y$ are roughly normally distributed, with a moderate positive skew.
One component of the device is known to introduce an error, which I'll call $\pm \delta$. I would like to be able to demonstrate that the device is accurate within that margin of error.
My Progress So Far
In response to a related question, someone suggested that a Two One-Sided T-test (TOST) procedure would be useful here. So that's what I've tried to do...
My Attempt at TOST
Null hypothesis: $\bar{x} - \bar{y} < -\delta$ or $\bar{x} - \bar{y} > \delta$.
Null hypothesis, first half: $\bar{x} - \bar{y} < -\delta$.
$t_0 = \frac{\bar{x} - \bar{y} + \delta}{s_{x, y}}$
Where $s_{x, y} = \sqrt{\frac{s_x^2 + s_y^2}{n}}$
Using my numbers, I get $t_0 \approx 10$. Given that the degrees of freedom is in the thousands, I guess this means we reject the null hypothesis?
Null hypothesis, second half: $\bar{x} - \bar{y} > \delta$.
$t_1 = \frac{\bar{x} - \bar{y} - \delta}{s_{x, y}}$
Again, using my numbers, I get $t_1 \approx 30$. Given that the degrees of freedom is in the thousands, I guess this means we reject the null hypothesis?
We rejected both halves of the null hypothesis, so I guess these means we can reject the null hypothesis overall?
What I Don't Understand
While there are probably a number of things I'm not grasping properly, the thing that I don't understand and which I know I don't understand is the sidedness of the t-test. I was always taught that the null hypothesis for a t-test was $\mu_a = \mu_b$. I don't understand how you can swap that equality for an inequality, and, once we make that swap, how we know that our t-test proves that $\mu_a - \mu_b < \delta$, and not $\mu_a - \mu_b > \delta$.
 A: $H_0: \vert\mu_x - \mu_y\vert = \delta$
$H_a: \vert\mu_x - \mu_y\vert < \delta$
In English, we want to show that the means of $X$ and $Y$ are not more than $\delta$ from one another.
Let's break $H_0$ and $H_a$ into two one-sided hypothesis tests.
$H_{0,1}: \mu_x - \mu_y = \delta$ (first null hypothesis)
$H_{a,1}: \mu_x - \mu_y < \delta$ (first alternative hypothesis)
$H_{0,2}: \mu_x - \mu_y = \delta$ (second null hypothesis)
$H_{a,2}: \mu_x - \mu_y > -\delta$ (second alternative hypothesis)
(In order for $H_a$ to be true, both $H_{a,1}$ and $H_{a,2}$ must be true, and if both both $H_{a,1}$ and $H_{a,2}$ are true, then $H_a$ is true.)
By rejecting $H_{0,1}$ in favor of $H_{a,1}$, you are saying that you believe $\mu_x - \mu_y < \delta$, so $\mu_x - \mu_y\in (-\infty, \delta)$.
By rejecting $H_{0,2}$ in favor of $H_{a,2}$, you are saying that you believe $\\mu_x - \mu_y > -\delta$, so $\mu_x - \mu_y \in (-\delta, \infty)$.
Since you believe that $\mu_x - \mu_y\in (-\infty, \delta)$ and $\mu_x - \mu_y\in (-\delta, \infty)$, you believe that $\mu_x - \mu_y\in (-\infty, \delta)\cap (-\delta, \infty) = (-\delta, \delta)$. In other words, you believe that the means of $X$ and $Y$ differ by no more than $\delta$.
$\square$
I really like a quote by gung. The bracketed parts are mine.

Very briefly, you select an interval within which you would consider that the true mean difference might as well be 0 for all you could care, then you perform a one-sided test to determine if the observed value is less than the upper bound of that interval [$H_{0,1}$ vs $H_{a,1}$], and another one-sided test to see if it is greater than the lower bound [$H_{0,2}$ vs $H_{a,2}$]. If both of these tests are significant, then you have rejected the hypothesis that the true value is outside the interval you care about. If one (or both) are non-significant, you fail to reject the hypothesis that the true value is outside the interval.

EXAMPLE
We have $X\sim N(\mu_x, 1)$ and $Y\sim N(\mu_y, 1)$. We want to show that $\vert \mu_x - \mu_y \vert < 2 $.
$H_0: \vert\mu_x - \mu_y\vert = 2$
$H_a: \vert\mu_x - \mu_y\vert < 2$
$H_{0,1}: \mu_x - \mu_y = 2$ (first null hypothesis)
$H_{a,1}: \mu_x - \mu_y < 2$ (first alternative hypothesis)
$H_{0,2}: \mu_x - \mu_y = 2$ (second null hypothesis)
$H_{a,2}: \mu_x - \mu_y > -2$ (second alternative hypothesis)
We collect $36$ observations from $X$ and $49$ observations from $Y$, so $n_x=38$ and $n_y=49$. The sample means are $\bar{x} = 3$ and $\bar{y} = 4$. Since we know the variance, we use a z-test for each one-sided test. Let's do the first test.
$H_{0,1}: \mu_x - \mu_y = 2$
$H_{a,1}: \mu_x - \mu_y < 2$
$$
Z = \dfrac{(3 - 4) - 2}
{
\sqrt{
\frac{1}{36} + \frac{1}{49}
}
}
=\dfrac{-3}{0.22}
=-13.6
$$
Since this is a "less than" hypothesis test, we find the lower tail probability.
1-scipy.stats.norm.cdf(-13.6)$\approx 0$
From this, we conclude that $\mu_x - \mu_y < 2$.
Let's do the second test.
$H_{0,1}: \mu_x - \mu_y = 2$
$H_{a,1}: \mu_x - \mu_y > -2$
$$
Z = \dfrac{(3 - 4) - (-2)}
{
\sqrt{
\frac{1}{36} + \frac{1}{49}
}
}
=\dfrac{1}{0.22}
=4.54
$$
Since this is a "greater than" hypothesis test, we find the upper tail probability.
1-scipy.stats.norm.cdf(4.54)$\approx 0$
From this, we conclude that $\mu_x - \mu_y > -2$.
Combining both tests, if $\mu_x - \mu_y$ has to be greater than $-2$ and has to be less than $2$, then $\vert \mu_x - \mu_y \vert <2$.
