# 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 $$\mu_a - \mu_b > \delta$$.

• I hope my answer helps you, but I do want to point out that hypothesis tests are of population parameters, not sample statistics. This is why my answer used $\mu_x$ and $\mu_y$ instead $\bar{x}$ and $\bar{y}$. – Dave Dec 9 '20 at 23:26

$$H_0: \vert\mu_x - \mu_y\vert = \delta$$

$$H_a: \vert\mu_x - \mu_y\vert < \delta$$

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 $$\vert \mu_x - \mu_y \vert <2$$.

• Your answer is helpful, but it doesn't get to the heart of what I don't understand, which is how the sidedness of each of the two t-tests makes the procedure slightly different for each one. As you say, $H_{0,1} : \mu_x - \mu_y = \delta$ and $H_{0,2} : \mu_x - \mu_y = \delta$, i.e. they're the same. Surely the alternative hypothesis shouldn't make any difference to how the test is carried out, only in how the result is interpreted, but then how do the two t-tests not end up being identical? – Tom Hosker Dec 10 '20 at 1:03
• For my own personal benefit, it would be really helpful if you could point me to a thorough worked example of TOST - I've definitely struggled to find one so far - or you could even carry out an example procedure in your answer. I think something like that would sweep away all my misconceptions in one stroke. (Sorry for bombarding you with unsolicited advice!) – Tom Hosker Dec 10 '20 at 1:09
• The sidedness checks if the difference is less than $\delta$ or greater than $-\delta$. The first one-sided test leads us to believe $\mu_x - \mu_y$ to be less than $\delta$, so the values above $\delta$ are ruled out as possibilities. The second one-sided test leads us to believe $\mu_x - \mu_y$ to be greater than $-\delta$, so the values below $-\delta$ are ruled out as possibilities. If e believe both of those, then we believe $-\delta < \mu_x - \mu_y < \delta$. – Dave Dec 10 '20 at 15:18
• I've always understood the purpose of the two tests. I've never understood how the procedure differs between the two. As I said above, a brief but thorough worked example would probably be the least painful means of clearing up my misconceptions. – Tom Hosker Dec 10 '20 at 15:44
• I do not understand what you mean about the procedure. You do each test as you would do any other one-sided test. – Dave Dec 10 '20 at 15:52