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I'm trying to test the difference of means of two groups, A and B, is larger than a specific value or not.

The two groups are considered as sampled from two binomial distributions and can be approximated to normal distribution. Let $a$ and $b$ be the value we calculate from the data and follow normal distribution ($a \sim \mathcal{N}(\mu_A, \sigma_A^2), b \sim \mathcal{N}(\mu_B, \sigma_B^2)$, thus $b - a \sim \mathcal{N}(\mu_B - \mu_A, \sigma_A^2 + \sigma_B^2)$).

With typical z-test, we use a null hypothesis $H_0: \mu_A = \mu_B$ and z-value which is defined as $z = \frac{b - a - (\mu_B - \mu_A)}{\sqrt{\sigma_A^2 + \sigma_B^2}} \sim \mathcal{N}(0, 1)$. Under $H_0$, $z = \frac{b - a}{\sqrt{\sigma_A^2 + \sigma_B^2}}$ and we compare this value with certain value based on $\alpha$.

However, what I would like to do now is to test if $\mu_B - \mu_A \gt 0.1 $. In this case, is it a correct way to test by using $H_0: \mu_A + 0.1 =\mu_B$ and use $z = \frac{b - a - 0.1}{\sqrt{\sigma_A^2 + \sigma_B^2}}$ and do one-sided test?

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This seems to be your second question about difference in binomials, so feel free to share if there's a more general theme to your question. To answer your question, you're correct, but I would approach it differently... your notation on your last paragraph is fine, but just want to make sure you understand it.

When you have two different independent groups, then you indeed have that asymptotically, $a \sim N(\mu_a,\sigma_a^2)$, $b \sim N(\mu_b,\sigma_b^2)$. Then, when thinking of $b-a$, that random variable is asymptotically the difference of two independent normal random variables, and you can show that it must be that $b-a \sim N(\mu_b - mu_a, \sigma_a^2 + \sigma_b^2)$ (as an aside, if the two groups are repeated measures from the same group, you then have that the difference itself asymptotically is normal, and then the variance is just of the difference, which is why paired t-tests have less variance).

Then, testing the null of $H_0 : \mu_a = \mu_b$ is equivalent to testing that $\mu_b -\mu_a = 0$, and from above, you know the distribution of $b-a$ (think of $b-a$ as a single random variable, let's call it $W$).

So then your hypothesis test is just testing if $W > .1$. That's just a standard one-sided test just like if you had one sample, and the test would accordingly be $\frac{\bar{W} - .1}{s_W/\sqrt{n}}$ (compare to one-side), where $\bar{W}$ is the sample mean of $b-a$, which by linearity of sample mean, is just $\bar{b}-\bar{a}$, and similar exercise for variance.

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