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Consider this problem in Montgomery's book:

Two types of plastic are suitable for use by an electronics component manufacturer. The breaking strength of this plastic is important. It is known that $\sigma_1=\sigma_2=1$ psi. From a random sample of size $n_1=10$ and $n_2=12$, we obtain $\bar x_1=162.5$ and $\bar x_2=155.0$. The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi.

(a) Based on the sample information, should it use plastic 1?

In the Solution manual, the hypotheses of the test are:

$H_0: \mu_1-\mu_2=10$ against $H_1:\mu_1-\mu_2>10$. Rejecting the null hypothesis means evidence to support the use of plastic 1.

In my interpretation, the company has to decide whether $\mu_1-\mu_2\geq 10$ holds ("at least 10"). It seems to me that both hypotheses corresponds to $\mu_1-\mu_2\geq 10$.

Question Why the alternative hypothesis is not $H_1:\mu_1-\mu_2<10$ ?

Celine Harumi
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  • "The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi." To be pedantic, in the context of hypothesis testing this statement is actually not enough to make a decision. The statistical analysis gives only a probability (or a confidence) and it can not answer the question conclusively whether the mean breaking strength of plastic 1 exceeds the mean breaking strength of plastic 2 with 10 psi. – Sextus Empiricus Dec 04 '22 at 23:31

2 Answers2

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If the null hypothesis is $\mu_1-\mu_2<10$, then rejecting the null indicates evidence that plastic 1 is not sufficiently strong to warrant use. This means that failing to reject the null puts you in a position where you lack evidence against the strength of plastic 1. However, this is not the same as having evidence against the strength of plastic 1; it means that you don’t know.

The hypothesis given as correct allows for appropriate skepticism about plastic 1. Unless you have considerable evidence that plastic 1 is stronger by at least the required 10 psi, the company will continue using plastic 2.

Dave
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  • "If the null hypothesis is $\mu_1−\mu_2<10$, then rejecting the null indicates evidence that plastic 1 is not sufficiently strong to warrant use." Shouldn't it be the other way around? "If the null hypothesis is $\mu_1−\mu_2<10$, then rejecting the null indicates evidence that plastic 1 is sufficiently strong to warrant use." – Sextus Empiricus Dec 04 '22 at 23:40
  • If $\mu_1=15$ and $\mu_2=10$ then $\mu_1-\mu_2=5$, meaning that plastic $1$ is not sufficiently strong to warrant use. @SextusEmpiricus – Dave Dec 04 '22 at 23:49
  • Yes $\mu_1 - \mu_2 = 5$ is not strong enough, but it wouldn't mean rejecting the null $\mu_1-\mu_2<10$. (on the other hand it can mean rejecting the null $\mu_1-\mu_2=10$) – Sextus Empiricus Dec 05 '22 at 00:06
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$H_0: \mu_1-\mu_2=10$ against $H_1:\mu_1-\mu_2>10$. Rejecting the null hypothesis means evidence to support the use of plastic 1.

In my interpretation, the company has to decide whether $\mu_1-\mu_2\geq 10$ holds ("at least 10"). It seems to me that both hypotheses corresponds to $\mu_1-\mu_2\geq 10$.

You could also formulate the hypothesis as

$H_0: \mu_1-\mu_2<10$ against $H_1:\mu_1-\mu_2 \geq 10$

effectively this results in the same criterium for rejecting the null hypothesis. That criterium is when $\bar{x}_1 - \bar{x}_2$ exceeds a certain value.

Why are they the same?

When you test a null hypothesis against some alternative then you choose a rejection criterium in such a way that it is sensitive to rejection when the alternative is true.

For a given type I error (false negative: rejecting the null when it is actually true) you want to minimize the type II error (false positive: the error of not rejecting when the alternative hypothesis is actually true).

When you test the hypothesis $H_0: \mu_1-\mu_2=10$ then you could have some two sided rejection criterium or a one sided rejection criterium. The one sided criterium has a sharper bound and will be more sensitive when the alternative hypothesis is true.

See for instance the graph from this question: Hypothesis testing: why $\mu > 0$ (or even $\mu > \epsilon$) "seems easier” to substantiate than $\mu \neq 0$?

power

For a given probability of rejecting the null hypothesis when it is actually true (in the image the 0.05 level where the two curves intersect), the one sided test is more sensitive when the true value of the parameter is on one side.

Sextus Empiricus
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