# Hypothesis testing without using test statistics

The 6 customers in the supermarket have done the taste tests of hams. One type of hams is hand-made and the other is factory-made.

5 out of 6 preferred hand-made hams.

$$~p:=~$$proportion of persons who prefer hand-made hams than factory-made ones, from the population of customers.

$$\begin{cases} H_0:p= {1 \over 2 }\\ H_1:p> {1 \over 2 } \end{cases}\\~~~~~(\alpha=0.05)$$

As $$~ H_0 ~$$ is true , the probability of 5 or more persons who prefer hand-made hams is derived with $$~ \mathcal B\left(6,{1 \over 2 }\right) ~$$

\begin{align} P(X\geq5)&=p(5)+p(6)\\&= 6 \left({1 \over 2 } \right)^1 \left({1 \over 2 } \right)^5+ \left({1 \over 2 } \right)^6\\&= {6 \over 2^6 }+ {1 \over 2^6 }\\&= {7 \over 64 }\\&=0.109375 \end{align}

This probability is larger than $$~ 0.05 ~$$ hence $$~ H_0 ~$$ can't be rejected with $$~ 0.05 ~$$(significance level)

Hence the conlusion was made that it can't be said that customers prefer hand-made hams than factory-made ones.

I can't understand what the above bold statement is saying.

Fisrt things to first, as we assume that the null hypothesis is true, then we can imagine a probability of 5 or more persons(from 6 customers) prefer handmade-hams is small. And we observed the probability as $$~0.109375~$$

As the null hypothesis is really true, the probability of reject $$~H_0~$$ should be less than or equal to $$~0.05~$$

But actually $$~0.109375\not\leq 0.05~$$is held. This suggests that a probability of $$~p>{1\over2}~$$ may be higher than a probability of $$~p={1\over2}~$$

So I think that $$~H_0~$$ should be rejected but my this claim seems actually wrong.

BTW I've written this add-section using smartphone so I skipped using many mathjax codes since it is time taking while with smartphone.

• Essentially you are going to reject the null hypothesis if the probability of seeing the result you saw or an equally or more extreme result is less than or equal to $\alpha$ supposing the null hypothesis to be true. The result you in fact saw was not extreme enough and so you did not reject the null hypothesis. Jul 27, 2022 at 8:22

With a significance level of 0.05, we require that: $$\mathbb P(\text{reject} | H_0) \leq 0.05\quad (A)$$
The power of this test is given by $$\mathbb P(\text{reject} | H_1)$$. Using the number of people who prefer handmade hams in a sample of size 6 (call this $$\hat n$$) as our test statistic, we reject $$H_0$$ when $$\hat n \geq t$$ where $$t$$ is any constant such that constraint $$(A)$$ is satisfied.
Since the power is decreasing with $$t$$, we want to choose the smallest $$t$$ such that our constraint $$(A)$$ is satisfied (this maximizes our power). Let's call the optimal value of $$t$$ as $$t^*$$, so that if $$\hat n \geq t^*$$ then we reject the null hypothesis.
The work in your answer shows that $$t=5$$ does not satisfy constraint $$(A)$$, since the probability of $$\hat n \geq 5$$ is larger than our significance level of $$0.05$$. Indeed, the only value of $$t$$ that satisfies the constraint is $$t=6$$. Since in your example we have $$\hat n = 5$$, then we cannot reject the null hypothesis.