Hypothesis testing without using test statistics

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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.

I need your help.

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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.

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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.