# Why reject, not accept $H_0$ when $p$ value less than $\alpha$? [duplicate]

Let's say ABC drug company claims their medicine cures 90% of patients. Now a doctor conducts a trial by taking a random sample of 15 patients. Result shows 11 of them are cured but 4 are not.

In this case, $H_0$ should be: $p=0.9$. And $H_1$ is: $p<0.9$. If we use $X$ to represent the number of people cured in the sample, then $X \sim B(15,0.9)$.

If significance level ($\alpha$) is set as 5%, the critical region in this case should be defined as:

$P(X<11)<\alpha$ where $\alpha=5\%$ , $H_0$ will be rejected if the expression is true.

This really confuses me. The smaller of $P(X<11)$ is, the larger of $P(X>11)$ should be. Then the more patients should be cured and more likely H0 should be accepted. Anything wrong with this?

To me, it seems to make more sense if $P(X>11) < \alpha$ proves to be true to reject $H_0$.

• I'd say that it actually isn't a duplicate of that other question since this questions is specifically concerned with why we don't accept the null. The only thing I could see in that other question touching on what we accept, what we reject, why it matters that was that we "accept the alternative hypothesis" (which I have huge issues with) and that we don't accept the null. But no explanation of why. I'd like to answer this question but I'll keep my answer in my mouth until this is decided on. – Nick Thieme Jun 29 '15 at 19:42
• For your stated $H_0$, $H_1$ is $p \neq 0.9$, not $H_1 < 0.9$. And so your following statements are not correct. – Affine Jun 29 '15 at 19:59
• @Affine Most people would understand such a claim to mean that the medicine cures at least 90% of patients, not exactly 90%. They would run afoul of truth-in-advertising laws when the rate is less than 90%, but not when the rate is greater than 90%. – whuber Jun 29 '15 at 20:17
• @whuber Right - which is where I suspected some of the confusion prompting the question comes from, and I wanted to point out the inconsistencies there. – Affine Jun 29 '15 at 20:23
• – Jason Sanchez Jun 29 '15 at 21:05

You are using your data to set your rejection region. This is incorrect.

If you want to test $H_0: p=0.9$ vs. $H_a: p<0.9$ using an iid sample of 15, then the cutoff is determined as follows:

1. Let $X_i=1$ if patient $i$ is cured, $0$ o/w
2. Assume $X_i \sim Ber(0.9) \implies T:=\sum X_i \sim Bin(15,0.9)$
3. We want a lower cutoff, so how few people can be cured and still be acceptable with $p=0.9$? From a binomial probability table, we get:

$$P(T\leq 11)=0.0556, P(T\leq 10) = 0.0127$$

Notice that you can't get your desired $5\%$ significance exactly due to discreteness.

Therefore, you will reject at the $5.56\%$ level if $T\leq 11$. In this case, it is, so you would reject. There is only approximately a $5\%$ chance that this few people would have been cured had the actual cure rate been $90\%$.

Now you can see that having more people cured would make you less likely to reject.

We know that $$P(\text{Type I Error}) = P(\text{rejecting} \ H_0|H_{0} \ \text{is true})$$

$$= 1-\sum_{x=x_c}^{15} \binom{15}{x} \left(\frac{9}{10} \right)^{x} \left(\frac{1}{10} \right)^{15-x}$$

For $\alpha = 0.05$, the rejection region is $\{x: x \leq 11 \}$. The p-value is the probability that we would get a result as extreme or more extreme than we did if the null hypothesis is true. This is:

$$\text{p-value} = 1-\sum_{x=11}^{15} \binom{15}{x} \left(\frac{9}{10} \right)^{x} \left(\frac{1}{10} \right)^{15-x} \approx 0.01$$