Null and Alternative Hypotheses

Two claims about the population

A hypothesis test weighs two claims about a population parameter, such as a proportion p or a mean $\mu$μ. The null hypothesis, $H_0$H₀, says there is no effect: the coin is fair, the mean is what it always was. The alternative hypothesis, $H_1$H₁, says there is an effect.

Both are written in symbols, about the parameter. For a coin suspected of landing heads too often, $H_0$H₀: p = 0.5 and $H_1$H₁: p > 0.5. For a machine that fills 500 g bags, suspected of underfilling, $H_0$H₀: $\mu =500$μ = 500 and $H_1$H₁: $\mu <500$μ < 500. For a die suspected of showing six too often, $H_0$H₀: $p=\frac{1}{6}$p = 1/6 and $H_1$H₁: $p>\frac{1}{6}$p > 1/6.

The claim the test is looking for evidence of, the bias or the underfilling, always goes in $H_1$H₁. The null is the position held until the data argue against it.

Why the null has one value

$H_0$H₀ names a single value of the parameter, and that is what makes it testable. If p = 0.5, the number of heads in 10 tosses is X ~ B(10, 0.5), one exact distribution, so the chance of any result can be worked out.

$H_1$H₁: p > 0.5 names no single value. It covers p = 0.51 and p = 0.9 alike, and gives no one distribution to work with. So a test assumes $H_0$H₀ and asks how surprising the data would be if it were true.

If $H_0$H₀: p = 0.5 is true, each of the 1,024 sequences of 10 tosses is equally likely. The bars count the sequences giving 0 to 10 heads: 252 give 5, and only 10 + 1 = 11 give 9 or more.

Evidence against the null

The coin lands heads 9 times in 10. If it is fair, the chance of 9 or more heads is $\frac{11}{1024}=0.0107$11/1024 = 0.0107, about 1%. A result that rare under $H_0$H₀ is evidence against it.

The usual rule is to fix a significance level before seeing the data, often 5%, and reject $H_0$H₀ if the probability of a result at least this extreme is below it. 0.0107 is below 0.05, so reject $H_0$H₀: there is evidence at the 5% level that the coin lands heads more often than tails.

Failing to reject is not proof

Suppose instead the coin lands heads 6 times. If it is fair, the chance of 6 or more heads is $\frac{210+120+45+10+1}{1024}=\frac{386}{1024}=0.377$(210 + 120 + 45 + 10 + 1)/1024 = 386/1024 = 0.377. That is not surprising, so $H_0$H₀ is not rejected.

That does not show the coin is fair. If heads had probability 0.6, the chance of 6 or more heads would be 0.633, so 6 heads fits that coin even better. The data cannot tell the two apart; they are simply not surprising enough to reject $H_0$H₀.

So the conclusion is worded with care: there is not enough evidence that the coin is biased. A test rejects $H_0$H₀ or fails to reject it, and never proves it true.

The usual mistakes

Putting the claim to be shown in $H_0$H₀. A drug company testing whether its drug works better writes $H_0$H₀: no difference and $H_1$H₁: the drug works better, because the effect needs the evidence.

Writing hypotheses about the sample. x̄ = 503 is a fact about the data, not a hypothesis. Hypotheses are about the population: $H_0$H₀: $\mu =500$μ = 500.

Saying that $H_0$H₀ is accepted or proved. Not rejecting it means only that the data were not surprising enough under it.

Answering neither. Every statement in a test takes a side: “the mean is unchanged” is $H_0$H₀, and “the mean has increased” is $H_1$H₁.