How to Know What the Null Hypothesis Is

Q: What is the null hypothesis for a paired t-test?

H₀: μ_d = 0, where μ_d is the mean of the population of differences. A paired design produces one sample of differences, so there is one parameter, not two. Writing H₀: μ₁ = μ₂ describes independent samples and leads to the wrong standard error.

By Jude Wallis · Published July 31, 2026

The null hypothesis sets a population parameter equal to one specific value: the value a claim states, a historical or theoretical value, or 0 when the question is whether any difference exists. Identify the parameter first, then that number, then write them equal.

Writing hypotheses is the "Setting Up a Test" step, which appears once per procedure in the Fall 2026 AP Statistics course: Unit 3 topics 3.5, 3.12, and 3.14 for categorical data, and Unit 4 topics 4.4 and 4.9 for means.

Three steps that work every time

The null hypothesis, written $H_0$ ("H-naught"), is the unexciting claim: the parameter equals one specific value, and there is nothing new to report. Finding it is a procedure, not an act of judgement.

Name the parameter. Read the question and decide what population quantity it is about. A percentage of a population is a proportion $p$ . An average is a mean $\mu$ ("mu"). A comparison of two groups is a difference, $p_1 - p_2$ or $\mu_1 - \mu_2$ . Before-and-after measurements on the same subjects give a mean difference $\mu_d$ .
Find the number it is compared to. Sometimes the problem states it ("the company claims 30%"). Sometimes it is a historical or published value. When the question is simply whether two groups differ, or whether a treatment does anything at all, that number is 0.
Set them equal. Write the parameter, an equals sign, and the number. That is the null. Whatever direction the question is really asking about goes into the alternative, $H_a$ .

The direction matters for $H_a$ only. "Is the new drug better?" and "Is the new drug different?" have the same null, $\mu_1 - \mu_2 = 0$ , and different alternatives. Pick one-sided or two-sided from the wording, and pick it before you look at the data. Null vs alternative hypothesis covers writing the pair together.

The null is about a parameter, never a statistic

This is the error that costs the most points, and it is easy to check for.

A parameter describes the population: $p$ , $\mu$ , $\sigma$ . A statistic describes your sample: $\hat{p}$ ("p-hat"), $\bar{x}$ ("x-bar"), $s$ . Hypotheses are always about parameters.

Writing $H_0: \hat{p} = 0.30$ is not a slightly worse version of the right answer; it is a claim with no content. You measured $\hat{p}$ . You know what it equals. There is nothing left to test. The unknown quantity, the one the whole test exists to reason about, is the population value $p$ . See parameter vs statistic if the distinction is still slippery.

The same logic explains the equals sign. The null has to name a single value so that there is one sampling distribution to compute a probability against. $H_0: p \geq 0.30$ names infinitely many values and pins down nothing, so it is not usable even though it reads sensibly in English. Why we assume the null hypothesis is true works through why the single value is non-negotiable.

One more habit: define your symbols in words. Write "where $p$ is the proportion of all registered voters in the district who support the measure," not just "where $p$ is the proportion." Naming the population is part of the answer.

The null for every AP inference procedure

Six procedures, one pattern. Memorize the middle column and you can write the null for any question on the exam.

Procedure	Null hypothesis	What the parameter means
One proportion	$H_0: p = p_0$	$p$ is the population proportion; $p_0$ is the claimed value
Two proportions	$H_0: p_1 = p_2$ , equivalently $p_1 - p_2 = 0$	the two population proportions are equal
One mean	$H_0: \mu = \mu_0$	$\mu$ is the population mean; $\mu_0$ is the claimed value
Paired (mean difference)	$H_0: \mu_d = 0$	$\mu_d$ is the mean of the population of differences
Two means	$H_0: \mu_1 = \mu_2$ , equivalently $\mu_1 - \mu_2 = 0$	the two population means are equal
Chi-square (homogeneity or independence)	stated in words, see below	no difference between distributions, or no association

The two-group rows have a claim about a difference hidden in them. $H_0: p_1 = p_2$ and $H_0: p_1 - p_2 = 0$ are the same statement, and either is acceptable. Write it the way that matches the interval you would build, and define which group is group 1.

The paired row is the one people get wrong. A paired design has one sample of differences, not two samples, so the parameter is $\mu_d$ and the null value is 0. If you write $H_0: \mu_1 = \mu_2$ for a before-and-after study you have described the wrong procedure entirely. When to use a paired t-test shows how to tell the designs apart, and paired vs two-sample t-test puts them side by side.

If you are unsure which row you are in, which statistical test to use sorts it out from the data type and the number of samples.

Chi-square nulls are sentences, not equations

The two chi-square tests on the Fall 2026 course, homogeneity and independence, break the equals-sign pattern in appearance but not in substance.

Homogeneity. $H_0$ : the distribution of the categorical variable is the same across all of the populations or treatments. $H_a$ : at least one of those distributions differs from the others.
Independence. $H_0$ : there is no association between the two categorical variables in the population. $H_a$ : there is an association.

Write these in words. There is no single parameter to set equal to a number, because the claim covers a whole table of cell proportions at once.

The substance is unchanged though: the null still pins down one complete model. It is what produces every expected count through $E = \frac{(\text{row total})(\text{column total})}{\text{grand total}}$ , and without those counts the statistic does not exist. Two details to get right in the alternative: say "at least one" rather than "all," and do not name which category differs, because the test does not tell you. Chi-square tests explained has the full setup.

One scope note. The chi-square goodness-of-fit test was removed from the AP exam starting with the Fall 2026 course, though it is still standard college content. Homogeneity and independence remain, as Unit 3 topics 3.14 and 3.15.

Where the null value comes from

Step 2 above is the only step with any judgement in it. The number in the null arrives from one of four places.

A stated claim. "The manufacturer says 8% of chips are defective" gives $H_0: p = 0.08$ . You are testing the manufacturer's number, so it goes in the null even though it is the thing you doubt.
A historical or published value. "Last year the mean commute was 27 minutes" gives $H_0: \mu = 27$ . The test asks whether this year differs from it.
Zero, for any comparison. Two groups, a treatment effect, a before-and-after change: if the question is whether there is any difference at all, the null value is 0. No difference is exactly what "nothing new to report" means here.
A theoretical value. A fair coin gives $H_0: p = 0.5$ . A fair six-sided die gives $p = 1/6$ for a chosen face.

The null takes the value even when you expect it to be wrong. That is the design: you assume the uninteresting claim and see whether the data can knock it down. What you must never do is read the sample statistic and then set the null to something convenient. The null comes from the question, before the data.

Mistakes that cost points

Check your hypotheses against this list before moving on.

Statistics in the hypotheses. $H_0: \bar{x} = 27$ or $H_0: \hat{p} = 0.08$ . Use $\mu$ and $p$ .
An inequality in the null. $H_0: \mu \geq 27$ . The null takes an equals sign; the inequality belongs in $H_a$ .
The claim you hope to support written as the null. If you want evidence that a treatment works, "it works" is the alternative. The null is that it does nothing.
Undefined symbols. Write out what $p$ or $\mu$ refers to, naming the population.
A mismatched pair. A one-sided alternative chosen after seeing which way the sample went. Direction is fixed by the question, not by the data.
Two samples where the design is paired, or the reverse. Ask whether each observation in one group is tied to a specific observation in the other. If it is, the parameter is $\mu_d$ .
Naming a category in a chi-square alternative. "At least one proportion differs" is right; "the proportion of seniors differs" is not what the test concludes.

When the hypotheses are right, everything downstream is mostly bookkeeping. Conditions for inference is the next step, and how to write a hypothesis test conclusion is the last one.

Two proportions: writing the null and carrying it through

A researcher randomly assigns 300 volunteers to two groups of 150. In the treatment group, 45 report improvement; in the control group, 30 do. Write the hypotheses and test at $\alpha = 0.05$ whether the treatment changes the improvement rate.

Name the parameters. Let $p_1$ be the proportion of all such volunteers who would improve under the treatment, and $p_2$ the proportion who would improve under the control.
Find the null value. The question is whether the treatment changes anything, so the comparison value is 0 and the claim of no effect is that the two proportions are equal.
Write the hypotheses: $H_0: p_1 = p_2$ and $H_a: p_1 \neq p_2$ . Equivalently $H_0: p_1 - p_2 = 0$ .
Compute the sample proportions: $\hat{p}_1 = \frac{45}{150} = 0.30$ and $\hat{p}_2 = \frac{30}{150} = 0.20$ .
Because the null says the two population proportions are equal, pool the samples into one estimate of that common value: $\hat{p}_c = \frac{45 + 30}{150 + 150} = \frac{75}{300} = 0.25$ .
Check the conditions. Random assignment is given, and the pooled counts are $150(0.25) = 37.5$ and $150(0.75) = 112.5$ in each group, all at least 10.
Standard error under the null: $SE = \sqrt{\hat{p}_c(1 - \hat{p}_c)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.25(0.75)\left(\frac{1}{150} + \frac{1}{150}\right)} = \sqrt{0.1875 \times 0.0133333} = \sqrt{0.0025} = 0.05$ .
Standardize: $z = \frac{0.30 - 0.20}{0.05} = 2.00$ .
Two-sided p-value: $p = 2P(Z \geq 2.00) = 0.0455$ .
Compare to $\alpha = 0.05$ : $0.0455 \leq 0.05$ , so reject $H_0$ .

$H_0: p_1 = p_2$ against $H_a: p_1 \neq p_2$ , giving $z = 2.00$ and $p = 0.0455$ . Reject $H_0$ at $\alpha = 0.05$ : there is convincing evidence the improvement rates differ. Notice that the pooled estimate 0.25 exists only because the null said the two proportions were the same value.

Paired data: why the null is a single mean difference

Six patients have their systolic blood pressure measured before and after four weeks on a new diet. Before: 145, 152, 138, 160, 149, 155. After: 138, 147, 135, 150, 144, 148. Write the hypotheses and test at $\alpha = 0.05$ whether the diet lowers blood pressure.

Identify the design. Each 'after' reading belongs to the same patient as a 'before' reading, so the two columns are not independent samples. This is a paired design with one sample of six differences.
Name the parameter: $\mu_d$ , the mean of the population of differences (before minus after) for patients on this diet.
Write the hypotheses. Lowering blood pressure means before minus after is positive, so $H_0: \mu_d = 0$ and $H_a: \mu_d > 0$ .
Compute the differences (before minus after): 7, 5, 3, 10, 5, 7.
Sample mean difference: $\bar{x}_d = \frac{7 + 5 + 3 + 10 + 5 + 7}{6} = \frac{37}{6} = 6.1667$ mm Hg.
Sum of squared deviations from 6.1667 is 28.8333, so $s_d = \sqrt{\frac{28.8333}{5}} = \sqrt{5.7667} = 2.4014$ .
Check conditions: the patients are treated as a random sample of such patients, and with $n = 6$ the differences must be plausibly normal with no strong skew or outliers. The six differences show none.
Standard error: $SE = \frac{s_d}{\sqrt{n}} = \frac{2.4014}{\sqrt{6}} = \frac{2.4014}{2.4495} = 0.9804$ .
Test statistic: $t = \frac{6.1667 - 0}{0.9804} = 6.290$ with $df = 6 - 1 = 5$ .
One-sided p-value: $p = 0.0007$ .
Compare to $\alpha = 0.05$ : reject $H_0$ .

$H_0: \mu_d = 0$ against $H_a: \mu_d > 0$ , giving $t = 6.290$ with $df = 5$ and $p = 0.0007$ . Reject $H_0$ : there is convincing evidence the mean drop in systolic blood pressure is greater than 0. Writing $H_0: \mu_1 = \mu_2$ here would describe two independent samples and would use the wrong standard error entirely.

Frequently asked questions

Does the null hypothesis always contain an equals sign?

Yes, in every AP procedure that uses symbols. The null must name one value so that there is a single sampling distribution to compute the p-value against. Chi-square nulls are written as sentences instead, but they still specify one complete model, which is what generates the expected counts.

How do I know whether the null value is 0 or some other number?

It is 0 whenever the parameter is a difference and the question is simply whether any difference exists. It is a specific number when the problem hands you a claim, a historical value, or a theoretical one such as 0.5 for a fair coin.

Can the null hypothesis be the claim I am trying to support?

No. The claim you want evidence for goes in the alternative. The test can only produce evidence against the null, so putting your claim there means a successful test would knock it down. Failing to reject the null is never support for it.

What is the null hypothesis for a paired t-test?

$H_0: \mu_d = 0$ , where $\mu_d$ is the mean of the population of differences. A paired design produces one sample of differences, so there is one parameter, not two. Writing $H_0: \mu_1 = \mu_2$ describes independent samples and leads to the wrong standard error.

How do I write the null for a chi-square test?

In words. For homogeneity: the distribution of the categorical variable is the same for all the populations or treatments. For independence: there is no association between the two variables in the population. The alternative says at least one distribution differs, or that an association exists, without naming which category.