How do you find a p value from a z score? You find the area under the standard normal curve beyond your z score, in the tail (or tails) your test is about. In practice that means one table lookup and one short formula: Φ(z) for a left-tailed test, 1 − Φ(z) for a right-tailed test, and 2 × (1 − Φ(|z|)) for a two-tailed test. The steps below walk through it by hand, with a full worked example, a check for the table mix-up that trips up many students, and a quick way to catch wrong answers.
Why finding a p value from a z score is hard
The arithmetic is easy. The confusion comes from three places.
First, there are three versions of the formula, and picking the wrong one gives an answer that looks perfectly reasonable. A right-tailed p value of 0.98 and a left-tailed p value of 0.02 come from the same z score.
Second, z tables aren't all the same. Some list the area to the left of z (cumulative), and some list the area between 0 and z. The NIST/SEMATECH e-Handbook of Statistical Methods, a US government reference, uses the second kind. Read one as if it were the other and every answer is off by 0.5.
Third, negative z scores need care. Many tables only print positive values, so you have to use the symmetry of the bell curve.
If you need a refresher on calculating z itself, our z score example problems with solutions cover six cases.
6 steps to find a p value from a z score
We'll use one example throughout. A juice company says its machine fills bottles with a mean of 500 ml. The population standard deviation is known to be 6 ml. A quality inspector samples 36 bottles and finds a sample mean of 497.9 ml. Is the machine underfilling?
1. Check that a z test fits
A p value from a z score is only valid if the test statistic follows a standard normal distribution. That's the case when:
- the population standard deviation (σ) is known, and
- the data are normal or the sample is large enough for the central limit theorem to apply.
The OpenStax Introductory Statistics textbook covers these conditions in its hypothesis testing chapter.
Example: σ = 6 ml is known and n = 36 is a reasonably large sample. A z test fits.
Why it matters: if σ is estimated from a small sample, you need the t distribution instead, and a z table will give a p value that is too small.
2. Calculate the z score
For a sample mean, the test statistic is:
z = (x̄ − μ₀) / (σ / √n)
Example: the standard error is 6 / √36 = 6 / 6 = 1. So z = (497.9 − 500) / 1 = −2.10.
Why it works: z measures how many standard errors the sample mean sits from the claimed mean. A z of −2.10 means the sample is 2.1 standard errors below 500 ml.
3. Pick the tail from your alternative hypothesis
Your alternative hypothesis (H₁) decides the tail, not the sign of z:
| Alternative hypothesis | Test type | P value is the area… |
|---|---|---|
| μ is less than μ₀ | Left-tailed | to the left of z |
| μ is greater than μ₀ | Right-tailed | to the right of z |
| μ ≠ μ₀ | Two-tailed | in both tails beyond z and −z |
Example: the inspector wants to know if the machine is underfilling, so H₁ says μ is less than 500. This is a left-tailed test.
Why it works: the p value answers "if H₀ were true, how likely is a result at least this extreme in the direction I care about?" The direction comes from the question, so you fix it before looking at the data.
4. Look up the area in a z table (and check which kind)
Find the row for the first decimal (2.1) and the column for the second (0.00).
Before you trust the number, check the table's heading or picture:
- Cumulative table (area to the left of z): Φ(2.10) = 0.9821
- Mean-to-z table (area from 0 to z, like NIST's): 0.4821. Add 0.5 to convert it to cumulative: 0.4821 + 0.5 = 0.9821.
Another reading example: for z = 1.87, go to row 1.8 and column 0.07. A cumulative table shows 0.9693. If your z score has more than two decimals, round it to two before you look it up; the difference in the p value is usually in the fourth decimal place. If you need more precision than that, use software instead of a printed table, since tables only give four or five digits.
Quick test: in a cumulative table, z = 0 reads 0.5000. In a mean-to-z table, z = 0 reads 0.0000.
For negative z, use symmetry: Φ(−z) = 1 − Φ(z). So Φ(−2.10) = 1 − 0.9821 = 0.0179.
Why it works: the standard normal curve is symmetric around 0, so the area to the left of −2.10 equals the area to the right of +2.10.
5. Turn the area into a p value
Apply the formula for your tail:
- Left-tailed: p = Φ(z)
- Right-tailed: p = 1 − Φ(z)
- Two-tailed: p = 2 × (1 − Φ(|z|))
Example (left-tailed): p = Φ(−2.10) = 0.0179.
If the inspector had asked "is the machine off target in either direction?", it would be two-tailed: p = 2 × (1 − 0.9821) = 0.0357.
Two more quick cases, so you can see the pattern:
| z score | Test | Calculation | P value |
|---|---|---|---|
| 1.50 | Right-tailed | 1 − 0.9332 | 0.0668 |
| 0.60 | Two-tailed | 2 × (1 − 0.7257) | 0.5485 |
| −2.10 | Left-tailed | Φ(−2.10) | 0.0179 |
| −2.10 | Two-tailed | 2 × (1 − 0.9821) | 0.0357 |
On a computer: in Excel or Google Sheets, =NORM.S.DIST(z, TRUE) returns Φ(z) (Microsoft documentation). Then apply the same formulas.
6. Compare with α and sanity-check
Compare the p value with your significance level, usually α = 0.05:
- p ≤ α: reject H₀
- p above α: fail to reject H₀
Example: 0.0179 is below 0.05, so the inspector rejects H₀. The sample gives evidence that the machine fills below 500 ml on average.
Then run two sanity checks:
- The 1.96 rule. A two-tailed z of ±1.96 gives p = 0.05 (NIST lists 1.96 as the critical value for a tail area of 0.025). If |z| is bigger than 1.96, your two-tailed p must be smaller than 0.05.
- The direction check. If z points the "wrong" way for a one-tailed test (for example, z = +2.10 when H₁ says μ is less than 500), the p value should be large, above 0.5. Here it would be 0.9821.
When a z score isn't enough
The z table method works for z tests, and only for z tests. Switch methods when:
- σ is unknown and the sample is small. Use a t test and the t distribution with n − 1 degrees of freedom.
- The data are very skewed and the sample is small. The normal approximation may not hold; a non-parametric test or exact method may fit better.
- You're testing a proportion with few successes or failures. The normal approximation to the binomial needs enough of both.
- You're running many tests at once. Some p values will fall below 0.05 by chance, so adjust for multiple comparisons.
And keep the meaning straight. The American Statistical Association's statement on p values (Wasserstein and Lazar, 2016) stresses that a p value is not the probability that the hypothesis is true, and that it doesn't measure the size or importance of an effect. A p of 0.0179 says the data would be unusual if the machine were filling correctly. It doesn't say how much it's underfilling; for that, report the sample mean or a confidence interval.
How Maths Problem can help
Hypothesis testing has a lot of small steps, and a slip in any of them (the wrong tail, a misread table, a sign error) changes the conclusion. Maths Problem solves statistics and math problems with detailed explanations, so after you've worked a question by hand you can check your answer and compare its explanation with your own working. It's most useful as a check on your work, not a replacement for it. Browse the Maths Problem blog for more worked statistics guides, or see our roundup of the best math solver apps for iPhone.
Frequently asked questions
Is the p value the same as the area under the curve?
Yes, for a z test the p value is the area under the standard normal curve in the tail or tails beyond your z score. Left-tailed tests use the area to the left, right-tailed tests the area to the right, and two-tailed tests add both tails together.
How do you find a two-tailed p value from a z score?
Take the absolute value of z, look up Φ(|z|) in a cumulative z table, subtract it from 1 to get one tail, then double it. For z = −2.10, Φ(2.10) = 0.9821, so one tail is 0.0179 and the two-tailed p value is 0.0357.
What z score gives a p value of 0.05?
For a two-tailed test, z = ±1.96 gives p = 0.05, because each tail holds 0.025. For a one-tailed test, z = 1.645 (or −1.645 for a left tail) gives p = 0.05. These are the critical values listed in the NIST standard normal table.
Can a p value be negative or greater than 1?
No. A p value is a probability, so it always falls between 0 and 1. If you get a negative number, you probably used the wrong tail formula, such as subtracting in the wrong order or forgetting the absolute value in a two-tailed test.
Should I use a z score or a t score to find the p value?
Use z when the population standard deviation is known or you're testing a proportion with a large enough sample. Use t when you estimate the standard deviation from a small sample. The t distribution has heavier tails, so it gives larger p values for the same test statistic.
Conclusion
To find a p value from a z score, confirm a z test fits, calculate z, choose the tail from your alternative hypothesis, read Φ(z) from a table you've checked is cumulative, and apply the matching formula. Then compare with α and run the 1.96 and direction checks. In our example, z = −2.10 gave a left-tailed p value of 0.0179 and a two-tailed p value of 0.0357. Work three or four practice problems this way and the method becomes routine.