Hypothesis Tests for a Mean: Lesson, Examples and Practice

1Learning goal: what you will be able to do

A hypothesis test asks a yes-or-no question of data: is what we observed too unusual to blame on chance? By the end of this lesson you will be able to set up a test about a population mean, calculate a test statistic and p-value, make a decision, and explain what the decision does and does not mean.

Time55 to 60 minutes

You needPen, paper, a calculator and a normal table

AssumedSampling distributions and confidence intervals for a mean

Learning outcomes

FoundationState a null and an alternative hypothesis for a claim about a mean, and choose between a one-tailed and a two-tailed test.
CoreCalculate the test statistic and the p-value for a one-sample z test, and decide using the significance level α.
CoreUse the critical-value method, relate a test to a confidence interval, and write a conclusion in context without overstating it.
ExtensionPerform a one-sample t test, and explain Type I and Type II errors and multiple testing.

Where this lesson fits

This is the first lesson of Module 5 in Introductory Statistics. It is the second major use of the standard error: the previous lesson used it to estimate, and this one uses it to decide. The next lesson, on proportions, repeats the method for yes-or-no data.

Module 4, Lesson 1: Sampling distributions and the Central Limit TheoremOpen
Module 4, Lesson 2: Confidence intervals for a meanOpen
Module 5, Lesson 1: Hypothesis tests for a meanThis lesson
Module 5, Lesson 2: Inference for proportionsPlanned

Where this topic appears in courses

Hypothesis testing is core content in a first university statistics course, in AP Statistics (the units on inference), and in the IB Mathematics higher level options, and it underpins empirical work in economics, medicine, psychology, engineering and data science. Courses differ in whether they lead with the p-value or the critical-value method; both are taught here. This lesson supports, and does not replace, your own syllabus and instructor.

2A familiar example: the suspicious coin

A friend flips a coin to decide who pays for lunch and it lands heads. Fine. It lands heads again. Still fine. After the sixth head in a row, you start to wonder whether the coin is fair. Why six, and not two?

Predict first: how surprising is six heads in a row from a fair coin?

Work out the chance of six heads in a row if the coin is fair.

About 1.6%. Each flip is heads with probability 1/2, so six in a row has probability (1/2)⁶ = 1/64 ≈ 0.0156. That is rare, but not impossible: about one fair coin in 64 would do it. Two heads in a row (25%) would not have made you suspicious at all.

You just ran a hypothesis test in your head. You started by assuming the coin is fair (the null hypothesis). You asked how likely a result this extreme would be if that assumption were true (the p-value). When that probability was small enough, you doubted the assumption. The formal version does this carefully, with a threshold chosen in advance so that you cannot move the goalposts after seeing the data.

For means the question is: "If the population mean really were the claimed value, how unusual is the sample mean we got?" The standard error provides the yardstick. Experiment with the visualiser below.

Try it: the test visualiser

The claim is H₀: μ = 100, with a known σ = 15. Move the sample mean and the sample size. The blue area is the p-value: the probability, if H₀ is true, of a result at least as extreme as yours. The red areas are the rejection region, whose total area equals α. If the dashed line falls in red, the test rejects H₀.

Sample mean x̄ =104

Sample size n =25

Alternative

Significance level α

Things to notice: the same sample mean gives a more extreme z when n is larger, because the standard error is smaller. A larger sample can turn a non-significant result into a significant one without any change in the sample mean.

3Clear explanation

The two hypotheses

The null hypothesis H₀ is the default claim, always stated with an equals sign: H₀: μ = μ₀. The alternative hypothesis H_a is what the evidence would support if the null is rejected. It has one of three forms:

Question being asked	H_a	Type of test
Is the mean different from μ₀ (either way)?	μ ≠ μ₀	Two-tailed
Is the mean larger than μ₀?	μ > μ₀	Right-tailed
Is the mean smaller than μ₀?	μ < μ₀	Left-tailed

Choose H_a from the question before looking at the data. Switching to a one-tailed test after seeing which way the sample mean fell is a form of cheating.

The test statistic

For a mean with known σ, the test statistic counts how many standard errors the sample mean is from the hypothesised mean:

z = (x̄ − μ₀) / (σ/√n)

The p-value

The p-value is the probability, assuming H₀ is true, of getting a test statistic at least as extreme as the one observed, in the direction of H_a.

Right-tailed: p = P(Z > z) = 1 − Φ(z).
Left-tailed: p = P(Z < z) = Φ(z).
Two-tailed: p = 2 × P(Z > |z|), the area in both tails.

What a p-value is not: it is not the probability that H₀ is true, and it is not the probability that the result happened by chance. It is the probability of data this extreme given that H₀ is true. A small p-value means the data are surprising if H₀ holds.

Making the decision

Fix a significance level α before collecting data (commonly 0.05; also 0.10 or 0.01). Then:

If p < α: reject H₀. The result is statistically significant at level α.
If p ≥ α: fail to reject H₀. We do not "accept" H₀; absence of evidence against it is not evidence for it.

Equivalent critical-value method: reject H₀ if the test statistic lies in the rejection region. For α = 0.05 the critical values are ±1.96 (two-tailed), 1.645 (right-tailed) and −1.645 (left-tailed). For α = 0.01 they are ±2.576 (two-tailed) and 2.326 (one-tailed).

Writing the conclusion

State the decision, link it to the claim, and mention the level, in context. For example: "At the 5% level there is statistically significant evidence that the mean fill is different from 500 ml." Avoid "proved" and avoid treating a significant result as an important one: with a huge sample, a tiny and useless difference can be significant, so always look at the size of the effect as well.

Link with confidence intervals

A two-tailed test at level α rejects H₀: μ = μ₀ exactly when μ₀ lies outside the (1 − α) confidence interval. The interval shows which values of the mean are plausible, which is more informative than a bare yes or no.

When σ is unknown: the t test

Replace σ with the sample standard deviation s and the normal table with the t distribution with n − 1 degrees of freedom: t = (x̄ − μ₀)/(s/√n). For 15 degrees of freedom the two-tailed 5% critical value is 2.131, a little larger than 1.96.

Errors

	H₀ is true	H₀ is false
Reject H₀	Type I error, probability α	Correct decision
Fail to reject H₀	Correct decision	Type II error, probability β

The power of a test, 1 − β, is the probability of correctly rejecting a false null. Power rises with sample size, with the size of the true effect, and with α. Running many tests inflates the chance of at least one false positive, which is why unplanned searching through data for "significant" results is unreliable.

Common mistakes

Mistake	Why it is wrong	Fix
"The p-value is the probability H₀ is true"	It is computed assuming H₀ is true	Say "if H₀ were true, data this extreme would occur with probability p"
"We accept H₀"	A non-significant result is not proof of no effect	Say "fail to reject"
Forgetting to double for a two-tailed test	Both tails count as extreme	Two-tailed p = 2 × the tail area
Choosing H_a after seeing the data	It inflates the chance of a false positive	Decide H₀, H_a and α first
Equating significant with important	Large samples detect trivial differences	Report the estimate and an interval too
Using σ instead of σ/√n	That tests an individual, not a mean	Always use the standard error

4Worked examples

Read each example with a pencil in hand. Cover the steps, attempt the first line yourself, then compare. Every test follows the same five steps: hypotheses, test statistic, p-value or critical value, decision, conclusion in context.

Example 1 Foundation 3 marks: writing the hypotheses

State H₀ and H_a for each situation. (a) A bakery says its loaves weigh 500 g on average, and an inspector suspects they weigh less. (b) A machine is set to cut 50 mm parts, and a technician wants to know whether it has drifted. (c) A school claims the average class test score is below 60, and a parent wants to challenge it by showing it is higher.

(a) H₀: μ = 500. H_a: μ < 500 (left-tailed, because the suspicion is "less").
(b) H₀: μ = 50. H_a: μ ≠ 50 (two-tailed, because drift could be in either direction).
(c) H₀: μ = 60. H_a: μ > 60 (right-tailed, because the challenge is that it is higher).

Answer: (a) left-tailed, (b) two-tailed, (c) right-tailed. The null always has the equals sign.

Example 2 Core 5 marks: a right-tailed z test

A manufacturer claims its light bulbs last 1,000 hours on average. The standard deviation is known to be 100 hours. A buyer tests 36 bulbs and finds a mean of 1,030 hours. Is there evidence at the 5% level that the mean life is greater than 1,000 hours?

Hypotheses. H₀: μ = 1000. H_a: μ > 1000. α = 0.05.
Test statistic. SE = 100/√36 = 16.67. z = (1030 − 1000)/16.67 = 1.80.
p-value. P(Z > 1.80) = 1 − 0.9641 = 0.0359.
Decision. 0.0359 < 0.05, so reject H₀.
Conclusion. At the 5% level there is statistically significant evidence that the mean bulb life exceeds 1,000 hours.

Answer: z = 1.80, p = 0.036, reject H₀. Note that at the 1% level the same data would not be significant, since 0.036 > 0.01. The level must be chosen before looking.

Example 3 Core 5 marks: a two-tailed z test

A filling machine should put 500 ml in each bottle, with a known σ of 4 ml. A random sample of 64 bottles has a mean of 499 ml. Test at the 5% level whether the mean fill differs from 500 ml.

Hypotheses. H₀: μ = 500. H_a: μ ≠ 500. α = 0.05.
Test statistic. SE = 4/√64 = 0.5. z = (499 − 500)/0.5 = −2.00.
p-value. Two-tailed: 2 × P(Z > 2.00) = 2 × 0.02275 = 0.0455.
Decision. 0.0455 < 0.05, so reject H₀.
Conclusion. At the 5% level there is statistically significant evidence that the mean fill differs from 500 ml; it appears to be lower.

Answer: z = −2.00, p = 0.0455, reject H₀ at 5% (but not at 1%).

Example 4 Core 4 marks: the critical-value method

A left-tailed test at α = 0.05 gives z = −1.20. Decide using the critical value, and find the p-value to confirm.

The left-tailed 5% critical value is −1.645. Reject H₀ only if z < −1.645.
−1.20 is not less than −1.645, so it is outside the rejection region. Fail to reject H₀.
Confirm with the p-value: Φ(−1.20) = 0.1151, which is greater than 0.05. Same decision.

Answer: fail to reject H₀. The two methods always agree.

Example 5 Core 4 marks: linking the test to an interval

Use the bottle data from Example 3 (x̄ = 499, SE = 0.5) to build a 95% confidence interval and check that it agrees with the test.

Margin = 1.96 × 0.5 = 0.98.
Interval = 499 ± 0.98 = (498.02, 499.98).
The hypothesised mean 500 is outside the interval, so a two-tailed test at 5% rejects H₀, as found in Example 3.
The interval adds information: the true mean fill is plausibly between 498.02 and 499.98 ml, a shortfall of up to about 2 ml.

Answer: (498.02, 499.98). It excludes 500, matching the rejection. Whether a shortfall of 0 to 2 ml matters is a practical question the test cannot answer.

Example 6 Extension 5 marks: a one-sample t test

A random sample of 16 items from an approximately normal population has x̄ = 54 and s = 8. Test H₀: μ = 50 against H_a: μ ≠ 50 at the 5% level.

σ is unknown, so use t with 16 − 1 = 15 degrees of freedom.
SE = 8/√16 = 2. t = (54 − 50)/2 = 2.00.
Critical value for 15 df, two-tailed 5%: 2.131. Since |2.00| < 2.131, the statistic is not in the rejection region.
The exact p-value is about 0.064, above 0.05, which agrees.
Conclusion. Fail to reject H₀: at the 5% level the data do not provide statistically significant evidence that the mean differs from 50.

Answer: t = 2.00, p ≈ 0.064, fail to reject H₀. With z and σ = 8 known, the same statistic 2.00 would have been significant (p = 0.0455): the extra uncertainty of estimating σ matters.

Example 7 Extension 3 marks: multiple tests

A researcher runs 20 independent tests at α = 0.05, and in every case the null hypothesis is actually true. What is the chance of at least one false positive?

Each test has probability 0.95 of correctly not rejecting.
All 20 correct: 0.95²⁰ = 0.3585.
At least one false positive: 1 − 0.3585 = 0.6415.

Answer: about 64%. This is why a single "significant" finding among many unplanned tests deserves caution, and why analyses should be planned and reported in full.

6Answer explanations

Attempt every question before you open these. A right answer with a wrong method is luck, so read each explanation even when you were correct. Questions 3 and 6 show their own worked solution as soon as you answer.

Question 1: writing the hypotheses

Answer: H₀: μ = 3 and H_a: μ > 3.

The null is the claim being tested and always contains the equals sign (a is reversed). The customers suspect deliveries take longer, so the alternative is one-sided and points upward; ≠ (c) would suit a question about any difference, and < (d) points the wrong way.

Question 2: the z statistic

SE = 10/√25 = 10/5 = 2. z = (52 − 50)/2 = 1.0.

A common slip is dividing by σ = 10 instead of the standard error, which gives 0.2 and understates the evidence.

Question 3: generated z statistic

Compute the standard error σ/√n, then divide the gap (x̄ − μ₀) by it. The sign matters: a sample mean below μ₀ gives a negative z. Use New numbers to practise.

Question 4: two-tailed p-value

The area beyond z = 2.00 in one tail is 0.02275. A two-tailed test counts both tails: 2 × 0.02275 = 0.0455.

Forgetting to double gives 0.0228, which would be the p-value of a one-tailed test.

Question 5: decision from a p-value

Answer: reject H₀; the result is statistically significant at the 5% level.

0.03 is smaller than 0.05, so the data would be unusual if H₀ were true. Option (b) reverses the rule. Options (c) and (d) misread the p-value as the probability that H₀ is true or false, which it is not. We also never "accept" H₀; we only fail to reject it.

Question 6: generated p-value

Find the tail area beyond |z|. Double it for a two-tailed test; use it as it stands for a one-tailed test in the direction of H_a. Use New numbers to practise both kinds.

Question 7: what a p-value means

Answer: if H₀ were true, data at least this extreme would occur with probability 0.02.

The p-value is a probability about the data, calculated assuming the null. It is not the probability that the null is true (a), nor that the alternative is true (b), and it says nothing about the size of the effect (d).

Question 8: the t statistic

SE = s/√n = 8/√16 = 8/4 = 2. t = (54 − 50)/2 = 2.0, with 15 degrees of freedom.

The arithmetic matches a z test, but the decision uses the t distribution, whose heavier tails make the same statistic less impressive.

Question 9: Type I error

Answer: rejecting a null hypothesis that is actually true; its probability is α.

It is a false alarm. Failing to reject a false null (a) is a Type II error. A small sample does not itself cause a Type I error (b), and an arithmetic mistake is simply a mistake (c).

7Short recap

A hypothesis test asks whether sample data are too unusual to be explained by chance if the null hypothesis is true.
H₀ always contains the equals sign. H_a is one-tailed (< or >) or two-tailed (≠), chosen before seeing the data.
Test statistic: z = (x̄ − μ₀)/(σ/√n). With σ unknown, use t with n − 1 degrees of freedom.
The p-value is the probability, assuming H₀ is true, of a result at least as extreme as the one observed. Double the tail area for a two-tailed test.
Reject H₀ when p < α; otherwise fail to reject. The critical-value method gives the same decision.
Failing to reject is not proof that H₀ is true, and a p-value is not the probability that H₀ is true.
A Type I error rejects a true null (probability α). A Type II error misses a real effect.
Statistical significance is not practical importance. Report an interval alongside the test, and beware many unplanned tests.

Tick what you can now do

I can state null and alternative hypotheses, one-tailed or two-tailed, for a given claim.
I can calculate a z statistic and its p-value, and decide at a stated significance level.
I can explain what a p-value does and does not mean, and write a conclusion in context.
I can describe Type I and Type II errors and run a one-sample t test.

If you ticked fewer than four, return to the matching worked example, redo it without looking, then try the generated questions again. A good next step is to read the same data through both lenses: a confidence interval and a test.

Related lessons and Encyclopedia entries

Confidence Intervals for a Mean: the previous lesson, whose intervals match the two-tailed tests here.
Sampling Distributions and the Central Limit Theorem: why the test statistic follows a normal curve.
p-value and Statistical significance: the Encyclopedia entries on these ideas.
Confidence interval and Margin of error: the companion approach to inference.
Nonresponse bias: a reason a test can be significant yet misleading.

References and further reading

Titles and publishers only. Search each site for the title to reach the exact page.

OpenStax, Introductory Statistics 2e, the chapters on hypothesis testing with one sample. Free and open textbook. https://openstax.org/
Diez, D. M., Barr, C. D. and Çetinkaya-Rundel, M., OpenIntro Statistics, the chapter on inference for numerical data. https://www.openintro.org/
Freedman, D., Pisani, R. and Purves, R., Statistics, the chapters on tests of significance. W. W. Norton.
MIT OpenCourseWare, 18.05 Introduction to Probability and Statistics, notes on null hypothesis significance testing. https://ocw.mit.edu/