Setting up hypotheses, significance levels, critical regions for binomial tests.
Statistics is about making decisions under uncertainty. A doctor wants to know if a new drug works better than a placebo. A manufacturer needs to check if a machine is producing items within specification. A teacher wants to know if a new teaching method improves results.
Hypothesis testing gives us a rigorous framework to answer these questions. Instead of guessing, we use probability to decide whether the data provides sufficient evidence to support a claim. This is one of the most widely-used tools in science, medicine, and business.
This lesson covers the foundations — setting up hypotheses correctly and carrying out binomial tests. In the next lesson, we extend these ideas to tests involving the normal distribution.
Every hypothesis test starts with two competing statements:
For example, if a coin is claimed to be biased towards heads:
H0:p=0.5H1:p>0.5
Use the explorer below to see how the significance level and test type affect the critical region:
Remember: we never say "accept H₀". We say "there is insufficient evidence to reject H₀". Absence of evidence is not evidence of absence.
The null hypothesis always contains an equals sign. The alternative hypothesis contains =, >, or <.
The significance level (α) is the probability of wrongly rejecting H0 when it is actually true (a Type I error).
Common significance levels are 1%, 5%, and 10%. A 5% significance level means we are willing to accept a 5% chance of incorrectly rejecting H0.
| Type | H1 | When to use |
|---|---|---|
| One-tailed (upper) | p>k | Testing if parameter has increased |
| One-tailed (lower) | p<k | Testing if parameter has decreased |
| Two-tailed | p=k | Testing if parameter has changed (either direction) |
Key difference: For a two-tailed test at 5% significance, you split the 5% between both tails — so each tail has 2.5%.
A manufacturer claims that 20% of their products are defective. A quality inspector believes the true rate is higher. State suitable hypotheses.
H0:p=0.2H1:p>0.2
This is a one-tailed test (upper tail) because the inspector specifically believes the rate is higher.
Answer at least 3 of 3 correctly to complete this section.
Use a binomial hypothesis test when:
Under H0, the test statistic X (number of successes) follows X∼B(n,p0), where p0 is the value stated in H0.
A fairground game claims that the probability of winning is 0.25. Out of 20 attempts, a player wins 9 times. Test at the 5% significance level whether the probability of winning is higher than claimed.
Step 1: H0:p=0.25, H1:p>0.25
Step 2: α=0.05 (one-tailed, upper tail).
Step 3: Under H0, X∼B(20,0.25).
Step 4: We need P(X≥9) under H0.
P(X≥9)=1−P(X≤8)=1−0.9591=0.0409
Step 5: 0.0409<0.05, so the result is significant.
Step 6: There is sufficient evidence at the 5% significance level to reject H0. The data suggests the probability of winning is higher than 0.25.
Exam tip: Show the comparison clearly — write ”0.0409<0.05 so reject H0” — and then give your conclusion in context.
A company claims that 30% of its customers are satisfied. A survey of 15 customers finds that 2 are satisfied. Test at the 5% level whether the satisfaction rate is lower than claimed.
H0:p=0.3, H1:p<0.3, α=0.05 (one-tailed, lower tail).
Under H0, X∼B(15,0.3). We observe X=2.
P(X≤2)=P(X=0)+P(X=1)+P(X=2)
=0.0047+0.0305+0.0916=0.1268
Since 0.1268>0.05, the result is not significant.
There is insufficient evidence at the 5% level to reject H0. The data does not provide sufficient evidence that the satisfaction rate is lower than 30%.
A gardener claims that 40% of seeds of a certain type germinate. From a packet of 20 seeds, find the critical region for a two-tailed test at the 10% significance level. The gardener then plants 20 seeds and 3 germinate. State your conclusion.
H0:p=0.4, H1:p=0.4, α=0.10 (two-tailed, so 5% in each tail).
Under H0, X∼B(20,0.4).
Lower tail: We need the largest c1 such that P(X≤c1)≤0.05.
P(X≤3)=0.0160 and P(X≤4)=0.0510.
So c1=3 (since 0.0160≤0.05 but 0.0510>0.05).
Upper tail: We need the smallest c2 such that P(X≥c2)≤0.05.
P(X≥12)=0.0565 and P(X≥13)=0.0210.
So c2=13 (since 0.0210≤0.05 but 0.0565>0.05).
Critical region: X≤3 or X≥13.
Since X=3 falls in the critical region (X≤3), we reject H0.
There is sufficient evidence at the 10% level to suggest that the germination rate differs from 40%.
Every conclusion should include:
Never say: “The mean has definitely decreased” or “We have proved that p>0.3.” Hypothesis tests provide evidence, not proof.
| Trap | How to avoid it |
|---|---|
| Writing “accept H0“ | Write “do not reject H0” or “insufficient evidence to reject H0“ |
| Conclusion not in context | Refer back to the scenario (seeds, bottles, scores, etc.) |
| Wrong tail | Re-read H1 — the direction tells you which tail |
| Forgetting to halve α for two-tailed | If H1 has =, divide α by 2 for each tail |
| Using P(X=k) instead of P(X≤k) or P(X≥k) | Always find the cumulative probability in the direction of the tail |
Answer at least 4 of 5 correctly to complete this section.
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