Hypothesis tests for the mean of a normal distribution, z-tests, interpreting results.
This continues from the binomial hypothesis testing lesson. There you tested proportions using the binomial distribution. Now we move to testing means using the normal distribution.
When a population is normally distributed with known variance, we can test whether a sample provides evidence that the population mean has changed. This is the foundation of quality control, clinical trials, and scientific experiments.
When the population is normally distributed with known variance σ2, we can test hypotheses about the mean μ.
If we take a random sample of size n with sample mean Xˉ, then under H0:μ=μ0:
Xˉ∼N(μ0,nσ2)
The test statistic is:
Z=σ/nXˉ−μ0
Under H0, Z∼N(0,1).
A machine fills bottles with a mean volume of 500 ml and standard deviation 8 ml. A sample of 25 bottles has a mean volume of 497 ml. Test at the 5% significance level whether the mean volume has decreased.
H0:μ=500, H1:μ<500, α=0.05 (one-tailed, lower).
Z=8/25497−500=1.6−3=−1.875
The critical value for a one-tailed test at 5% is z=−1.6449.
Since −1.875<−1.6449, the test statistic falls in the critical region.
There is sufficient evidence at the 5% significance level to reject H0. The data suggests the mean volume has decreased from 500 ml.
The critical value is the boundary of the rejection region. The critical region contains all values of the test statistic that lead to rejection of H0.
| Test type | Critical region at 5% | Critical region at 1% |
|---|---|---|
| One-tailed (upper) | z>1.6449 | z>2.3263 |
| One-tailed (lower) | z<−1.6449 | z<−2.3263 |
| Two-tailed | ∥z∥>1.96 | ∥z∥>2.5758 |
The IQ scores in a population are known to follow N(100,225). A sample of 36 students from a particular school has a mean IQ of 105. Test at the 5% significance level whether this school’s mean IQ differs from the population mean.
H0:μ=100, H1:μ=100, α=0.05 (two-tailed).
Z=15/36105−100=2.55=2.0
For a two-tailed test at 5%, the critical values are z=±1.96.
Since 2.0>1.96, the test statistic falls in the critical region.
There is sufficient evidence at the 5% level to reject H0. The data suggests the mean IQ at this school differs from the population mean of 100.
Two-tailed note: We compare ∣z∣=2.0 against 1.96. Equivalently, we compare z=2.0 against +1.96 (since it is positive). Both approaches work.
Answer at least 3 of 3 correctly to complete this section.
When deciding which test to use, ask yourself:
| Situation | Test type |
|---|---|
| Testing a proportion p with fixed trials | Binomial test |
| Testing a mean μ with known σ2 | Normal (Z) test |
A factory produces bolts with a mean length of 50 mm and standard deviation 2 mm. After recalibrating the machine, a sample of 40 bolts has a mean length of 50.8 mm. Test at the 1% level whether the mean length has changed.
H0:μ=50, H1:μ=50, α=0.01 (two-tailed).
Z=2/4050.8−50=0.31620.8=2.530
For a two-tailed test at 1%, the critical values are z=±2.5758.
Since ∣2.530∣=2.530<2.5758, the test statistic does not fall in the critical region.
There is insufficient evidence at the 1% significance level to reject H0. The data does not provide sufficient evidence that the mean bolt length has changed from 50 mm.
Note: At the 5% level, 2.530>1.96, so we would reject H0. The conclusion depends on the significance level chosen.
Every conclusion should include:
Never say: “The mean has definitely decreased” or “We have proved that μ=50.” Hypothesis tests provide evidence, not proof.
Answer at least 3 of 3 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.