Imagine you're comparing exam scores of two large classes. You know the population variance, you have plenty of data, and you want to check if a difference really exists or if it’s just random noise. The simplest tool for this situation is the Z-test. It’s often the first parametric test taught in statistics because of its clarity and practical application.
Purpose of Z-Test
The Z-test is designed to check whether a sample mean or proportion differs significantly from the population mean or a hypothesized value. It builds directly on the Central Limit Theorem, which tells us that with large sample sizes, the sampling distribution of the mean approximates a normal distribution.
In short, it answers the question:
“Is this observed difference statistically significant, or is it due to chance?”
When to Use a Z-Test
- Known population variance (σ²): If the population standard deviation (σ) is given.
- Large sample size (n ≥ 30): Ensures normal approximation holds well.
- Data type: Continuous data for mean testing, or categorical data for proportion testing.
For example, if you know the average IQ of a population is 100 with a standard deviation of 15, and you take a sample of 50 students, you can test whether their mean IQ differs significantly from 100.
Formula for Z-Test
a. Z-Test for Mean
When testing whether the sample mean differs from a population mean:
Z = (X̄ – μ) / (σ / √n)
- X̄ = sample mean
- μ = population mean (hypothesized)
- σ = population standard deviation
- n = sample size
b. Z-Test for Proportion
When testing a sample proportion against a population proportion:
Z = (p̂ – p) / √[p(1 – p) / n]
- p̂ = sample proportion
- p = population proportion (hypothesized)
- n = sample size
Example
Problem: A company claims that the average battery life of its smartphones is 20 hours. A consumer group tests 64 phones and finds the average battery life to be 19.2 hours. Assume the population standard deviation is 2 hours. Test at 5% significance whether the company’s claim holds true.
Step 1: State the Hypotheses
H₀: μ = 20 (no difference)
H₁: μ ≠ 20 (there is a difference)
Step 2: Collect Information
- Sample mean (X̄) = 19.2
- Population mean (μ) = 20
- σ = 2
- n = 64
- Significance level (α) = 0.05
Step 3: Compute Z
Z = (19.2 – 20) / (2 / √64)
= (–0.8) / (2 / 8)
= –0.8 / 0.25
= –3.2
Step 4: Compare with Critical Value
At 5% significance (two-tailed), critical Z = ±1.96.
Since –3.2 < –1.96, we reject H₀.
Step 5: Conclusion
The data provides strong evidence that the company’s claim of 20 hours is not valid. The average battery life is significantly lower.
Interpretation of p-Value
Instead of comparing Z with critical values, you can look at the p-value:
- If p ≤ α: reject H₀ (evidence against null is strong).
- If p > α: fail to reject H₀ (no significant evidence against null).
In our example, the p-value for Z = –3.2 is about 0.0014. Since this is much less than 0.05, the result is highly significant.
Merits of Z-Test
- Simple and straightforward for beginners.
- Works well for large samples.
- Widely used in practice for quick statistical checks.
Demerits of Z-Test
- Requires population standard deviation, which is often unknown.
- Not suitable for small sample sizes (then we use t-test).
At first glance, the Z-test looks like just another formula. But in reality, it’s a gateway into statistical thinking. It teaches you how to judge whether numbers carry meaning or are just noise. Next time you see averages or percentages in reports, ask yourself
“Is this difference real, or is it just chance?”
That curiosity is exactly what transforms statistics from a subject into a powerful decision-making tool.