Hypothesis Testing: Z-Test

April 12, 2026

Problem

A company claims mean battery life is 500 hours. A sample of 36 gives mean 490, σ = 30. Test at α = 0.05.

Explanation

Setting up the hypothesis test

Step 1 — State hypotheses:

H0:μ=500H_0: \mu = 500 (battery life is as claimed)

Ha:μ500H_a: \mu \neq 500 (battery life differs from claim)

Step 2 — Significance level: α=0.05\alpha = 0.05 (two-tailed).

Computing the test statistic

Step 3 — Calculate zz:

z=xˉμ0σ/n=49050030/36=105=2.0z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{490 - 500}{30 / \sqrt{36}} = \frac{-10}{5} = -2.0

Making the decision

Step 4 — Find critical values. For two-tailed at α=0.05\alpha = 0.05: zcrit=±1.96z_{\text{crit}} = \pm 1.96.

Step 5 — Compare: z=2.0>1.96|z| = 2.0 > 1.96. The test statistic falls in the rejection region.

Step 6 — p-value: P(Z>2.0)=2×0.0228=0.0456<0.05P(|Z| > 2.0) = 2 \times 0.0228 = 0.0456 < 0.05.

Reject H0. Evidence suggests battery life differs from 500 hours.\boxed{\text{Reject } H_0. \text{ Evidence suggests battery life differs from 500 hours.}}

Interpretation

At the 5% significance level, the sample mean of 490 is significantly different from the claimed 500. The company's claim is not supported by this data.

Try it in the visualization

The normal curve shows the rejection regions (red shaded tails). The test statistic is marked. If it falls in the red zone → reject H0H_0.

Interactive Visualization

Parameters

500.00
490.00
30.00
36.00
0.05
Two-tailed
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