One-Sample T-Test
Hypothesis test for a mean when σ is unknown, from data or summary statistics.
Results
Reject H₀: t = 1.8454, p-value = 0.0396
The p-value (0.0396) is less than or equal to α = 0.05, so the data give enough evidence to conclude that the population mean is greater than 30.
With n = 22, this test assumes the population is approximately normal.
Results
| Test statistic t | 1.8454 |
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| p-value | 0.0396 |
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| df | 21 |
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| x̄ | 32.4 |
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| n | 22 |
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| Standard error | 1.3005 |
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| Critical value(s) | 1.721 |
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| Decision at α = 0.05 | Reject H₀ |
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On the TI-84
T-Test 30, 32.4, 6.1, 22, >
Show steps
- Hypotheses. H₀: μ = 30 Ha: μ > 30
- Test statistic. t = (x̄ − μ0) ÷ (s ÷ √n) = (32.4 − 30) ÷ (6.1 ÷ √22) = 2.4 ÷ 1.3005 = 1.8454, with df = 21
- p-value. Right-tailed test: area to the right of the test statistic: p-value = 0.0396. Since 0.0396 ≤ 0.05, reject H₀.
- Critical-value method. At α = 0.05, the rejection region is t ≥ 1.721. The test statistic t = 1.8454 is in the rejection region, so we reject H₀. (Same decision as the p-value method.)
- Conclusion. The p-value (0.0396) is less than or equal to α = 0.05, so the data give enough evidence to conclude that the population mean is greater than 30.
About this calculator
When to use a t-test
This is the usual test for one mean: you have a sample, its mean x̄ and standard deviation s, and a claimed value μ0. The test statistic follows a t-distribution with df = n − 1.
Conditions
The data should come from a random sample. If n is less than about 30, the population should be roughly normal: make a histogram or boxplot and look for strong skew or outliers.
Writing the conclusion
Always answer in context. If you reject H0, there is enough evidence for the alternative claim. If you fail to reject, there isn't enough evidence, which is not the same as proving H0 is true.
More inference calculators: CI for a Mean · CI for a Proportion · Sample Size · Z-Test (mean) · 1-Prop Z-Test · 2-Sample T-Test · Paired T-Test · 2-Prop Z-Test