- Sample mean (x̄)
- 52
- Hypothesised population mean (μ)
- 50
- Sample standard deviation (s)
- 5
- Sample size (n)
- 25
2.000000
Open with these values2.000000
Result: 2.000000The t-statistic counts how many standard errors your sample mean lies from a hypothesised mean. The divisor is the standard error s ÷ √n, not the standard deviation itself — that square root is what makes a large sample more convincing than a small one. Judge it against a t-table with n − 1 degrees of freedom.
2.000000
Open with these values1.825742
Open with these values-2.400000
Open with these valuest = (x̄ − μ) ÷ (s ÷ √n)
The divisor is s ÷ √n. A mean of 105 against 100 with s = 15 and n = 30 gives 5 ÷ 2.738613 = 1.825742; dividing by 15 itself would return 0.333333.
A one-sample test scores t against n − 1 degrees of freedom, so n = 25 gives 24. The shape of the t-distribution changes with that number, and the critical value changes with it.
For a two-tailed test at 24 degrees of freedom the critical value is about 2.06, so a t of 2 falls just short. A larger magnitude is stronger evidence against the null hypothesis.
A z-score uses a known population standard deviation, while t uses the sample one in its place. That extra uncertainty gives the t-distribution heavier tails and ties it to the sample size through the degrees of freedom.
t = (x̄ − μ) ÷ s.
The divisor is the standard error s ÷ √n, not the standard deviation. The two coincide only at n = 1, and this test needs at least two observations.
A t of 2 always clears the significance bar.
That depends on the degrees of freedom. At 24 of them the two-tailed critical value is about 2.06, so a t of exactly 2 falls just short.
A negative t means something went wrong.
It only says the sample mean lies below the hypothesised one: 48 against 50 with s = 5 and n = 25 gives −2. It is the magnitude that goes into the table.
| Mean, μ, s, n | Reading | t |
|---|---|---|
| 48, 50, 5, 25 | Two standard errors below μ | -2 |
| 100, 100, 5, 9 | No difference at all | 0 |
| 52, 50, 5, 25 | Two standard errors above μ | 2 |
| 105, 100, 15, 30 | Just under two standard errors above | 1.825742 |
| 47.5, 50, 6, 36 | Two and a half standard errors below | -2.5 |
Subtract the hypothesised population mean from the sample mean, then divide by the standard error: t = (x̄ − μ) ÷ (s ÷ √n). A sample of 25 with a mean of 52, a hypothesised mean of 50 and a standard deviation of 5 gives (52 − 50) ÷ (5 ÷ √25) = 2 ÷ 1 = 2.
It is a standardised measure of how far a sample mean sits from a hypothesised value, expressed in units of standard error. It is the test statistic of the one-sample t-test and follows a Student t-distribution when the data are roughly normal. That lets you decide whether an observed difference is likely real or just sampling noise.
Because the question is about the mean, not about a single observation. The sample mean varies less than the data do, and its own spread is the standard error s ÷ √n. Dropping the square root would understate t whenever n is larger than 1.
Both count standard units away from a reference value, but a z-score uses the known population standard deviation while a t-statistic uses the sample standard deviation as an estimate. That extra uncertainty is why the t-distribution has heavier tails than the normal curve, and why it depends on the sample size through its degrees of freedom.
For a one-sample t-test they equal n − 1, one less than the sample size, so n = 25 gives 24. You use that number to look up the critical value in a t-table or to find the p-value. The shape of the t-distribution changes with the degrees of freedom.
Compare its absolute value with the critical value from a t-table at your significance level and your degrees of freedom. For a two-tailed test with 24 degrees of freedom the critical value is about 2.06, so a t of 2 would fall just short. A larger magnitude means stronger evidence against the null hypothesis.
Information, not professional advice.
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