Case 1
- Sample mean (x̄)
- 100
- Known population SD (σ)
- 15
- Sample size (n)
- 30
- Z critical value (z)
- 1.96
94.6323 – 105.3677 (±5.3677)
- Upper limit
- 105.3677
- Margin of error
- 5.3677
Change any value to update the result.
94.6323 – 105.3677 (±5.3677)
Confidence interval: 94.6323 – 105.3677 (±5.3677)With a sample mean of 100, known population σ = 15, n = 30 and z = 1.96, the calculator returns 94.6323 to 105.3677, with margin ±5.3677. This z interval requires known population σ; an estimated sample standard deviation needs a t method.
Held fixed: Sample mean (x̄) 100.0000, Known population SD (σ) 15.0000, Sample size (n) 30.
| Z critical value (z) | Confidence interval |
|---|---|
| 1.000 | 97.2614 – 102.7386 (±2.7386) |
| 1.500 | 95.8921 – 104.1079 (±4.1079) |
| 1.960Your value | 94.6323 – 105.3677 (±5.3677) |
| 2.000 | 94.5228 – 105.4772 (±5.4772) |
| 2.500 | 93.1535 – 106.8465 (±6.8465) |
| 3.000 | 91.7842 – 108.2158 (±8.2158) |
94.6323 – 105.3677 (±5.3677)
67.02 – 68.98 (±0.98)
0.8809 – 1.1671 (±0.1431)
lower = x̄ − z × σ ÷ √n; upper = x̄ + z × σ ÷ √n
This z interval requires known population σ. A sample SD and the Student-t method are not supported.
If the sampling were repeated many times, about 95 percent of the intervals built this way would capture the true mean. It does not say that the true mean lies inside this one interval with 95 percent probability.
The result gives the lower and upper limits and the ± margin. The limits are the sample mean minus and plus z × σ ÷ √n.
A higher confidence level uses a larger z: 1.645 at 90 %, 1.96 at 95 %, 2.576 at 99 %. With σ = 15 and n = 30 the margin grows from 4.5050 to 5.3677 to 7.0547.
This is a z interval using known population σ. If σ is estimated from the sample, use a t interval with n − 1 degrees of freedom; a large n alone does not make s and σ identical.
MisconceptionA 95 % interval has a 95 % chance of containing the true mean.
CorrectionThe 95 % belongs to the procedure: repeat the sampling and about 95 % of such intervals capture it. This one either contains the true mean or it does not.
MisconceptionDoubling the sample halves the margin of error.
CorrectionThe margin divides σ by √n, so halving it takes four times the sample. Going from n = 30 to n = 60 only shrinks a margin of 5.3677 to 3.7955.
MisconceptionThe sample standard deviation s is interchangeable with population σ here.
CorrectionThis z interval assumes known population σ. With sample s and unknown σ, use a t interval that accounts for the sample size.
| Mean, σ, n, z | Level | Lower to upper |
|---|---|---|
| 68, 3, 36, 1.645 | 90 % | 67.1775 to 68.8225 |
| 68, 3, 36, 1.96 | 95 % | 67.02 to 68.98 |
| 1.024, 0.337, 30, 2.326 | 98 % | 0.8809 to 1.1671 |
| 100, 15, 30, 1.96 | 95 % | 94.6323 to 105.3677 |
Find the margin of error E = z × σ ÷ √n, then subtract it from and add it to the sample mean: the interval is x̄ ± E. A mean of 100 with σ = 15, n = 30 and z = 1.96 gives a margin of 5.3677 and an interval from 94.6323 to 105.3677.
The z value follows from the confidence level: 1.645 for 90 %, 1.96 for 95 % and 2.576 for 99 %. A higher level uses a larger z and widens the interval — you trade precision for certainty.
It is a range built around a sample estimate that is likely to contain the true population value. A 95 % interval means that if the sampling were repeated many times, about 95 % of the intervals built this way would capture the true mean.
Increase the sample size: the margin divides σ by √n, so quadrupling the sample roughly halves it. Accepting a lower confidence level or reducing the variability in the data shrinks it too.
Use a t interval when population σ is unknown and you estimate it with sample s. The t critical value depends on n − 1 degrees of freedom. This calculator only uses known population σ and a z critical value.
Information, not professional advice.
Diese Seite gibt es auch auf Deutsch.
Zu Deutsch wechseln