Case 1
- Standard deviation (σ)
- 15
- Sample size (n)
- 30
- Z critical value (z)
- 1.96
5.3677
Open with these valuesConfidence Interval Calculator
5.3677
Result: 5.3677The interval is your sample mean plus and minus one number, and that number is what this calculator returns. With a standard deviation of 15, a sample of 30 and z = 1.96, the margin is 5.3677 — a mean of 100 becomes 94.6323 to 105.3677.
The numbers at a glance
Held fixed: Standard deviation (σ) 15.0000, Sample size (n) 30.
| Z critical value (z) | Result |
|---|---|
| 1.000 | 2.7386 |
| 1.500 | 4.1079 |
| 1.960Your value | 5.3677 |
| 2.000 | 5.4772 |
| 2.500 | 6.8465 |
| 3.000 | 8.2158 |
Worked examples
Case 2
- Standard deviation (σ)
- 10
- Sample size (n)
- 100
- Z critical value (z)
- 1.96
1.9600
Open with these valuesCase 3
- Standard deviation (σ)
- 3.2
- Sample size (n)
- 16
- Z critical value (z)
- 2.576
2.0608
Open with these valuesHow it's calculated
E = z × σ ÷ √n, interval = x̄ ± E
- StepEnter the standard deviation of the population or sample.
- StepEnter how many observations the sample holds.
- StepEnter the z value for your confidence level: 1.645, 1.96 or 2.576.
- ResultAdd the margin to your sample mean and subtract it — that is the interval.
What this number means
The 95 % describes the method, not this interval
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 output is the margin, not the two bounds
This calculator returns E; the interval is your sample mean plus and minus that number. A mean of 100 with a margin of 5.3677 runs from 94.6323 to 105.3677.
Certainty is paid for in width
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.
The z method assumes known σ or a large sample
Use z when the population standard deviation is known or the sample is large, roughly n ≥ 30. For a small sample with an unknown standard deviation the t distribution is more accurate, and this calculator does not use it.
Commonly misread
A 95 % interval has a 95 % chance of containing the true mean.
The 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.
Doubling the sample halves the margin of error.
The 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.
The sample mean has to be entered to get the margin.
E = z × σ ÷ √n contains no mean, which is why there is no field for one. The mean enters only when you place the margin around it.
Reference table
| σ, n, z | Confidence level | Margin of error |
|---|---|---|
| 15, 30, 1.645 | 90 % | 4.5050 |
| 15, 30, 1.96 | 95 % | 5.3677 |
| 15, 30, 2.576 | 99 % | 7.0547 |
| 10, 100, 1.96 | 95 %, larger sample | 1.9600 |
| 3.2, 16, 2.576 | 99 %, small sample | 2.0608 |
Questions
How do I calculate a confidence interval?
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.
Which z critical value should I use?
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.
What is a confidence interval?
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.
How do I make the interval narrower?
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.
When should I use a t value instead?
Use z when the population standard deviation is known or the sample is large, roughly n ≥ 30. For a small sample with an unknown population standard deviation the t distribution is more accurate, because its tails are heavier. This calculator uses the z method.
Sources and last check
Information, not professional advice.