What is the point estimate?
Type successes and trials, your data, or a summary (size, mean and standard deviation). The point estimate calculator gives the best single estimate of the population value and the confidence interval around it.
- Point estimate
- 0.13
The point estimate is 0.13, with a 95% confidence interval from 0.0902820228 to 0.1836635187.
- Lower limit
- 0.0902820228
- Upper limit
- 0.1836635187
- Margin of error
- 0.04669074796
- Wilson center
- 0.1369727708
- Standard error
- 0.0237802439
- Critical value
- 1.95996
- Sample size
- 200
Point estimate: 0.13. The point estimate is 0.13, with a 95% confidence interval from 0.0902820228 to 0.1836635187.
How to calculate
Computes the point estimate of a proportion (successes ÷ trials) with its Wilson confidence interval, or of a mean from data or a summary with its t confidence interval and margin of error.
Example with the default inputs (Estimate Proportion, Successes (x) 26, Trials (n) 200, Confidence level 95%): The point estimate is 0.13, with a 95% confidence interval from 0.0902820228 to 0.1836635187.
Method: Proportion: p̂ = x ÷ n; Wilson limits (p̂ + z²/2n ± z√(p̂(1 − p̂)/n + z²/4n²)) ÷ (1 + z²/n). Mean: x̄ ± t(1 − α/2, N − 1) × s ÷ √N.
- The sample is random and the trials or values are independent.
- The t interval assumes the values come from a roughly normal population, or a large sample.
- z and t are the two-sided critical values for the confidence level: α = 1 − level, upper quantile α/2.
Worked examples
Each example is checked against the calculator on every build.
- Estimate Proportion, Successes (x) 26, Trials (n) 200, Confidence level 90% gives Point estimate 0.13, Lower limit 0.095773.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §7.2.4.1 Confidence intervals (p̂ = N_d ÷ N; the Wilson score limits). https://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm (p̂ = 0.13, N = 200: p ≥ 0.09577)
- Estimate Mean from summary, Sample size (N) 195, Sample mean (x̄) 9.26146, Sample standard deviation (s) 0.022789, Confidence level 95% gives Point estimate 9.26146, Lower limit 9.258241, Upper limit 9.264679.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.5.2 Confidence Limits for the Mean (Ȳ ± t₁₋α/₂,N₋₁ × s ÷ √N). https://www.itl.nist.gov/div898/handbook/eda/section3/eda352.htm (N = 195, mean 9.261460, s = 0.022789, t = 1.9723: interval 9.258242 to 9.264679)
- Estimate Mean from data, Data 12.1, 11.8, 12.4, 12, 12.2, Confidence level 95% gives Point estimate 12.1, Standard deviation (s) 0.223607, Margin of error 0.277645.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.5.2 Confidence Limits for the Mean (Ȳ ± t₁₋α/₂,N₋₁ × s ÷ √N). https://www.itl.nist.gov/div898/handbook/eda/section3/eda352.htm
How it works
With confidence level C (in percent), α = 1 − C ÷ 100.
Proportion (x successes in n trials, x ≤ n):
- Point estimate p̂ = x ÷ n.
- z = the upper α/2 quantile of the standard normal distribution (1.959964 at 95%).
- Wilson limits (NIST): (p̂ + z²/(2n) ± z √(p̂(1 − p̂)/n + z²/(4n²))) ÷ (1 + z²/n), each held to the range 0 to 1.
- Wilson center = (p̂ + z²/(2n)) ÷ (1 + z²/n); margin of error = z √(p̂(1 − p̂)/n + z²/(4n²)) ÷ (1 + z²/n).
- Standard error = √(p̂(1 − p̂) ÷ n).
- More successes than trials has no answer.
Mean (from data, or from a summary of size N, mean x̄ and standard deviation s):
- From data: N = the number of values, x̄ = their mean, s = the sample standard deviation (N − 1 in the denominator).
- Point estimate = x̄.
- t = the upper α/2 quantile of Student's t with N − 1 degrees of freedom.
- Standard error = s ÷ √N; margin of error = t × s ÷ √N; limits = x̄ ± margin.
Every value shows to 10 significant figures (the critical value to 6), rounded half up. All arithmetic is in double precision.
Assumptions
- A random sample of independent trials or values. The t interval assumes a roughly normal population or a large sample.
- Confidence from 50% to 99.99%; trials up to 10⁹; data from 2 to 10,000 values.
Worked examples by hand
26 defects in 200 items, 90% (NIST). p̂ = 26 ÷ 200 = 0.13; z = 1.644854; z²/n = 0.013528; center = (0.13 + 0.006764) ÷ 1.013528 = 0.134938; half-width = 1.644854 × √(0.0005655 + 0.0000169) ÷ 1.013528 = 0.039165; lower = 0.09577, NIST's one-sided 95% limit.
NIST's 195 measurements, 95%. x̄ = 9.261460; s ÷ √195 = 0.022789 ÷ 13.964 = 0.0016320; t = 1.97227; margin 0.0032187; interval 9.258241 to 9.264679.
Five measurements: 12.1, 11.8, 12.4, 12.0, 12.2, 95%. x̄ = 60.5 ÷ 5 = 12.1; squared deviations 0, 0.09, 0.09, 0.01, 0.01, sum 0.2; s = √(0.2 ÷ 4) = 0.2236; standard error 0.2236 ÷ √5 = 0.1; t(0.975, 4) = 2.776445; margin 0.27764.
Other questions people ask
What is a point estimate?
It is one number from a sample used as the best guess of a population value. The sample proportion p̂ = x ÷ n estimates a population proportion, and the sample mean x̄ estimates a population mean.
How do I calculate the point estimate of a proportion?
Divide the successes by the trials: p̂ = x ÷ n. If 26 of 200 items are defective, p̂ = 26 ÷ 200 = 0.13.
How do I find the point estimate from a confidence interval?
For a mean, the point estimate is the middle of the interval: (lower + upper) ÷ 2, and the margin of error is (upper − lower) ÷ 2. A Wilson interval for a proportion is not centred on p̂, so use x ÷ n instead.
Why does the proportion use the Wilson interval?
The simple interval p̂ ± z√(p̂(1 − p̂)/n) can run below 0 or above 1 and is too narrow for small samples. NIST recommends the Wilson score interval, which always stays between 0 and 1.
What is the difference between the point estimate and the margin of error?
The point estimate is the single best guess; the margin of error says how far the true value may be from it at the chosen confidence level. The interval is the estimate minus and plus the margin (for a mean).
Should I use z or t for a mean?
When the population standard deviation is unknown and you use the sample standard deviation s, use t with N − 1 degrees of freedom, as NIST does. For large samples t is close to z.