Confidence interval calculator

Confidence interval calculator for a mean (t or z), a proportion (Wilson or Wald) or a difference of two means, with the margin of error and both bounds.

更新日 検証済みの例:7

%
試す
Margin of error (±)
Margin of error (±): 2.740646
小数点以下の最大桁数:6;最も近い値へ、等距離ならゼロから遠い値へ
Lower bound
69.659354
Upper bound
75.140646
Point estimate
72.4
Standard error
1.35
Critical value
2.030108
Degrees of freedom
35

We can be 95% confident the true mean lies between 69.6594 and 75.1406 (72.4 ± 2.7406). The confidence describes the method — about 95% of intervals built this way contain the true value — not a probability for this one interval.

95% interval for the mean: 72.4 ± 2.7406

00.10.26870727476μDensity69.659475.1406
計算方法 S
  1. Standard error

    SE=sn=8.136=1.35SE = \frac{s}{\sqrt n} = \frac{8.1}{\sqrt{36}} = 1.35
  2. Critical value

    t0.975, 35=2.030108t_{0.975,\ 35} = 2.030108
  3. Margin of error

    E=2.030108×1.35=2.740646E = 2.030108 \times 1.35 = 2.740646
  4. Interval

    72.4±2.740646=[69.659354, 75.140646]72.4 \pm 2.740646 = [69.659354,\ 75.140646]

Confidence interval calculatorについて

A confidence interval is a point estimate plus or minus a margin of error, and the margin is a critical value times the standard error. For a mean the standard error is s/√n and the critical value comes from Student's t with n − 1 degrees of freedom, or from the normal distribution when σ is known. For a proportion the Wilson score interval is the default, with the Wald formula p̂ ± z√(p̂(1 − p̂)/n) offered for comparison.

Survey results, lab measurements and A/B tests are reported this way. The default sample of 36 with mean 72.4 and standard deviation 8.1 gives SE = 1.35, t = 2.030 on 35 df and a 95% interval of 69.66 to 75.14, or 72.4 ± 2.74.

The 95% describes the procedure: across many samples, about 95% of intervals built this way contain the true value. Any single interval either contains it or does not.

計算例

Mean with sample SD, n = 36 (defaults)

Interval for
Mean, σ unknown (t)
Enter
Summary statistics
Sample 1 mean
72.4
Sample standard deviation s₁
8.1
Sample size n₁
36
Confidence level
95%
Critical value
2.030108
Margin of error (±)
2.740646
Lower bound
69.659354
Upper bound
75.140646

照合元:t₀.₉₇₅,₃₅ = 2.030 (t table); Python bisection on the A&S 26.7.3 closed form gives 2.0301079283; margin = t·8.1/6

Mean with known σ

Interval for
Mean, σ known (z)
Enter
Summary statistics
Sample 1 mean
72.4
Sample size n₁
36
Known population σ
8
Confidence level
95%
Critical value
1.959964
Margin of error (±)
2.613285

照合元:z₀.₉₇₅ = 1.959964 (z table; Python NormalDist().inv_cdf(0.975)); margin = z·8/6

Proportion 540/1000, Wilson

Interval for
Proportion
Successes x
540
Sample size n
1000
Method
Wilson score
Confidence level
95%
Lower bound
0.509015
Upper bound
0.570679
Margin of error (±)
0.030832

照合元:Wilson score formula evaluated in Python with z = NormalDist().inv_cdf(0.975)

Proportion 540/1000, Wald

Interval for
Proportion
Successes x
540
Sample size n
1000
Method
Wald (p̂ ± z·SE)
Confidence level
95%
Lower bound
0.50911
Upper bound
0.57089
Margin of error (±)
0.03089

照合元:p̂ ± z√(p̂(1−p̂)/n) in Python

よくある質問

What does a 95% confidence interval mean?

The method captures the true value in 95% of repeated samples. For the default data the interval is 69.66 to 75.14, but there is not a 95% probability that the true mean lies in that particular range: the true mean is fixed, and this interval either contains it or not. Nor does the interval hold 95% of individual values, which spread far wider (SD 8.1); that needs a prediction or tolerance interval.

How do you calculate the margin of error?

Multiply the critical value by the standard error. For a mean, E = t × s/√n: with s = 8.1 and n = 36, E = 2.030 × 1.35 = 2.74. For a proportion, E ≈ z√(p̂(1 − p̂)/n): 54% of 1,000 respondents gives 1.96 × 0.0158 ≈ 0.031, or ±3.1 percentage points. Because n sits under a square root, quadrupling the sample size halves the margin.

When should I use a z interval instead of a t interval?

Only when the population standard deviation σ is known, which is rare outside textbook problems and long-running process data. With σ estimated from the sample, use t. Its 95% critical value is larger for small samples, 2.262 for n = 10 against 1.960 for z, and approaches z as n grows: 2.030 at n = 36 and 1.984 at n = 101.

Why use the Wilson interval for a proportion?

The Wald interval p̂ ± z·SE covers the true proportion less often than stated when n is small or p̂ is near 0 or 1; Brown, Cai and DasGupta (2001) found its coverage can fall far below 95% even with hundreds of observations. Wilson stays close to the stated level and never leaves the range 0 to 1. With 0 successes in 20 trials Wald gives the zero-width interval [0, 0], while Wilson gives 0 to 0.161.

If a confidence interval for a difference excludes 0, is the result significant?

Yes, for the matching test: a 95% interval for a difference in means excludes 0 exactly when a two-sided test at α = 0.05 using the same method (Welch or pooled) rejects no difference. The default two-group summaries give 3.5 ± 4.0, from −0.5 to 7.5, which includes 0, so the difference is not significant at 5%. The interval also shows how large the effect could plausibly be, which a p-value does not.

「Confidence interval calculator」の精度はどのくらいですか?

精度は入力値と計算方法の前提に依存します。十進演算には有効数字50桁を使いますが、推定、数値計算手法、元データの精度はそれより低い場合があります。表示の丸め処理でこれらの制約がなくなるわけではありません。 独立した出典の解答と照合した計算例:7。 例えば、「Mean with sample SD, n = 36 (defaults)」はt₀.₉₇₅,₃₅ = 2.030 (t table); Python bisection on the A&S 26.7.3 closed form gives 2.0301079283; margin = t·8.1/6と照合しています。

この計算方法の出典は何ですか?

NIST/SEMATECH e-Handbook of Statistical Methods, §7.2.2.1 Confidence limits for the mean; NIST/SEMATECH e-Handbook, §7.2.4.1 Confidence intervals for a proportion (Wilson and normal approximation); Brown, Cai & DasGupta (2001). Interval estimation for a binomial proportion. Statistical Science 16(2), 101–133.

この計算機について

xˉ±t1−α/2, n−1sn,Wilson: p^+z22n±zp^(1−p^)n+z24n21+z2/n\bar x \pm t_{1-\alpha/2,\,n-1}\frac{s}{\sqrt n},\qquad \text{Wilson: } \frac{\hat p + \frac{z^2}{2n} \pm z\sqrt{\frac{\hat p(1-\hat p)}{n} + \frac{z^2}{4n^2}}}{1 + z^2/n}

出典

  1. NIST/SEMATECH e-Handbook of Statistical Methods, §7.2.2.1 Confidence limits for the mean
  2. NIST/SEMATECH e-Handbook, §7.2.4.1 Confidence intervals for a proportion (Wilson and normal approximation)
  3. Brown, Cai & DasGupta (2001). Interval estimation for a binomial proportion. Statistical Science 16(2), 101–133

出典と照合済み

この計算機には、独立した出典の解答を使った計算例が 7 件あります。テストに組み込まれており、ここでも実行できます。

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