# 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.

परस्परसंवादी आवृत्ती: https://www.calcopenly.com/mr/statistics/confidence-interval-calculator
विषय: सांख्यिकी आणि संभाव्यतेचे कॅल्क्युलेटर

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.

## इनपुट

- **Interval for** (पर्याय: Mean, σ unknown (t), Mean, σ known (z), Proportion, Difference of two means)
- **Enter** (पर्याय: Summary statistics, Raw data)
- **Sample 1**
- **Sample 2**
- **Sample 1 mean**
- **Sample standard deviation s₁**
- **Sample size n₁**
- **Known population σ**
- **Sample 2 mean**
- **Sample standard deviation s₂**
- **Sample size n₂**
- **Variances** (पर्याय: Unequal (Welch), Equal (pooled))
- **Successes x**
- **Sample size n**
- **Method** (पर्याय: Wilson score, Wald (p̂ ± z·SE))
- **Confidence level**

## निकाल

- Margin of error (±) — मुख्य निकाल
- Lower bound
- Upper bound
- Point estimate
- Standard error
- Critical value
- Degrees of freedom

## सूत्र

$$
\bar x \pm t_{1-\alpha/2,\,n-1}\frac{s}{\sqrt n},\qquad \text{Wilson: } \frac{\hat p + \frac{z^2}{2n} \pm z\sqrt{\frac{\hat p(1-\hat p)}{n} + \frac{z^2}{4n^2}}}{1 + z^2/n}
$$

## सोडवलेली उदाहरणे

### 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**
- पडताळणीचा स्रोत: 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**
- पडताळणीचा स्रोत: 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**
- पडताळणीचा स्रोत: 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**
- पडताळणीचा स्रोत: p̂ ± z√(p̂(1−p̂)/n) in Python

### Difference of means, Welch

- Interval for: Difference of two means
- Enter: Summary statistics
- Sample 1 mean: 72.4
- Sample standard deviation s₁: 8.1
- Sample size n₁: 36
- Sample 2 mean: 68.9
- Sample standard deviation s₂: 9.4
- Sample size n₂: 40
- Variances: Unequal (Welch)
- Confidence level: 95%
- **Point estimate: 3.5**
- **Degrees of freedom: 73.8705**
- **Margin of error (±): 4.000864**
- पडताळणीचा स्रोत: Welch–Satterthwaite df in Python fractions; t quantile by bisection on Gauss–Legendre quadrature of the t density (pyref.t_sf_numeric)

### Mean from raw data, 90%

- Interval for: Mean, σ unknown (t)
- Enter: Raw data
- Sample 1: 68, 75, 71, 80, 66, 73, 77, 70, 74, 69
- Confidence level: 90%
- **Point estimate: 72.3**
- **Margin of error (±): 2.505254**
- पडताळणीचा स्रोत: Python statistics.mean/stdev; t₀.₉₅,₉ = 1.833 (t table), 1.8331129327 by bisection on the A&S closed form

## प्रश्न

### 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” किती अचूक आहे?

अचूकता तुमच्या इनपुटवर आणि पद्धतीच्या गृहीतकांवर अवलंबून असते. दशांश गणना 50 सार्थ अंक वापरते, पण अंदाज, संख्यात्मक पद्धती आणि मूळ डेटा कमी अचूक असू शकतात; दाखवलेल्या मूल्यांचे पूर्णांकन केल्याने या मर्यादा दूर होत नाहीत. स्वतंत्र स्रोतांतील सोडवलेल्या उदाहरणांशी पडताळणी: ७. उदाहरणार्थ, “Mean with sample SD, n = 36 (defaults)” ची पडताळणी 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 याच्याशी केली आहे.

### या पद्धतीचा स्रोत कोणता?

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.

## स्रोत

- [NIST/SEMATECH e-Handbook of Statistical Methods, §7.2.2.1 Confidence limits for the mean](https://www.itl.nist.gov/div898/handbook/prc/section2/prc221.htm)
- [NIST/SEMATECH e-Handbook, §7.2.4.1 Confidence intervals for a proportion (Wilson and normal approximation)](https://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm)
- Brown, Cai & DasGupta (2001). Interval estimation for a binomial proportion. Statistical Science 16(2), 101–133