When using the sample data we know the sample's statistic but we don't know the true value of the population's measures. Instead, we may treat the population's measure as a random variable. The confidence interval is the range that is likely to contain the true value with a probability of the confidence level.

The mean's confidence interval is based on the sample average. When you know the population's standard deviation (σ) you should use the normal distribution $$\bar{x}\pm Z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$$ When you don't know the population's standard deviation you should use the sample standard deviation(S) and the normal distribution $$\bar{x}\pm T_{(1-\frac{\alpha}{2},n-1)}\frac{S}{\sqrt{n}}$$

The population's confidence interval is based on the sample standard deivation. The following statistic distribute Χ^{2}_{(n-1)} $$\frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{(n-1)}$$ You may extrac the σ based on Χ^{2} α/2 percentile and 1-α/2 percentile. $$\frac{(n-1)S^2}{\chi^2_{(1-\frac{\alpha}{2},n-1)}}\le \sigma \le \frac{(n-1)S^2}{\chi^2_{(\frac{\alpha}{2},n-1)}}$$

The population's confidence interval is based on the sample proportion.

Normal approximation For large enough sample size the sample proportion distributes normaly: $$\hat{p}=\frac{successes}{n} \sim Z(\hat{p},\hat{p}(1-\hat{p}))$$ Following the confidence interval formula: $$\hat{p}\pm Z \sqrt{\frac{\hat{p}(1-\hat{p}}{n}}$$ The normal approximation dosn't support good result for edge proportions, near 0 or 1.

Wilson score interval The wilson score interval support better results than the normal approximation, especially for small sample sizes and for edge proportions, near 0 or 1. $$\frac{\hat{p}+\frac{Z^2}{2n}}{1+\frac{Z^2}{n}}\pm \frac{Z}{1+\frac{Z^2}{n}} \sqrt{\frac{\hat{p}(1-\hat{p}}{n}+\frac{Z^2}{4n^2}}$$