# 17.3.7.2 Algorithms (One Sample Proportion Test)

## Contents

### Hypotheses

Let $n\!$ be the sample size and $n_{1}\!$be the number of events or successes. Then the sample proportion $\tilde{p}\!$ can be expressed: $\tilde{p}=\frac{n_{1}}{n}$

Let $p\!$ be the sample proportion and the $p_{0}\!$ is the hypothetical proportion, this function tests the hypotheses: $H_0:p=p_{0}\!$ vs $H_1:p \ne p_{0}\!$, for a two-tailed test. $H_0: p\ge p_{0}\!$ vs $H_1:p < p_{0}\!$, for a lower-tailed test. $H_0:p\le p_{0}\!$ vs $H_1:p > p_{0}\!$, for an upper-tailed test.

### Normal approximation

#### P Value

When $n_{1}\ge10\!$ and $n-n_{1}\ge10\!$, we can compute a p-value using a normal approximation of a binomial distribution. To perform the test, compute the $z\!$ and $p_{value}\!$ value by: $z=\frac{\tilde{p}-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}\!$ $p_{value}=2p(Z>|z||p=p_{0})\!$,for two tailed test $p_{value}=p(Z\le z|p=p_{0})\!$,for upper tailed test $p_{value}=p(Z\ge z|p=p_{0})\!$for lower tailed test

#### Confidence Interval

confidence level is equal to $1-\alpha$, the confidence interval for the sample proportion can be generated by:

Null Hypothesis Confidence Interval $H_0:p=p_{0}\!$ $\left[\tilde{p}- Z_{\frac{\alpha}{2}}\sqrt{\frac{\tilde{p}(1-\tilde{p})}{n}}, \tilde{p}+ Z_{\frac{\alpha}{2}}\sqrt{\frac{\tilde{p}(1-\tilde{p})}{n}}\right]$ $H_0: p\ge p_{0}\!$ $\left[\tilde{p}- Z_{\alpha}\sqrt{\frac{\tilde{p}(1-\tilde{p})}{n}}, 1\right]$ $H_0:p\le p_{0}\!$ $\left[0, \tilde{p}+ Z_{\alpha}\sqrt{\frac{\tilde{p}(1-\tilde{p})}{n}}\right]$

### Binomial Test

#### Exact P_value

In Origin, the exact test for one proportion is based on the Binomial Test . $H_0: p\ge p_{0}\!$ $P_{value}=p(X\le n_{1}|p_0)$ $H_0:p\le p_{0}\!$ $P_{value}=p(X\ge n_{1}|p_0)$ $H_0:p=p_{0}\!$:

Let $M=n*p_0\!$,

when $n_1=M\!$ $P_{value}=1\!$

when $n_1\le M\!$ $P_{value}=P(X\le n_1)+P(X\ge n-y+1)$, where y is the count for z such that $P(z)\le p(n_1)$ and $n\ge z\ge \left \lfloor M \right \rfloor+1$

when $n_1\ge M\!$ $P_{value}=P(X\le y-1)+P(X\ge n_1)$, where y is the count for z such that $P(z)\le p(n_1)$ and $0\le z\le \left \lfloor M \right \rfloor$

#### Exact Confidence Interval

Exact Confidence interval: confidence levels is $1-\alpha$

Null Hypothesis Confidence Interval $H_0:p=p_{0}\!$ $\left[QBETA_{(1 - \alpha/2, n_1 + 1, n - n_1)}, QBETA_{(\alpha/2, n_1, n - n_1 + 1)}\right]$ $H_0: p\ge p_{0}\!$ $\left[QBETA_{(1 - \alpha, n_1 + 1, n - n_1)}, 1\right]$ $H_0:p\le p_{0}\!$ $\left[0, QBETA_{(\alpha, n_1, n - n_1 + 1)}\right]$

where $QBETA$ denotes the quantile function of Beta distribution.