/math-2fccdd06e4c2b79e4f65f2f532fdf9ec.png?v=0 "n_{1}\!")
be the size of sample 1 and /math-fdad14c036fde9f986da06150b87a44c.png?v=0 "x_{1}\!")
be the number of event or success ,then the sample proportion /math-2b49ceafb7c17ba633238ebd9932468b.png?v=0 "\tilde{p_{1}}\!")
can be expressed:/math-a31d5b1c9315097b39138038b35a733d.png?v=0 "\tilde{p_{1}}=\frac{x_{1}}{n_{1}}")
.
Let be the size of sample 1 and be the number of event or success ,then the sample proportion can be expressed:.
Similarly,for another sample , sample size is /math-524b3b5be113e2ddfb80c81946f7acd9.png?v=0 "n_{2}\!")
and /math-32846ccc5f26d802a0fe5ea9bbe7f00d.png?v=0 "x_{2}\!")
is the number of event,then sample proportion /math-bdf797b2281b2325c6b84e5fbbdd0419.png?v=0 "\tilde{p_{2}}=\frac{x_{2}}{n_{2}}")
Let /math-7b808e5178ceb5dc7d7cd3055dd492db.png?v=0 "p_{1}\!")
and be the true population proportion for sample 1 and 2. and the /math-bd9eee3ffcc47aa2c4ff7032d9e24ee9.png?v=0 "d_{0}\!")
is the hypothesized difference between the population proportions.
/math-c607e29efc94ad70d2c3c422748aaaff.png?v=0 "H_0:p_{1}-p_{2}=d_{0}\!")
for two tailed test
/math-b34f0ce1b16f76275e5ca5bb2095309a.png?v=0 "H_0:p_{1}-p_{2}\ge d_{0}\!")
for One-tailed test
/math-eff4b11b5021a77c9b957a60ef7d7839.png?v=0 "H_0:p_{1}-p_{2}\le d_{0}\!")
for One-tailed test
we can perform normal approximation test with assumptions :
/math-1ac2183e2b27a97935f0ce1eac821c81.png?v=0 "x_{1}\ge10\!")
and /math-7e77cc4bc8f9de81bd613971cca6bdbe.png?v=0 "n_{1}-x_{1}\ge10\!")
,
/math-7db56eb2a970f6af0b5d2a8ec332adf9.png?v=0 "x_{2}\ge10\!")
and /math-9c47106b3141ec18de2b358627ea9010.png?v=0 "n_{2}-x_{2}\ge10\!")
.
To perform the test, calculates the /math-814b9f64b478b6caf94e723a0e6814c5.png?v=0 "z\!")
and /math-c619dfb3f0ada810733c791d3e461ade.png?v=0 "p_{value}\!")
value :
/math-a83acc8877291a1769abf9c63cac0c06.png?v=0 "z=\frac{\tilde{p_{1}}-\tilde{p_{2}} -d_{0}}{\sqrt{\frac{\tilde{p_{1}}(1-\tilde{p_{1}})}{n_{1}}}+\frac{\tilde{p_{2}}(1-\tilde{p_{2}})}{n_{2}}}
\!")
.
A special case is that when /math-2a3499732983c7813c587075c92c0990.png?v=0 "d_{0}")
is zero, Origin can use a pooled estimate of p for the test if you check the "pooled" box to do this:
/math-c8de0be6d90bda1b287888d08b796b4e.png?v=0 "z=\frac{\tilde{p_{1}}-\tilde{p_{2}}}{\sqrt{\tilde{p_{0}}(1-\tilde{p_{0}})({\frac{1}{n_{1}}+ \frac{1}{n_{2}}}})}\!")
, where/math-581efaf63802fe89785cddcf19e96cb4.png?v=0 "p_{0}=\frac{x_{1}+x_{2}}{n_{1}+n_{2}}")
The p-values for each hypotheses are given by:
,/math-29f7013ed3172dc36f431f0b8482b50b.png?v=0 "p_{value}=2P(Z_{1}\ge|z|)\!")
,for two tailed test
,/math-10a0f892ce645224d3e01a17f2e69302.png?v=0 "p_{value}=P(Z_{1}\le z)\!")
,for upper tailed test
,/math-3bf80569261b1154f5d7acdf9d0f6cad.png?v=0 "p_{value}=P(Z_{1}\ge z)\!")
for lower tailed test
For a given confidence level/math-378df00d12056b404d7d02aef9d8650b.png?v=0 "1-\alpha")
,the confidence interval for the sample proportion can be generated by:
| Null Hypothesis | Confidence Interval |
|---|---|
|
|
![\left[(\tilde{p_{1}}-\tilde{p_{2}})- Z_{\frac{\alpha}{2}}\sqrt{\frac{\tilde{p_{1}}(1-\tilde{p_{1}})}{n_{1}}+ \frac{\tilde{p_{2}}(1-\tilde{p_{2}})}{n_{2}}}, (\tilde{p_{1}}-\tilde{p_{2}})+ Z_{\frac{\alpha}{2}}\sqrt{\frac{\tilde{p_{1}}(1-\tilde{p_{1}})}{n_{1}}+ \frac{\tilde{p_{2}}(1-\tilde{p_{2}})}{n_{2}}}\right]](/origin-help/en/images/Algorithm_(Two_sample_proportion_test)/math-ea62046992830b5407422a0bc50eb5e4.png?v=0 "\left[(\tilde{p_{1}}-\tilde{p_{2}})- Z_{\frac{\alpha}{2}}\sqrt{\frac{\tilde{p_{1}}(1-\tilde{p_{1}})}{n_{1}}+ \frac{\tilde{p_{2}}(1-\tilde{p_{2}})}{n_{2}}}, (\tilde{p_{1}}-\tilde{p_{2}})+ Z_{\frac{\alpha}{2}}\sqrt{\frac{\tilde{p_{1}}(1-\tilde{p_{1}})}{n_{1}}+ \frac{\tilde{p_{2}}(1-\tilde{p_{2}})}{n_{2}}}\right]")
|
|
|
![\left[(\tilde{p_{1}}-\tilde{p_{2}})- Z_{\frac{\alpha}{2}}\sqrt{\frac{\tilde{p_{1}}(1-\tilde{p_{1}})}{n_{1}}+ \frac{\tilde{p_{2}}(1-\tilde{p_{2}})}{n_{2}}}, 1\right]](/origin-help/en/images/Algorithm_(Two_sample_proportion_test)/math-25f19502677d1d4ee361f6bceae2271a.png?v=0 "\left[(\tilde{p_{1}}-\tilde{p_{2}})- Z_{\frac{\alpha}{2}}\sqrt{\frac{\tilde{p_{1}}(1-\tilde{p_{1}})}{n_{1}}+ \frac{\tilde{p_{2}}(1-\tilde{p_{2}})}{n_{2}}}, 1\right]")
|
|
|
![\left[-1, (\tilde{p_{1}}-\tilde{p_{2}})+ Z_{\frac{\alpha}{2}}\sqrt{\frac{\tilde{p_{1}}(1-\tilde{p_{1}})}{n_{1}}+ \frac{\tilde{p_{2}}(1-\tilde{p_{2}})}{n_{2}}}\right]](/origin-help/en/images/Algorithm_(Two_sample_proportion_test)/math-912083364694e7cf293a0a2b314d5ba2.png?v=0 "\left[-1, (\tilde{p_{1}}-\tilde{p_{2}})+ Z_{\frac{\alpha}{2}}\sqrt{\frac{\tilde{p_{1}}(1-\tilde{p_{1}})}{n_{1}}+ \frac{\tilde{p_{2}}(1-\tilde{p_{2}})}{n_{2}}}\right]")
|
Fisher's exact test can be used for all sample sizes when /math-a6b4ba4f8e6e0c7aa62a6321e42f50c5.png?v=0 "d_{0} \!")
is zero.
Let p(x) denote the probility of hypergeometric distribution when X=x.
/math-87ee7c011ddd771e42be5b24401b7b38.png?v=0 "P(X=x)=\frac{\begin{pmatrix}x_{1}+x_{2} \\{x}\end{pmatrix}\begin{pmatrix}{n_{1}+n_{2}-x_{1}-x_{2}}\\{n_{1}-x}\end{pmatrix}}{\begin{pmatrix}{n_{1}+n_{2}}\\{n_{1}}\end{pmatrix}}")
Let M denote hypergeometric distribution mode:
/math-a5c2eef81958cd52e29970d65dd26510.png?v=0 "M=\left \lfloor \frac{(n_1+1)(x_1+x_2+1)}{n_1+n_2+2}\right \rfloor")
The p-values for each hypothesis are given below:
/math-e8e42b6377357ab28f8d67401a32dcbb.png?v=0 "H_0:p_{1}\ge p_{2}\!")
, /math-ac0d5ea358618a9ad75aa424e562381e.png?v=0 "p_{value}=P(x\le x_{1})\!")
/math-2d65f2eb45f1ed6987af501d89e0c698.png?v=0 "H_0:p_{1}\le p_{2}\!")
, /math-11412e997c86e471799bb7527f8951e5.png?v=0 "p_{value}=P(x\ge x_{1})\!")
When /math-9bc36a86a3c356596eaded763aafa716.png?v=0 "H_0:p_{1}= p_{2}\!")
:
/math-4291cab60dfd9a45d2b2e27ca563fd02.png?v=0 "a:x_{1} < M\!")
:
/math-1de87365a19dd27513e6b7373b9522d3.png?v=0 "p_{value} = P(X\le x_{1}) + P(X\ge y)")
where y is the smallest integer /math-8fcaeb2e7c7c9d85baf692b14f25bba1.png?v=0 "\ge M")
such that /math-a204b7185fc011518c001fd57d7c9a32.png?v=0 "p(y) \le p(x_1)\!")
.
/math-60a92aacb57e6b5495c9d18453da11c8.png?v=0 "b:x_{1} = M\!")
/math-76be9e9d8c2c9e081f9a0a4f75723372.png?v=0 "p_{value} = 1.0\!")
/math-1cc2cb9480b59d99e1d6d338f9348f95.png?v=0 "c: x_1 > M\!")
/math-242d0497051cf5e0c14dc7013dbeb161.png?v=0 "p_{value} = P(X\ge x_{1}) + P(X\le y)")
where y is the largest integer /math-cdc6f4e3821a03f2e3d5bff6b42afc7d.png?v=0 "\le M")
such that .