Another section shows how to use a test based on the normal distribution to see whether a sample proportion (p) differs significantly from a population proportion (π) . This section shows how to conduct a test of the same null hypothesis using a test based on the chi square distribution.

The two tests always yield identical results. The advantage of the test based on the chi square distribution is that it can be generalized to more complex situations. In the other section, an example was given in which a researcher wished to test whether a sample proportion of 62/100 differed significantly from an hypothesized population value of 0.5. The test based on z resulted in a z of 2.3 and a probability value of 0.0107. To compute the significance test using chi square, the following table is formed:

Succeeded |
62 |

Failed |
38 (50) |

The number of people falling in a specified category is listed as the first line in each cell (62 succeeded, 38 failed). The second line in each cell (in parentheses) contains the number expected to succeed if the null hypothesis is true. Since the null hypothesis is that the proportion that succeed is 0.5, (0.5)(100) = 50 are expected to succeed and (0.5)(100) = 50 are expected to fail.