Complete the following steps to interpret a 2 proportions test. Key output includes the estimate of the difference, the confidence interval, and the p-value.

First, consider the difference in the sample proportions, and then examine the confidence interval.

The difference is an estimate of the difference in the population proportions. Because the difference is based on sample data and not on the entire population, it is unlikely that the sample difference equals the population difference. To better estimate the population difference, use the confidence interval for the difference.

The confidence interval provides a range of likely values for the difference between two population proportions. For example, a 95% confidence level indicates that if you take 100 random samples from the population, you could expect approximately 95 of the samples to produce intervals that contain the population difference. The confidence interval helps you assess the practical significance of your results. Use your specialized knowledge to determine whether the confidence interval includes values that have practical significance for your situation. If the interval is too wide to be useful, consider increasing your sample size. For more information, go to Ways to get a more precise confidence interval.

Difference | 95% CI for Difference |
---|---|

0.0992147 | (0.063671, 0.134759) |

In these results, the estimate of the population difference in proportions in summer employment for male and female students is 0 approximately 0.099. You can be 95% confident that the ratio of population standard deviations is between approximately 0.06 and 0.13.

To determine whether the difference between the population proportions is statistically significant, compare the p-value to the significance level. Usually, a significance level (denoted as α or alpha) of 0.05 works well. A significance level of 0.05 indicates a 5% risk of concluding that a difference exists when there is no actual difference.

- P-value ≤ α: The difference between the proportions is statistically significant (Reject H
_{0}) - If the p-value is less than or equal to the significance level, the decision is to reject the null hypothesis. You can conclude that the difference between the population proportions does not equal the hypothesized difference. If you did not specify a hypothesized difference, Minitab tests whether there is no difference between the proportions (Hypothesized difference = 0). Use your specialized knowledge to determine whether the difference is practically significant. For more information, go to Statistical and practical significance.
- P-value > α: The difference between the proportions is not statistically significant (Fail to reject H
_{0}) - If the p-value is greater than the significance level, the decision is to fail to reject the null hypothesis. You do not have enough evidence to conclude that the difference between the population proportions is statistically significant. You should make sure that your test has enough power to detect a difference that is practically significant. For more information, go to Power and Sample Size for 2 Proportions.

Minitab uses the normal approximation method and Fisher's exact method to calculate the p-values for the 2 proportions test. If the number of events and the number of nonevents is at least 5 in both samples, use the smaller of the two p-values. If either the number of events or the number of nonevents is less than 5 in either sample, the normal approximation method may be inaccurate. Fisher's exact method is valid for all samples, but tends to be conservative. A conservative p-value understates the evidence against the null hypothesis.

Sample | N | Event | Sample p |
---|---|---|---|

Sample 1 | 802 | 725 | 0.903990 |

Sample 2 | 712 | 573 | 0.804775 |

Null hypothesis | H₀: p₁ - p₂ = 0 |
---|---|

Alternative hypothesis | H₁: p₁ - p₂ ≠ 0 |

Method | Z-Value | P-Value |
---|---|---|

Normal approximation | 5.47 | 0.000 |

Fisher's exact | 0.000 |

In these results, the null hypothesis states that there is no difference in the proportion of male and female students who get a summer job. The number of events and nonevents for both samples is at least 5, so both p-values are valid. Because the p-values for both methods are less than 0.0001, which is less than the significance level of 0.05, the decision is to reject the null hypothesis and conclude that the proportion of students who get a summer job differs for males and females.