# 9.1.2.2 - Minitab: Difference Between 2 Independent Proportions

9.1.2.2 - Minitab: Difference Between 2 Independent ProportionsWhen conducting a hypothesis test comparing the proportions of two independent proportions in Minitab two p-values are provided. If \(np \ge 10\) and \(n(1-p) \ge 10\), use the p-value associated with the normal approximation method. If this assumption is not met, use the p-value associated with Fisher's exact method.

##### Minitab Note!

When conducting a hypothesis test for a difference between two independent proportions in Minitab you need to remember to change the "test method" to "use the pooled estimate of the proportion." This is because the null hypothesis is that the two proportions are equal.

##
Minitab^{®}
– Testing Two Independent Proportions using Raw Data

Let's compare the proportion of females who have tried weed to the proportion of males who have tried weed.

- Open Minitab file: class_survey.mpx
- Select
*Stat > Basic Statistics > 2 Proportions* - Select
*Both samples are in one column*from the dropdown - Double click the variable
*Try Weed*in the box on the left to insert the variable into the*Samples*box - Double click the variable
*Biological Sex*in the box on the left to insert the variable into the*Sample IDs*box - Under the
*Options*tab change the*Test method*to*Use the pooled estimate of the proportion* - Click OK and OK

This should result in the following output:

##### Method

Event: Try_Weed = Yes

\(p_1\): proportion where Try_Weed = Yes and Biological Sex = Female

\(p_2\): proportion where Try-Weed = Yes and Biological Sex = Male

Difference: \(p_1\)-\(p_2\)

##### Descriptive Statistics: Try Weed

Biological Sex | N | Event | Sample p |
---|---|---|---|

Female | 127 | 56 | 0.440945 |

Male | 99 | 62 | 0.626263 |

##### Estimation for Difference

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

-0.185318 | (-0.313920, -0.056716) |

*CI based on normal approximation*

##### Test

Null hypothesis | \(H_0\): \(p_1-p_2=0\) |
---|---|

Alternative hypothesis | \(H_1\): \(p_1-p_2\neq0\) |

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

Normal approximation | -2.77 | 0.006 |

Fisher's exact | 0.007 |

*The test based on the normal approximation uses the pooled estimate of the proportion (0.522124)*

## Five Step Hypothesis Testing Procedure: Weed Example

**Step 1:**

\(H_0 : p_f - p_m =0\)

\(H_a : p_f - p_m \neq 0\)

Check assumptions:

\(n_f p_f = 56\)

\(n_f (1-p_f) = 127-56 = 71\)

\(n_m p_m = 62\)

\(n_m (1-p_m) = 99-62 = 37\)

All \(np\) and \(n(1-p)\) are at least ten, therefore we can use the normal approximation method.

**Step 2:**

From output, \(z=-2.77\)

**Step 3:**

From output, \(p=0.005\)

**Step 4:**

\(p \leq \alpha\), reject the null hypothesis

**Step 5:**

There is evidence that in the population the proportion of females who have tried weed is different from the proportion of males who have tried weed.

##
Minitab^{®}
– Testing Two Independent Proportions using Summarized Data

Let's compare the proportion of Penn State World Campus graduate students who have children to the proportion of Penn State University Park graduate students who have children. In a representative sample there were 120 World Campus graduate students; 92 had children. There were 160 University Park graduate students; 23 had children.

- In Minitab, select
*Stat > Basic Statistics > 2 Proportions* - Change
*Both samples are in one column*to*Summarized data*in the dropdown - For
*Sample 1*next to*Number of events*enter*92*and next to*Number of trials*enter*120* - For
*Sample 2*next to*Number of events*enter*23*and next to*Number of trials*enter 160 - Under the
*Options*tab change the*Test method*to*Use the pooled estimate of the proportion* - Click OK and OK

This should result in the following output:

##### Method

\(p_1\): proportion where Sample 1 = Event

\(p_2\): proportion where Sample 2 = Event

Difference: \(p_1\)-\(p_2\)

##### Descriptive Statistics

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

Sample 1 | 120 | 92 | 0.766667 |

Sample 2 | 160 | 23 | 0.143750 |

##### Estimation for Difference

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

0.622917 | (0.529740, 0.716093) |

*CI based on normal approximation*

##### Test

Null hypothesis | \(H_0\): \(p_1-p_2=0\) |
---|---|

Alternative hypothesis | \(H_1\): \(p_1-p_2\neq0\) |

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

Normal approximation | 10.49 | 0.000 |

Fisher's exact | 0.000 |

*The test based on the normal approximation uses the pooled estimate of the proportion (0.410714)*

## Five Step Hypothesis Testing Procedure: Parents

**Step 1:**

\(H_0 : p_w - p_u =0\)

\(H_a : p_w - p_u \neq 0\)

Check assumptions:

\(n_w p_w = 92\)

\(n_w (1-p_w) = 120-92 = 28\)

\(n_u p_u = 23\)

\(n_u (1-p_u) = 160-23 = 137\)

All \(np\) and \(n(1-p)\) are at least ten, therefore we can use the normal approximation method.

**Step 2:**

From output, \(z=10.49\)

**Step 3:**

From output, \(p=0.000\)

**Step 4:**

\(p \leq \alpha\), reject the null hypothesis

**Step 5:**

There is evidence that in the population the proportion of World Campus students who have children is different from the proportion of University Park students who have children.

# 9.1.2.2.1 - Example: Dating

9.1.2.2.1 - Example: Dating## Example: Dating

This example uses the course survey dataset:

A random sample of Penn State University Park undergraduate students were asked, "Would you date someone with a great personality even if you did not find them attractive?" Let's compare the proportion of males and females who responded "yes" to determine if there is evidence of a difference.

We are looking for a "difference," so this is a two-tailed test.

\(H_{0} \colon p_1 - p_2 =0\)

\( H_{a} \colon p_1 - p_2 \neq 0 \)

Check assumptions

\(n_f p_f = 367\)

\(n_f (1-p_f) = 571 - 367 = 204\)

\(n_m p_m = 148\)

\(n_m (1 - p_m) = 433 - 148 = 285\)

All of these counts are at least 10 so we will use the normal approximation method.

From output, \(z=9.45\)

Event: DatePerly = Yes |

\(p_1\): proportion where DatePerly = Yes and Gender = Female |

\(p_2\): proportion where DatePerly = Yes and Gender = Male |

Difference: \(p_1-p_2\) |

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

Female | 571 | 367 | 0.642732 |

Male | 433 | 148 | 0.341801 |

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

0.300931 | (0.241427, 0.360435) |

Null hypothesis | \(H_0\): \(p_1-p_2=0\) |
---|---|

Alternative hypothesis | \(H_1\): \(p_1-p_2\neq0\) |

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

Fisher's exact | <0.0001 | |

Normal approximation | 9.45 | <0.0001 |

*The pooled estimate of the proportion (0.512948) is used for the tests.*

From output, \(p<0.0001\)

\(p \leq \alpha\), reject the null hypothesis

There is evidence that in the population of all Penn State University Park undergraduate students the proportion of men who would date someone with a great personality even if they did not find them attractive is different from the proportion of women who would date someone with a great personality even if they did not find them attractive.