# 8.1.2.2.1 - Minitab: 1 Proportion z Test, Raw Data

8.1.2.2.1 - Minitab: 1 Proportion z Test, Raw DataIf you have data in a Minitab worksheet, then you have what we call "raw data." This is in contrast to "summarized data" which you'll see on the next page.

In order to use the normal approximation method both \(np_0 \geq 10\) and \(n(1-p_0) \geq 10\). Before we can conduct our hypothesis test we must check this assumption to determine if the normal approximation method or exact method should be used. This must be checked manually. **Minitab will not check assumptions for you.**

In the example below, we want to know if there is evidence that the proportion of students who are male is different from 0.50.

\(n=226\) and \(p_0=0.50\)

\(np_0 = 226(0.50)=113\) and \(n(1-p_0) = 226(1-0.50)=113\)

Both \(np_0 \geq 10\) and \(n(1-p_0) \geq 10\) so we can use the normal approximation method.

##
Minitab^{®}
– Conducting a One Sample Proportion z Test: Raw Data

**Research question:** Is the proportion of students who are male different from 0.50?

- Open Minitab file:
- In Minitab, select
*Stat > Basic Statistics > 1 Proportion* - Select
*One or more samples, each in a column*from the dropdown - Double-click the variable
*Biological Sex*to insert it into the box - Check the box next to
*Perform hypothesis*test and enter*0.50*in the*Hypothesized proportion*box - Select
*Options* - Use the default
*Alternative hypothesis*setting of*Proportion ≠ hypothesized**proportion*value - Use the default
*Confidence level*of 95 - Select
*Normal approximation method* - Click OK and OK

The result should be the following output:

#### Method

Event: Biological Sex = Male

p: proportion where Biological Sex = Male

Normal approximation is used for this analysis.

N | Event | Sample p | 95% CI for p |
---|---|---|---|

226 | 99 | 0.438053 | (0.373368, 0.502738) |

Null hypothesis | H _{0}: p = 0.5 |
---|---|

Alternative hypothesis | H _{1}: p ≠ 0.5 |

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

-1.86 | 0.063 |

## Summary of Results

We could summarize these results using the five-step hypothesis testing procedure:

\(np_0 = 226(0.50)=113\) and \(n(1-p_0) = 226(1-0.50)=113\) therefore the normal approximation method will be used.

\(H_0\colon p = 0.50\)

\(H_a\colon p \ne 0.50\)

From the Minitab output, \(z\) = -1.86

From the Minitab output, \(p\) = 0.0625

\(p > \alpha\), fail to reject the null hypothesis

There is NOT evidence that the proportion of all students in the population who are male is different from 0.50.