Example: Overweight Section
The following example uses a scenario in which we want to know if the proportion of college women who think they are overweight is less than 40%. We collect data from a random sample of 129 college women and 37 said that they think they are overweight.
First, we should check assumptions to determine if the normal approximation method or exact method should be used:
\(np_0=129(0.40)=51.6\) and \(n(1-p_0)=129(1-0.40)=77.4\) both values are at least 10 so we can use the normal approximation method.
Minitab® – Performing a One Proportion z Test with Summarized Data
To perform a one sample proportion z test with summarized data in Minitab:
- In Minitab, select Stat > Basic Statistics > 1 Proportion
- Select Summarized data from the dropdown
- For number of events, add 37 and for number of trials add 129.
- Check the box next to Perform hypothesis test and enter 0.40 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: Event proportion
Normal approximation is used for this analysis.
N | Event | Sample p | 95% Upper Bound for p |
|---|---|---|---|
129 | 37 | 0.286822 | 0.352321 |
Null hypothesis | H 0: p = 0.4 |
|---|---|
Alternative hypothesis | H 1: p < 0.4 |
Z-Value | P-Value |
|---|---|
-2.62 | 0.004 |
Summary of Results Section
We could summarize these results using the five-step hypothesis testing procedure:
\(np_0=129(0.40)=51.6\) and \(n(1-p_0)=129(1-0.40)=77.4\) both values are at least 10 so we can use the normal approximation method.
\(H_0\colon p = 0.40\)
\(H_a\colon p < 0.40\)
From output, \(z\) = -2.62
From output, \(p\) = 0.004
\(p \leq \alpha\), reject the null hypothesis
There is convincing evidence that the proportion of women in the population who think they are overweight is less than 40%.