8.1.2.2.2 - Minitab: 1 Sample Proportion z test, Summary Data
8.1.2.2.2 - Minitab: 1 Sample Proportion z test, Summary DataExample: Overweight
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
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 evidence that the proportion of women in the population who think they are overweight is less than 40%.
8.1.2.2.2.1 - Minitab Example: Normal Approx. Method
8.1.2.2.2.1 - Minitab Example: Normal Approx. MethodExample: Gym membership
Research question: Are less than 50% of all individuals with a membership at one gym female?
A simple random sample of 60 individuals with a membership at one gym was collected. Each individual's biological sex was recorded. There were 24 females.
First we have to check the assumptions:
np = 60 (0.50) = 30
n(1-p) = 60(1-0.50) = 30
The assumptions are met to use the normal approximation method.
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 24 and for number of trials add 60.
- 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: Event proportion
Normal approximation is used for this analysis.
N | Event | Sample p | 95% Upper Bound for p |
---|---|---|---|
60 | 24 | 0.400000 | 0.504030 |
Null hypothesis | H 0: p = 0.5 |
---|---|
Alternative hypothesis | H 1: p < 0.5 |
Z-Value | P-Value |
---|---|
-1.55 | 0.061 |
We could summarize these results using the five-step hypothesis testing procedure:
\(np_0=60(0.50)=30\) and \(n(1-p_0)=60(1-0.50)=30\) both values are at least 10 so we can use the normal approximation method.
\(H_0\colon p = 0.50\)
\(H_a\colon p < 0.50\)
From output, \(z\) = -1.55
From output, \(p\) = 0.061
\(p \geq \alpha\), fail to reject the null hypothesis
There is not enough evidence to support the alternative that the proportion of women memberships at this gym is less than 50%.