8.1.2.2.2.1 - Minitab Example: Normal Approx. Method

Example: Gym membership Section

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:

  1. In Minitab, select Stat > Basic Statistics > 1 Proportion
  2. Select Summarized data from the dropdown
  3. For number of events, add 24 and for number of trials add 60.
  4. Check the box next to Perform hypothesis test and enter 0.50 in the Hypothesized proportion box
  5. Select Options
  6. Use the default Alternative hypothesis setting of Proportion < hypothesized proportion value 
  7. Use the default Confidence level of 95
  8. Select Normal approximation method
  9. Click OK and OK

The result should be the following output:

Method

Event: Event proportion
Normal approximation is used for this analysis.

Descriptive Statistics
N Event Sample p 95% Upper Bound for p
60 24 0.400000 0.504030
Test
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:

1. Check assumptions and write hypotheses

\(np_0=60(0.50)=30\) and \(n(1-p_0)=60(1-0.50)=30\) both values are at least 10 so we can use the normal approximation method.

\(H_0\colon p = 0.50\)

\(H_a\colon p < 0.50\)

2. Calculate the test statistic

From output, \(z\) = -1.55

3. Determine the p-value

From output, \(p\) = 0.061

4. Make a decision

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

5. State a "real world" conclusion

There is evidence that the proportion of women memberships at this gym is less than 50%.