The z-test for proportions is a statistical method used to determine whether there is a significant difference between the proportions of two groups. It is particularly useful when dealing with categorical data and can help in making informed decisions based on sample data. Below are three practical examples that illustrate the application of the z-test for proportions in different scenarios.
In a recent survey conducted among 1,000 eligible voters, researchers found that 540 preferred Candidate A while 460 preferred Candidate B. The objective is to determine if the proportion of voters favoring Candidate A significantly differs from 50%.
The null hypothesis (H0) is that the proportion of voters preferring Candidate A is 0.5, while the alternative hypothesis (H1) is that it is not equal to 0.5.
The z-test statistic can be calculated using the formula:
[ z = \frac{(p - p0)}{\sqrt{\frac{p0(1 - p0)}{n}}} ]
Substituting the values:
[ z = \frac{(0.54 - 0.5)}{\sqrt{\frac{0.5(1 - 0.5)}{1000}}} \approx 1.265 ]
Using a significance level of 0.05, we compare the calculated z-value against the z-critical values (±1.96). Since 1.265 is less than 1.96, we fail to reject the null hypothesis. This indicates that the preference for Candidate A does not significantly differ from 50%.
A pharmaceutical company wants to test the effectiveness of a new drug designed to reduce cholesterol levels. In a clinical trial with 200 patients, 120 showed a reduction in cholesterol after taking the drug. The company wants to know if this proportion is significantly higher than the known success rate of 50% for existing treatments.
Here, the null hypothesis (H0) states that the proportion of patients experiencing a reduction in cholesterol is 0.5, while the alternative hypothesis (H1) claims it is greater than 0.5.
We can calculate the z-test statistic:
[ z = \frac{(0.6 - 0.5)}{\sqrt{\frac{0.5(1 - 0.5)}{200}}} \approx 2.828 ]
With a significance level of 0.01, the z-critical value is approximately 2.33. Since 2.828 is greater than 2.33, we reject the null hypothesis. This suggests that the new drug is significantly more effective than existing treatments.
A retail store conducted a customer satisfaction survey and found that out of 300 surveyed customers, 210 reported being satisfied with their shopping experience. The store wants to determine if the satisfaction rate differs from the industry standard of 70%.
The null hypothesis (H0) posits that the satisfaction rate is equal to 0.7, while the alternative hypothesis (H1) states that it is not equal to 0.7.
To find the z-test statistic, we use the formula:
[ z = \frac{(0.7 - 0.7)}{\sqrt{\frac{0.7(1 - 0.7)}{300}}} = 0 ]
In this case, the z-value of 0 means we are exactly at the industry standard. For a significance level of 0.05, the z-critical values are ±1.96. Since 0 is within this range, we fail to reject the null hypothesis, indicating that the store’s satisfaction rate does not significantly differ from the industry standard.
These examples illustrate the versatility of the z-test for proportions across various fields, from political surveys to medical research and market analysis.