Model Answer
0 min readIntroduction
Hypothesis testing is a crucial statistical method used to determine whether there is enough evidence to reject a null hypothesis in favor of an alternative hypothesis. In organizational settings, it’s frequently employed to assess differences between groups, such as comparing the income levels of different employee segments. The question presents a scenario where we need to determine if the income of local workers is significantly higher than the average income of all workers in the company. This requires a one-tailed z-test, given the large sample size and known population standard deviation.
Formulating Hypotheses
First, we need to define our null and alternative hypotheses:
- Null Hypothesis (H0): The mean income of local workers is equal to the mean income of all workers. (μlocal = μtotal = ₹5,000)
- Alternative Hypothesis (H1): The mean income of local workers is greater than the mean income of all workers. (μlocal > μtotal = ₹5,000)
Calculating the Test Statistic (Z-score)
The z-score is calculated using the following formula:
Z = (x̄ - μ) / (σ / √n)
Where:
- x̄ = Sample mean income of local workers (₹5,500)
- μ = Population mean income (₹5,000)
- σ = Population standard deviation (₹1,200)
- n = Sample size (144)
Plugging in the values:
Z = (5500 - 5000) / (1200 / √144) = 500 / (1200 / 12) = 500 / 100 = 5
Determining the Critical Value
Given the significance level (α) is 0.05 and it’s a one-tailed test (we are only interested in whether the local workers’ income is *higher*), we need to find the critical z-value. Using a standard z-table or statistical software, the critical z-value for α = 0.05 (one-tailed) is approximately 1.645.
Decision Rule
We will reject the null hypothesis if the calculated z-score is greater than the critical z-value.
Comparing the Test Statistic and Critical Value
Our calculated z-score is 5, and the critical z-value is 1.645. Since 5 > 1.645, we reject the null hypothesis.
Conclusion
Based on the hypothesis test, we can conclude that there is statistically significant evidence at the 0.05 significance level to suggest that the mean income of local workers is significantly higher than the mean income of all workers in the company.
Conclusion
In conclusion, the statistical analysis strongly supports the suspicion that local workers have a higher average income than the overall workforce. The calculated z-score of 5 significantly exceeds the critical value of 1.645, leading to the rejection of the null hypothesis. This finding has implications for workforce management, potentially indicating localized pay scales or skill premiums. Further investigation into the factors contributing to this income disparity could be beneficial for the company.
Answer Length
This is a comprehensive model answer for learning purposes and may exceed the word limit. In the exam, always adhere to the prescribed word count.