A company is interested in knowing if there is a difference in the average salary received by foremen in two divisions. Accordingly samples of 12 foremen in the first division and 10 in the second division are selected at random. The data is given as follows.
Second division
10
980
74
Test the significance difference between means at 5% Level of significance
12
1050
68
First division
Sample size
Average monthly salary(in Rs)
Standard deviation of salaries (in Rs)
To test the significance of the difference between the means of the two divisions, we can perform a two-sample t-test.
First, let's calculate the means and standard deviations for each division:
First Division:
- Sample size (n1) = 12
- Average monthly salary = Rs 1050
- Standard deviation = Rs 68
Second Division:
- Sample size (n2) = 10
- Average monthly salary = Rs 74
- Standard deviation = Rs 235
Next, we can use the following formula to calculate the t-statistic:
t = (x̄1 - x̄2) / √((s1^2/n1) + (s2^2/n2))
where x̄1 and x̄2 are the sample means, s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.
Plugging in the values, we get:
t = (1050 - 74) / √((68^2/12) + (235^2/10))
t = 976 / √(364 + 547)
t = 976 / √911
t = 976 / 30.18
t ≈ 32.33
Next, we need to determine the critical t-value at a 5% level of significance with degrees of freedom equal to 12 + 10 - 2 = 20. Using a t-distribution table or statistical software, we find the critical t-value to be approximately ±2.086.
Since our calculated t-value (|32.33|) is much larger than the critical t-value (2.086), we can reject the null hypothesis that there is no significant difference between the average salaries of foremen in the two divisions. This means that there is a significant difference in the average salary received by foremen in the two divisions.