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Wk3 - Assignment Example

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Moreover the ANOVA test has been employed assuming equal variances. Therefore, the output summary is given below.
Where µA shows the average salaries for…
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Download file to see previous pages table, is found to be less than 0.05, therefore, we cannot accept null hypothesis which states that the average salaries of employees of each grade is equal. It can thus be concluded that significant difference exists in the average salaries of employees at different grades.
The first null hypothesis that states that average salaries for all grades are equal is accepted as the associated P value is found to be equal to 0.07 which is greater than 0.05. The second null hypothesis that states that average salaries for both male and female employees are equal is, however, rejected as the associated P value is found to be equal to 1.5 x 10-10 which is quite less than the significance level 0.05. The third null hypothesis that tests whether the interaction is statistically significant is found to be statistically significant as the associated P value is found to be 0.17 which is significantly greater than the significance level of 0.05. It means that the grades and gender do have an interaction that impacts the equality of pays amongst employees on average. Moreover it shows that the mean salaries are equal amongst different grades while they are found to be unequal amongst male and female employees without considering their grades. Moreover in order to avoid biasness in the results the observations have been chosen at random so that the sample results can be employed for the interpretation of population parameters. Therefore mean value in this case can be used to interpret the results of mean values of population.
In order to test that the compa values in the population are equal by grade and/or gender, and are independent of each factor the ANOVA has been done using Data Analysis Tool pack of MS Excel 2007. The values are chosen on convenience basis for each category. The summary of outputs thus obtained is mentioned below. The two factor ANOVA has been performed with replication.
All three null hypotheses are rejected as the associated P values are found to be less ...Download file to see next pagesRead More
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