In: Math
The Excel file Salary reports the monthly salaries for 93 randomly and independently selected employees of a bank; there are 32 salaries of male employees and 61 salaries of female employees.
Let um = the mean monthly salary for all male bank employees, and uf = the mean monthly salary for all female bank employees. Your objective is to find some evidence of um > uf, that is, the female employees are discriminated against.
Provide descriptive statistical summaries of the data sets for male and female employees. What are your primary observations concerning the two data sets? Calculate the 99% confidence intervals for um and uf, and interpret them. Do these intervals overlap?
Formulate a hypothesis test for supporting um > uf. What is the distribution of the test statistic? What is the value of the test statistics? What is the p-value of the test? What is your conclusion and its interpretation when the test is conducted under the 0.01 significance level?
Do your findings support a discrimination suit against the employer?
Instructions
For Task 1, apply “Descriptive Statistics” in Data Analysis of Excel (see instructions on pages 354-356). Summarize the obtained relevant statistics. Recall that in the row “Confidence Level (99.0%)” of the Descriptive Statistics output you actually see the margin of error of the confidence interval for the corresponding population mean.
To complete Task 2, formulate the null and alternative hypotheses, and apply “t-Test; Two-sample Assuming Unequal Variances” in Data Analysis of Excel with α = 0.01 (see instructions on pages 446-447). Note. In Excel, e.g., 2.71E-06 is 2.71(10^-6), which is practically zero.
Feel free to express your opinion.
Use Microsoft Word to write a managerial report with your name shown on the first page. The report should include all your Excel outputs (copy and paste them), so do not attach any separate Excel files. Hint. You may assume that you are an intern working for a branch of the bank and your boss, who has a very limited knowledge about business statistics, asked you to conduct a statistical analysis concerning the comparison of the salaries of male and female employees. You report may look like a letter written to your boss in which you present your findings.
Male Salary | Female Salary |
4620 | 3900 |
5040 | 4020 |
5100 | 4290 |
5100 | 4380 |
5220 | 4380 |
5400 | 4380 |
5400 | 4380 |
5400 | 4380 |
5400 | 4440 |
5400 | 4500 |
5700 | 4500 |
6000 | 4620 |
6000 | 4800 |
6000 | 4800 |
6000 | 4800 |
6000 | 4800 |
6000 | 4800 |
6000 | 4800 |
6000 | 4800 |
6000 | 4800 |
6000 | 4800 |
6000 | 4800 |
6000 | 4980 |
6000 | 5100 |
6300 | 5100 |
6600 | 5100 |
6600 | 5100 |
6600 | 5100 |
6840 | 5100 |
6900 | 5160 |
6900 | 5220 |
8100 | 5220 |
5280 | |
5280 | |
5280 | |
5400 | |
5400 | |
5400 | |
5400 | |
5400 | |
5400 | |
5400 | |
5400 | |
5400 | |
5400 | |
5400 | |
5400 | |
5520 | |
5520 | |
5580 | |
5640 | |
5700 | |
5700 | |
5700 | |
5700 | |
5700 | |
6000 | |
6000 | |
6120 | |
6300 | |
6300 |
Task 1
Following is the descriptive statisitcs obtained using Excel for male and female salary data:
From the above data, we can see that mean salary of male is greater than mean salary of females. Also, it can be seen that both maximum and minimum salaries of males are greater than maximum and minimum salary of females. The standard deviation of male salary is higher than female salary. This shows that there is greater variation in male salary in comparison to female salary.
99% Confidence Interval for Male Salary = 5956.88 + 335.06 = (5621.81, 6291.94)
99% Confidence Interval for female Salary = 5138.85 + 183.89 = (4954.96, 5322.74)
We can see that there is no overlap between 99% confidence interval for Male and Female Salary. This mean that we are 99% confident that population mean of male salary is higher than population mean of female salary. This we will also prove with the hypothesis testing in Task 2.
Task 2
Hypothesis testing has been done using two sample t test assuming unequal variance. This is because both the sample are independent. It is a one tail (right) test as we are interested in knowing whether mean monthly salary for all male bank employees is greater than salary for all female bank employees. Following is the output obtained using Excel for two sample t test assuming unequal variance:
Step 1: Set up null and alternative hypotheses.
Ho: μm = μf There is no
difference in the mean monthly salary for all male and female bank
employees
Ha: μm > μf The mean monthly
salary for all male bank employees is higher than the mean monthly
salary for all female bank employees
Step 2: Determine α (level of significance of hypothesis test)
α = 0.01 (99% Confidence Interval)
Step 3: Calculate Test Statistics
From the excel output, we can see that test statistic is 5.830. The p-value corresponding to test statistics is calcualted using t distribution.
Step 3: Make Decision
P-value = 0.000 (Check the excel output)
α = 0.01
So in this case P-value is less than level of significance α
If P-value is less than level of significance, we reject null hypothesis.
Interpretation
There is enough evidence to support the claim that the mean monthly salary for all male bank employees is higher than the mean monthly salary for all female bank employees.
Yes our findings support a discrimination suit against the employer.