Your organization, XYZ Company, notes several hurdles to attracting and retaining staff in the post-pandemic climate. One example is the employer’s inability to adjust to the remote work arrangement, and another is a lack of pay equity commitment. In keeping with XYZ Company’s goal to treat employees well, management is looking for feedback to measure the company’s performance as the employer of choice. You work as the Manager of Workforce Analytics within the Human Resources department. The company is growing rapidly, and your boss, Jane, who is the Chief People Officer (CPO), wants to make sure that the company is treating its employees equitably. She is analysis-driven, so she taps you to work on several projects to uncover any potential issues. Her objectives are to: Describe the current state of salary data using the measures of central tendency and variability. Give a point estimate and construct a confidence interval of the number of employees who want to work remotely. Condu
Hypothesis Testing – The Mean Salary for Engineering Managers at XYZ Company
In order to test the claim that the mean salary for engineering managers at XYZ Company is at least $170,000, a one-tailed hypothesis test with a significance level of 0.05 was conducted. The null hypothesis (H0) and alternative hypothesis (Ha) are the following.
- Null Hypothesis (H0): The mean salary for engineering managers at XYZ Company (μ (mu)) is greater than or equal to $170,000.
- H0: mu >= $170,000
- Alternative Hypothesis (Ha): The mean salary for engineering managers at XYZ Company (μ) is less than $170,000.
- Ha: mu < $170,000
Next, it was essential to find the sample mean salary for engineering managers in the company. From the provided data, the sample mean salary is:
- Sample Mean Salary = $174,339
Next, the confidence interval to test the claim was calculated. Since we conducted a one-tailed test with a significance level of 0.05, we had to find the critical value corresponding to this significance level. For a one-tailed test, one looks for the z-value that leaves 5% of the data in the tail (95% confidence level). Consulting the standard normal distribution table, the critical z-value is approximately -1.645.
The confidence interval was constructed as follows:
- CI = Sample Mean ± (Critical Value) * (Standard Error)
The standard error can be calculated as follows:
- Standard error = Standard Deviation / √Sample Size (Zach, 2022)
From the given data related to employee salaries:
- Standard Deviation = $95,378
- Sample Size, n = 225
- Standard Error = $95,377.67 / √225 ≈ $6,352.59
Now, we can calculate the confidence interval as follows:
- CI = $174,338.75 ± (-1.645) * $6,352.59
Therefore,
- CI ≈ $174,338.75 – $10,457 to $174,339
As the calculated confidence interval includes only positive values, we fail to reject the null hypothesis. There is not enough evidence to suggest that the mean salary for engineering managers at XYZ Company is less than $170,000. Therefore, based on the data and the hypothesis test, one can conclude that, with 95% confidence, the mean salary for engineering managers at XYZ Company is at least $170,000.
Therefore, the leadership’s claim that the mean salary for engineering managers is at least $170,000 is supported by the data, and there is no significant evidence to suggest otherwise. XYZ Company’s engineering managers’ salaries appear to be on par with industry standards, which aligns with the company’s goal to treat its employees well and remain competitive in the post-pandemic climate.
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References
Zach. (2022). Standard deviation vs. standard error: What’s the difference? Statology. https://www.statology.org/standard-deviation-vs-standard-error/
Your organization, XYZ Company, notes several hurdles to attracting and retaining staff in the post-pandemic climate. One example is the employer’s inability to adjust to the remote work arrangement, and another is a lack of pay equity commitment. In keeping with XYZ Company’s goal to treat employees well, management is looking for feedback to measure the company’s performance as the employer of choice. You work as the Manager of Workforce Analytics within the Human Resources department. The company is growing rapidly, and your boss, Jane, who is the Chief People Officer (CPO), wants to make sure that the company is treating its employees equitably. She is analysis-driven, so she taps you to work on several projects to uncover any potential issues. Her objectives are to: Describe the current state of salary data using the measures of central tendency and variability. Give a point estimate and construct a confidence interval of the number of employees who want to work remotely. Condu
Estimates and Confidence Intervals – XYZ Company
The point estimate of the number of respondents who answered “office” can be calculated by multiplying the sample size by the sample proportion:
- Number of respondents who answered “office” = 1,003 x 0.372 ≈ 373.
To construct a 95% confidence interval estimate of the true proportion of all employees who would like to work at the office, we use the formula:
- Confidence interval = sample proportion ± (critical value) x (standard error).
Given information:
- Sample proportion (p̂) = 0.372
- Sample size (n) = 1,003
- Critical value (z) for a 95% confidence level = 1.96
The standard error is calculated as follows:
- Standard error (SE) = sqrt[(p̂ * (1 – p̂)) / n]
- Standard error (SE) = sqrt[(0.372 * 0.628) / 1,003] ≈ 0.018
Now, we can construct the confidence interval:
- Confidence interval = 0.372 ± (1.96 * 0.018)
- Confidence interval ≈ (0.337, 0.407) rounded to one decimal digit.
So, the 95% confidence interval estimate of the true proportion of employees who would like to work in the office is approximately (33.7%, 40.7%).
To test whether the majority (more than 50%) of employees would like to come back to the office, we can set up the following null and alternative hypotheses:
- Null Hypothesis (H0): The true proportion of employees who would like to work in the office is less than or equal to 50%.
H0: p ≤ 0.5
- Alternative Hypothesis (Ha): The true proportion of employees who would like to work in the office is greater than 50%.
Ha: p > 0.5
We will use a significance level of 0.05 for the hypothesis test.
Next, we calculate the test statistic (z-score):
z = (sample proportion – hypothesized proportion) / standard error
where the hypothesized proportion is 0.5 (50%), and the standard error is 0.018.
Plugging in the values, we get:
z = (0.372 – 0.5) / 0.018 ≈ -7.11
Using the standard normal distribution table, we find that the p-value is less than 0.0001.
Therefore, since the p-value is less than the significance level of 0.05, we reject the null hypothesis. We conclude that there is strong evidence to suggest that most employees (more than 50%) would like to return to the office for the upcoming year.
In conclusion, based on the survey results and the hypothesis test, it appears that a significant proportion of employees at XYZ Company prefer to work in the office rather than continue working from home. The leadership should consider these findings while making decisions about the work arrangement policies for the upcoming year and ensure that the transition back to the office is carried out with proper planning and employee engagement.