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The Salary and Finances of the Electrical Department - Research Proposal Example

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In the paper “The Salary and Finances of the Electrical Department” the various sections of the Electrical Department are analyzed and their summary measures are presented. The percentage of each of the salaries, when compared to the total in the section, is also presented…
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The Salary and Finances of the Electrical Department
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Managing Information Assignment Task The various sections of the Electrical Department are analyzed and their summary measures are presented in the following section. Accessories: The data for ‘Accessories’ section are extracted from the given raw data and summarized below. The percentage of each of the salaries when compared to the total in the section is also presented in the table. Salary ($ K) Percentage 3 19.48% 3.8 24.68% 3.7 24.03% 3.9 25.32% 1 6.49% Total = 15.4 Mean Salary for ‘Accessories’ = Sum of the salaries / Number of entries = 15.4 / 5 = $ 3.08 K Standard Deviation = √(Sum of squared differences with the mean / n) = $ 1.22 K Mode refers to the most frequently appearing value in the set of data. In this case, there is no definite mode, as all the values appear only once. Median refers to the central value when arranged in ascending or descending order. In this case, the Median is given by the (n + 1) / 2, i.e., 6/2 = 3 rd value in the series, which is $ 3.70 K. Range is the difference between the maximum and minimum value in the series. Range = Maximum Value – Minimum Value = $ 3.9 K - $ 1 K = $ 2.9 K From the normal distribution curve, 95% of the area is covered within 1.96 standard deviations of the mean. Hence to arrive at the confidence interval of the mean for a 95% significance level, the standard error is computed. Standard Error = Standard Deviation / √n = $ 1.22 K / √5 = $ 0.54 K Confidence Interval = {Mean + 1.96 Std. Error, Mean - 1.96 Std. Error} = {$ 3.08 K + (1.96) (0.54), $ 3.08 K - (1.96) (0.54)} = {$ 4.15 K, $ 2.01 K} Consumer Product 1: The data for ‘Consumer Product 1’ section are extracted from the given raw data and summarized below. The percentage of each of the salaries when compared to the total in the section is also presented in the table. Salary ($ K) Percentage 3.3 20.63% 3 18.75% 3 18.75% 3 18.75% 3.7 23.13% Total = 16 Mean Salary = Sum of the salaries / Number of entries = 16 / 5 = $ 3.20 K Standard Deviation = √(Sum of squared differences with the mean / n) = $ 0.31 K Mode refers to the most frequently appearing value in the set of data. In this case, ‘$ 3 K’ occurs three times and hence Mode is $ 3 K. Median refers to the central value when arranged in ascending or descending order. In this case, the Median is given by the (n + 1) / 2, i.e., 6/2 = 3 rd value in the series, which is $ 3 K. Range is the difference between the maximum and minimum value in the series. Range = Maximum Value – Minimum Value = $ 3.7 K - $ 3 K = $ 0.70 K From the normal distribution curve, 95% of the area is covered within 1.96 standard deviations of the mean. Hence to arrive at the confidence interval of the mean for a 95% significance level, the standard error is computed. Standard Error = Standard Deviation / √n = $ 0.31 K / √5 = $ 0.14 K Confidence Interval = {Mean + 1.96 Std. Error, Mean - 1.96 Std. Error} = {$ 3.20 K + (1.96) (0.14), $ 3.20 K - (1.96) (0.14)} = {$ 3.47 K, $ 2.93 K} Consumer Product 2: The data for ‘Consumer Product 1’ section are extracted from the given raw data and summarized below. The percentage of each of the salaries when compared to the total in the section is also presented in the table. Total = 32.8 Mean Salary = Sum of the salaries / Number of entries = 32.8 / 10 = $ 3.28 K Standard Deviation = √(Sum of squared differences with the mean / n) = $ 0.26 K Mode refers to the most frequently appearing value in the set of data. In this case, ‘$ 3 K’ occurs four times and hence Mode is $ 3 K. Median refers to the central value when arranged in ascending or descending order. In this case, the Median is given by the (n + 1) / 2, i.e., 11/2 = 5.5 (Mean of 5th and 6th values) in the series, which is $ 3.35 K. Range is the difference between the maximum and minimum value in the series. Range = Maximum Value – Minimum Value = $ 3.6 K - $ 3 K = $ 0.60 K From the normal distribution curve, 95% of the area is covered within 1.96 standard deviations of the mean. Hence to arrive at the confidence interval of the mean for a 95% significance level, the standard error is computed. Standard Error = Standard Deviation / √n = $ 0.26 K / √10 = $ 0.08 K Confidence Interval = {Mean + 1.96 Std. Error, Mean - 1.96 Std. Error} = {$ 3.28 K + (1.96) (0.08), $ 3.28 K - (1.96) (0.08)} = {$ 3.44 K, $ 3.12 K} Mainframe Systems: The data for ‘Mainframe Systems’ section are extracted from the given raw data and summarized below. The percentage of each of the salaries when compared to the total in the section is also presented in the table. Salary ($ K) Percentage 3.1 5.43% 3.6 6.30% 3 5.25% 2 3.50% 3 5.25% 3.7 6.48% 2.7 4.73% 1 1.75% 3 5.25% 3.2 5.60% 3 5.25% 3.1 5.43% 3.1 5.43% 3.2 5.60% 3.1 5.43% 3.4 5.95% 3.3 5.78% 3.4 5.95% 3.2 5.60% Total = 57.1 Mean Salary = Sum of the salaries / Number of entries = 57.1 / 19 = $ 3.01 K Standard Deviation = √(Sum of squared differences with the mean / n) = $ 0.60 K Mode refers to the most frequently appearing value in the set of data. In this case, ‘$ 3.10 K’ occurs four times and hence Mode is $ 3.10 K. Median refers to the central value when arranged in ascending or descending order. In this case, the Median is given by the (n + 1) / 2, i.e., 20/2 = 10th in the series, which is $ 3.10 K. Range is the difference between the maximum and minimum value in the series. Range = Maximum Value – Minimum Value = $ 3.7 K - $ 1 K = $ 2.70 K From the normal distribution curve, 95% of the area is covered within 1.96 standard deviations of the mean. Hence to arrive at the confidence interval of the mean for a 95% significance level, the standard error is computed. Standard Error = Standard Deviation / √n = $ 0.60 K / √19 = $ 0.14 K Confidence Interval = {Mean + 1.96 Std. Error, Mean - 1.96 Std. Error} = {$ 3.01 K + (1.96) (0.14), $ 3.01 K - (1.96) (0.14)} = {$ 3.28 K, $ 2.73 K} Electronic Systems: The data for ‘Electronic Systems’ section are extracted from the given raw data and summarized below. The percentage of each of the salaries when compared to the total in the section is also presented in the table. Salary ($ K) Percentage 3 8.77% 4.1 11.99% 2.9 8.48% 3 8.77% 3 8.77% 3 8.77% 3.1 9.06% 3 8.77% 3 8.77% 3.1 9.06% 3 8.77% Total = 34.2 Mean Salary = Sum of the salaries / Number of entries = 34.2 / 11 = $ 3.11 K Standard Deviation = √(Sum of squared differences with the mean / n) = $ 0.33 K Mode refers to the most frequently appearing value in the set of data. In this case, ‘$ 3 K’ occurs seven times and hence Mode is $ 3 K. Median refers to the central value when arranged in ascending or descending order. In this case, the Median is given by the (n + 1) / 2, i.e., 12/2 = 6th in the series, which is $ 3 K. Range is the difference between the maximum and minimum value in the series. Range = Maximum Value – Minimum Value = $ 4.1 K - $ 2.9 K = $ 1.20 K From the normal distribution curve, 95% of the area is covered within 1.96 standard deviations of the mean. Hence to arrive at the confidence interval of the mean for a 95% significance level, the standard error is computed. Standard Error = Standard Deviation / √n = $ 0.33 K / √11 = $ 0.10 K Confidence Interval = {Mean + 1.96 Std. Error, Mean - 1.96 Std. Error} = {$ 3.11 K + (1.96) (0.10), $ 3.11 K - (1.96) (0.10)} = {$ 3.31 K, $ 2.91 K} Power Systems: The data for ‘Power Systems’ section are extracted from the given raw data and summarized below. The percentage of each of the salaries when compared to the total in the section is also presented in the table. Salary ($ K) Percentage 3 11.19% 3 11.19% 1 3.73% 2 7.46% 3 11.19% 3 11.19% 2.9 10.82% 3 11.19% 3 11.19% 2.9 10.82% Total = 26.8 Mean Salary = Sum of the salaries / Number of entries = 26.8 / 10 = $ 2.68 K Standard Deviation = √(Sum of squared differences with the mean / n) = $ 0.67 K Mode refers to the most frequently appearing value in the set of data. In this case, ‘$ 3 K’ occurs six times and hence Mode is $ 3 K. Median refers to the central value when arranged in ascending or descending order. In this case, the Median is given by the (n + 1) / 2, i.e., 11/2 = 5.5 (Mean of 5th and 6th values) in the series, which is $ 3 K. Range is the difference between the maximum and minimum value in the series. Range = Maximum Value – Minimum Value = $ 3 K - $ 1 K = $ 2 K From the normal distribution curve, 95% of the area is covered within 1.96 standard deviations of the mean. Hence to arrive at the confidence interval of the mean for a 95% significance level, the standard error is computed. Standard Error = Standard Deviation / √n = $ 0.67 K / √10 = $ 0.21 K Confidence Interval = {Mean + 1.96 Std. Error, Mean - 1.96 Std. Error} = {$ 2.68 K + (1.96) (0.21), $ 2.68 K - (1.96) (0.21)} = {$ 3.09 K, $ 2.27 K} Consolidated Data – Electrical Department: The salary information from all the sections are combined to gather information about the department on the whole. Section Total Salary ($ K) Percentage Accessories 15.40 8.45% Consumer Product 1 16.00 8.78% Consumer Product 2 32.80 17.99% Mainframe Systems 57.10 31.32% Electrical Systems 34.20 18.76% Power Systems 26.80 14.70% Total 182.30 100.00% Mean Salary of the Department = Sum of the salaries / Number of entries = 182.30 / 60 = $ 3.04 K Standard Deviation = √(Sum of squared differences with the mean / n) = $ 0.59 K Mode refers to the most frequently appearing value in the set of data. In this case, ‘$ 3 K’ occurs twenty five times and hence Mode is $ 3 K. Median refers to the central value when arranged in ascending or descending order. In this case, the Median is given by the (n + 1) / 2, i.e., 61/2 = 30.5 (Mean of 30th and 31st values) in the series, which is $ 3 K. Range is the difference between the maximum and minimum value in the series. Range = Maximum Value – Minimum Value = $ 4.1 K - $ 1 K = $ 3.1 K From the normal distribution curve, 95% of the area is covered within 1.96 standard deviations of the mean. Hence to arrive at the confidence interval of the mean for a 95% significance level, the standard error is computed. Standard Error = Standard Deviation / √n = $ 0.59 K / √60 = $ 0.08 K Confidence Interval = {Mean + 1.96 Std. Error, Mean - 1.96 Std. Error} = {$ 3.04 K + (1.96) (0.08), $ 3.04 K - (1.96) (0.08)} = {$ 3.20 K, $ 2.88 K} Task 2: Short explanations including the limitations on the significance of the above calculations in relation to the monthly income: The central tendency measures indicate how the data is distributed in the various sections of the electrical department and also the electrical department as a whole. The mean along with standard deviation is a valid measure to identify the dispersion of the salaries within the data. Mode, on the other hand, indicates the salary earned by a majority of the employees in a section. Median along with range is a clear indication of the highest and lowest salaries in the section along with their dispersion within the data. The main calculation in the above task is that of the confidence interval, which is a significant measure, indicating that 95% of the population’s mean can be derived from the data given within the limits computed. One of the major limitations on the significance of the calculations is that it does not provide any credible information about the level of skill required by the employees in various sections. For instance, there might be a mixture of highly skilled and low skilled employees in a section. In this case, the central tendency measures, such as the mean and standard deviation, mode, median and range are not credible estimates regarding the sample taken. Moreover, the confidence intervals also become obsolete, as the sample mean is not credible. The calculations are based on the salary data from a single month, i.e., September 2006. It is highly possible that the data might be skewed, as the salaries may not be consistent throughout the year. Hence the above calculations may not actually represent the entire population of salaries over the whole year. Task 3: There are several selection methods that can be used while recruiting a person for the new job. The further sections will deal with the advantages and disadvantages of various initial selection processes. - Face to Face Interviews: There are several advantages of face to face interviews. Here the company can adapt the questions as necessary and can clarify any doubts. Also it allows for ensuring responses are clear and properly understood. This can also include repeating and rephrasing the questions. This form of interview also allows avoiding any form of discomfort, stress, problems that might be faced by the respondents. Also for the interviewers, there is a chance for them to check for the body language and unconsciously exhibited habits of the person. Also it is easier to understand a person and their character based on a face to face meeting rather than any other form. However, one of the biggest disadvantages of face to face interviews is the amount of costs involved in arranging for the meetings. The interviewers and company need to spend a lot to find the right person for the job. Other disadvantages also include the geographical limitations, costs of training the interviewers to ensure reduced bias and no discrimination, also in the case of the respondents there is a chance of anonymity issues when in interviews of this kind. - Panel Interviews: This is another form of interview that can be adopted by the company. This is relatively a less common form of interview. There are several advantages of this form of interview. Firstly, the panellist can focus on the answers instead of constant thinking of what next has to be asked. There can be a varied choice of questions as the perspectives of different people are different at all times. Also this form of interview saves time and costs and does not require having multiple interviews for the candidates. There are several disadvantages that are present for this form of interview as well. These include, that it can be very intimidating to the candidates, can confuse the candidates since the goal of different interviewer and panellist is different for the interviews. Also it can be difficult to have followed – up questions to the answers by the candidates. It is also relatively difficult to get a good understanding of the candidates because, each panellist is allowed one question and this cannot provide an understanding of the consistency in the answers by the candidates. It is also difficult to see how the candidate responds to different people and styles of conversations as well. Also it is not as effective as a face to face, one on one interview as the questions are directly fired at the candidate and there is no clear conversation here. It can also be very costs and time waste to have several people leave their work and participate in conducting the interviews. This will reduce the money otherwise the people could have earned for the company as well. - Postal: This is one of the least famous forms of interviews. This is more of the initial levels of the interviews. Here the company can send out questionnaires and can also gain curriculum vitae from the post. Companies can also use this to gain many applicants to respond back to the advertisements. This is a very cost effective mode to send gain as many people as possible. However this is also faced with a number of issues. Firstly, it would be difficult to check for the genuinely of the applications. Secondly, if a company does decide to check for each application and to check if it is genuine, then the cost involved will be exorbitant. It is relatively better for the company to use a face to face interview and pay for the expenses of that rather than the extra costs. This is relatively not a very well accepted form of initial choosing of people. - Email: With the growth of internet and use of technology, it has been noted that most companies use emails as a form of initial short listing of candidates. The main advantage of this is that it breaks all barriers and allows people to apply for jobs without having to consider the limitation in terms of the locations. Also it allows people to have constant access at all times and leaves them a chance to check the mails constantly irrespective of where they are. This is also one of the fastest forms of communication and it is preferred to any other form as it allows the candidate to stay in touch constantly at all times. The main disadvantage of this is that it does become difficult to know whether the questionnaires are filled by the candidate themselves or by any other person. Also it can be misused by candidates and applications can come from any part of world. It can be abused to a great extent as well. - Telephonic Interviews: This is a growingly choice of most interviewers. These form a very strong form of initial interviewing as the companies can call the candidate and speak to them over the phone directly. Here the company can ask the person for a number of questions which include examples of their previous job as well as other questions as required by the job. The advantages of this method are many. The cost of making calls to the candidates is relatively much lower than inviting the candidates in person for the interviews. Also, the telephonic interviews allow the company to gain a strong understanding of the candidate communications skills. Also it allows the company to interview people from different cities as well. This however can be disadvantageous to some extent. It does not allow to get a clear perspective of the candidates and to view their body language. All the information that is gained from interviews and from the curriculum vitae need to be kept confidential. This is essential as the curriculum vitae and interviews generally involve a number of personal details being shared with the human resources team. Hence the only people who can have access to the information are the owners of the company, the human resources team executive handling the case of the candidate and the respective managers for whom the interviews are being conducted. Any mistakes or errors in handling the information can lead to the company being charged under the data protection act. It is the responsibility of the company to ensure that all the information and data that is available, not only of the employees but the customers as well, is completely confidential and no unauthorised personnel should have access to this information. The information which is generally available can be used for a number of illegal activities. These include identity theft, credit card frauds and many more. Also it can lead to a number of issues within the company as well. Task 4: The information that has been discussed above deals with the salaries of the various departments in the company. The salary and finances of a company are very sensitive data. The information needs to be accessed by the only the intended users, here it would mean only the finance departments and the owners of the business alone. This is essential as any leak of information or any errors made in these calculations has a direct impact on the business both financially as well as legally. The salary information is legally allowed to be used only by the finance department of the company. It is essential that even managers of the company are not aware of the salary being earned by the people and there is no personal information of the individual that are also available to anyone else within the organisation. This is majorly because the data protection act binds all companies to follow the rules and to comply with no sharing of any personal information, this would include, everything including name, address, phone numbers, email ids, salary, security numbers and anything that could be compromised. Identity theft is when personal identification information is stolen for the illegal use of opening bank accounts, stealing money, pension payouts or to gain any other benefits. Here the accused generally pretends to be someone else for his own benefit at the cost of his victim. For instance in US the personal information such as name, address, date of birth, phone number and social security number are the most essential and are required for opening bank accounts. Thus loss of any one of these pieces of information could open a chance for id theft. Identity theft could seem to be a simple problem but the consequences could be drastic depending on the kind of fraud done by the thief. It could lead the victim to have a bad credit report, or even lead to arresting the victim for any other drastic crime. Identifying an identity theft is a very long process and can take even as long as twelve months by the time very large damages can have been made. Task 5: Memorandum To: Section and Department Heads CC: Chairman, CEO From: Personnel Officer – Electrical Department Date: 8/29/2016 Re: Meeting – Reg. Salary and Selection of new Employees Meeting Details This is to request you to be a part of the meeting that will be held at Conference Room on 27th August, 2009 at 3 PM. The meeting is to discuss the various aspects related to the salary and selection of new employees for the team. The information regarding Electrical Team salaries will also be analyzed. Meeting Agenda The meeting will begin with a short presentation on the salary information of the employees in the Electrical Department. Following the presentation, a brief discussion on the issues related to selection process and the confidentiality issues will take place. Also, a decision needs to be made for the selection process to be adopted for the new employees. The meeting will be wrapped up by 4.30 PM. Hope to see you all there. Please make sure you attend this meeting as the issues are of prime importance. Read More
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