FTE and Variance Assignment

FTE and Variance

Section One: Calculation of Full-Time Equivalents

Personnel Budget Case Study

Question One

There are a number of items that are required in order to ensure that a budget is effectively prepared. According to lectures of this course, several items that are required include: patient acuity, patient days, length of a patient stay, number of the vacancies that are anticipated in each level, staffs’ educational needs in the in the year that follows, costs associated with staff benefits, as well as non-productive staff time that is anticipated. The background information/data about 1 West unit is as follows:

Patient Data: Average Daily Census (30); Unit Capacity (32); Average HPPD (8.8); and Total Care Hours (96,360)

Staff Data: Productive hours/employee/year (1,780); Nonproductive hours/employee/year (300); and Total Hours/employee/year (2,080)

Skill Mix: RNs (80%); LVNs (10%); and Nurse Aides (10%)

Question Two

Calculating the number of the productive FTEs to be needed:

The initial step would be to consider the subsequent year’s workload, figuring in mind that the anticipated number of patient days will be 10,950 [obtained by multiplying average daily census (30) by 365 days].

Step 1:

Calculating Workload = HPPD x Number of patient days = 8.8 x 10,950 = 96,360

Step 2:

Calculating Productive time = Subtracting total nonproductive hours from total FTE hours; 2,080 – 300 = 1,780 hours

Step 3:

Calculating FTEs = Workload divided by productive time: 96,360/1,780 = 54.3 productive FTEs needed to staff the unit.

Question Three

Determining the number of RNs, LVNs, as well as nurse aides that will be required to ensure that the unit is well staffed based on the assumption that there will be a 12-hours working shifts for the staff. In other words, this requires the calculation of the number of persons-shifts to be need during periods of 24-hour shifts.

Since the staffs are shared 50% for day shift and 50% for night shift in the following mix RNs (80%); LVNs (10%); and Nurse Aides (10%), the numbers are as follows:

In a 12-hour shift:

RNs = 80/100 * 54.3 = 43.44

LVNs = 10/100 * 54.3 = 5.43

Nurse aides = 10/100 * 54.3 = 5.43

Question Four

Assigning the staff by shift and by type considering that:

Day Shift                    50%

Night Shift                  50%

Then,

RNs = 80/100 * 54.3 = 43.44 each shift

LVNs = 10/100 * 54.3 = 5.43 each shift

Nurse aides = 10/100 * 54.3 = 5.43 each shift

Question Five

Coverting staff positions to full-time Equivalents or FTE positions involves converting to 24/7 as follows:

Total FTEs = 54.3 x 1.4 = 76.02 FTEs needed to staff the unit 24/7, which is equivalent to 76.02 FTE positions

This will help convert staff positions to FTE positions needed for a 24/7 shift as follows:

RNs = 80/100 * 76.02 = 60.8

LVNs = 10/100 * 76.02 = 7.6

Nurse aides = 10/100 * 76.02 = 7.6

Section Two: Variance Analysis

Variance Analysis Case Study

1. Comparing the original budget to the flexible budget reveals that, the original one had no significant unfavorable variances compared to the flexible one, which is characterized by unfavorable variances not only in terms of volume but also in terms of price and quantity.  For instance, a volume variance is experienced when the actual volume is higher or lower than the budgeted volume and may be expressed in terms of FTEs or patient days.  To determine volume variance, the calculation is as follows:

(Budgeted Volume – Actual Unit Volume)(Budgeted Rate) = Volume Variance

In order to get the budgeted rate, the following formula is used:

Budgeted Rate = Budget Allocation/Budgeted Volumes

Thus,

Budgeted Rate = \$40 *5 * 340/340 = \$200

Volume Variance = (340 – 400) (200) = – 12,000

A comparison between the flexible budget and actual budget shows that, there are unfavorable variations in terms of price and quantity. This is because employees’ average hourly rate has increased from \$40.00 to \$45.00, whereas hours per care per patient have increased from 5.0 to 5.6. These two changes are indicative of an unfavorable variation. Price variance or unit cost variance determination can be calculated using the equation shown below:

(Budgeted Unit Price – Actual Unit Cost)(Actual Volume) = Unit Cost Variance

Price variance = [(40 * 5) – (45*5.6) (400)] = – \$20,800

In addition, quantity variance determination can be calculated using the equation shown below:

(Budgeted Use – Actual Use)(Budgeted Unit Cost) = Quantity Variance

Quantity variance = [(340 – (400) (40 * 5)] = -12,000

• There are several factors that led to the differences in variances including both external and internal factors. In particular, the internal factors include changes in staff efficiency which has led to an increase in the hours per care per patient from 5.0 to 5.6. Changes in technology as well as nature of surgeries may have also be other internal factors that have led to these variations.
• This is because all of these factors can collaboratively combine to delay the rate at which patients are operated eventually increasing the overall average time required to provide care to each patient in hours per visit. On the other hand, external factors that may have caused the variations include type of staff available, census changes as well as price changes, all of which can be attributed to an increase in the hours per care per patient from 5.0 to 5.6, number of visits from 340 to 400 as well as employees’ average hourly rate from \$40.00 to \$45.00. The unfavorable variations occurred because the above discussed factors combined to negatively impact the budgeted figures.

References

Zimmermann, P. G. (2002). Nursing management secrets, Issue 974; Volume 13 of Secrets series. Elsevier Health Sciences. p. 55.

Generating a Hypothesis

Inferential Statistics and Findings: Generating a Hypothesis

The aim of the paper is to generate a hypothesis with the use of research questions and two variables- dependent variable: sales/revenue and independent variable: interest rates. In addition, the paper uses a statistical technique for testing the hypothesis at 95% confidence interval while interpreting the results. XYZ Corporation is a US-based company that majors in home improvement and construction.

Moreover, the company operates in various big-box format stores across the United States. Nevertheless, XYZ Corporation wants to determine the effects of interest rates and sales on its future profitability, therefore, the company approached a research team that suggested the below research question.

Research question

Is there a relationship between federal (Feds) directed interest rate (ID) and sales /revenue (DV) on future profit of XYZ Corporation?

Hypothesis
H0 = XYZ Corporation’s future profits are not determined by interest rates and sales/revenue
H1 = XYZ Corporation’s future profits are determined by interest rates and sales/revenue

Hypothesis tests with a 95% confidence level

The veracity of the null hypothesis depends on data  as well as the objective of significance tests to estimate the likelihood, hence the P-value, which underscores that the occurrence is an issue of chance. In this case, if the Pearson Correlation is less than 5% it means that  the null hypothesis is rejected.

However, if the P-value is greater than 5% then it means the statistical test fails to reject the null hypothesis (Gupta, 2012).  Data was indiscriminately selected from 382 observations with respect to interest rates and sales of XYZ Corporation.

Interpretation of the results and findings

Statistical analysis demonstrates that this is a normal distribution, with a mean of 10.97644 and standard deviation of 10.14628 for sales (Table 1). In this case, sales were observed between \$0.00 and \$89.00.  Changes in the federal interest rates could help determine sales of XYZ Corporation.

In addition, the mean and standard deviation for interest rates were observed at 9.695538 and 7.836266 respectively (Table 2).  High-interest rates are highly likely to inhibit any profit making from lack of sales.  While low-interest rates by the federal government may lead to increased sales.

Again, it demonstrated a 95% confidence interval (sales:1.020716, interest rates: 0.789369) this implies that XYZ Corporation can be about 95 percent confident that sales and interest rates will lead to an increased future a profit. Because the confidence interval of sales corresponds to the value of alpha, which is 1 indicates that there is no significant correlation between sales and future profit of XYZ Corporation.

However, the results are empirically reliable. When it comes to interest rates the confidence interval is less than 1, statistically the test fails to refute the null hypothesis, an aspect that demonstrates statistical significance of the test (Monette, Sullivan & DeJong, 2011).

Table 1. Descriptive statistics showing XYZ Corporation Sales

Table 2. Descriptive statistics showing XYZ Corporation Interest Rates

References

Gupta, S. K. (2012). The relevance of confidence interval and P-value in inferential. Indian Journal of Pharmacology. Retrieved on  June 9, 2016, from statistics http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3271529/

Monette, D. R., Sullivan, T. J., & DeJong, C. R. (2011). Applied Social Research: A Tool for the Human Services (8th ed.). Belmont, CA: Brooks/Cole.

Descriptive statistics and inferential statistics

Descriptive statistics and inferential statistics

1. Introduction

This report is an analysis of the data generated using SPSS and presented using charts and tables. The report firstly presents the results of selected descriptive statistical analyses. Subsequently, the report summarises the numerical results with descriptive statistics analysis tables or graphs, including the interpretation of these tables and graphs. The fourth section or the report is a presentation of the data regarding numerical results of the inferential statistics. This is followed by a discussion of the same, before a summative conclusion is presented in the last section.

• Selected descriptive statistics

Descriptive statistics refers to the kinds of data that analysts and researchers use in presenting the characteristics of the sample used in a study. According to Kothari (2004), they are used in checking whether the variables that the researcher has chosen to use violate any assumptions that the researcher might have made, which might be consequential to the findings. Another important function of descriptive statistics used in this section is that they help to answer the core research questions.

In the present study, the descriptive statistics selected are for public use micro data area code (PUMA), house weight (WHTP), state code, (ST), numbering of persons (NP), rooms (RMS), bedrooms, (BDS), and household income (HINCP). The data retrieved was as presented in table 1 below

Table 1: PUMA, ST, BDS, RMS, mean, median, and standard deviation

Table 2: RMS, BDS, ST, and PUMA, frequency table

From the data in table 1 above, a number of observations are blatant and clear. The first is that the means of RMS, BDS, ST and PUMA are 4.87, 2.61, 15, and 248.05 respectively. For rooms, the number of rooms, the median score was 5, where the scores varied from 1 to 9. This means that the majority of respondents have about 5 rooms.

When it comes to the number of bedrooms, the median score was 3, whereas the mean was 2.61, this shows that the majority of respondents have 3 rooms. The state code was 15 for all respondents whereas the mean for public use of micro data area code was 248.05. The mean was 302, whereas the minimum and maximum scores were 100 and 307 respectively.

From table 2, a number of assertions can also be made, and the first is about PUMA. From the table, the evidence shows that for public use of micro data area code, 19.4% of the respondents scored category 100, which made it the highest selected category, whereas 15.9% of the respondents checked 200, making it the second most selected category. Comparatively, 301 was the least selected category at 8.7%.

Additionally, for number of bedrooms, a majority of the respondents said that they had three bedrooms in their houses, and this represented 3.9% of all responses, closely followed by those with two bedrooms at 24.6%. At the same time, the number of people living in houses with no bedrooms or five bedrooms was the least with a score of 4.3% and 6.3% respectively.

This data is in line with the data about rooms, which shows that 22% of respondents stay in a five-roomed apartment, followed by 18% and 15%, who stay in four and five roomed houses respectively. Because of the number of rooms and bedrooms in their houses, it is plausible to conclude that a majority of the respondents stay with other people or expect other people to visit often, which are why they have extra rooms in the house, as well as extra bedrooms in the house.

Additionally, from the data, it is obvious that a majority of the people are in the middle between the rich and the poor, as those who stay in studio apartments are as marginal as those who stay in luxury apartments that can contain at least five bedrooms. .

• Selected inferential statistical analyses

Inferential statistics refer to the data analysis methods where the researcher or analyst uses a given set of data to determine whether there is a link between given variables being studied. By using inferential statistics, the researcher can tell whether the relationship that seems to exist between variables is a fact, or whether it is not a fact. According to Kothari (2004), a number of measures and techniques can be used to accomplish inferential statistics. The two types of inferential statistics used in this report are correlation and regression analyses.

Correlation was conducted using the Pearson correlation analysis. Pearson correlation analysis is employed to measure the linear relationship between two or more variables. The value of Pearson correlation ranges between -1 and +1, with -1 indicating negative correlation, 0 indicating no correlation and +1 indicating positive correlation between the variables.  Besides, the closer the value is to +1, the stronger the relationship between the variables (Saunders, Lewis & Thornhill, 2007). For this study, the data is as shown below.

According to table 4-20, Sig. (2-tailed) =0.000, and all the four variables have a significant correlation at the 0.01 significant level. Pearson correlation between PUMA and NP is .110, whereas the relation between PUMA and BDS and RMS is .042 and .067 respectively. This shows that there is a weak but positive relationship between PUMA and all the independent variables, although the weakest relationship is that between PUMA and BDS.

Table 3: Correlations

**  Correlation is significant at the 0.01 level (2-tailed).

Regression analysis helps estimate and investigate the association between variables. R Square is used to show the degree of relationship between the dependent and independent variables. R Square value ranges between 0 and 1, and the closer the value is to 1, the stronger the relationship between the variables further indicating the greater degree to which variation in independent variable explains the variation in dependent variable (Seber and Lee, 2012).

Based on the model summary table 4-21, R stand for the correlation coefficient and it depicts the association between dependent variable and independent variables. It is evident that a positive relationship exists between the dependent variable and independent variables as shown by R value (0.126).

However, the relationship is a very weak one. Besides, it can be seen that the variation in the three independent variables (RMS, BDS and NP) explain 1.6% variation of PUMA as represented by the value of R Square. Therefore, it means that other factors that are not studied on in this study contribute 98.4% of the PUMA programs. This means that the other factors are very important and thus need to be put into account in any effort to enhance PUMA. Additionally, this research therefore identifies the three independent variable studied on in this research as the non-critical determinants of PUMA boundaries.

Table 4: regression analysis results

Model Summary

a  Predictors: (Constant), RMS, NP, BDS

Further, this research established through the analysis f variance that the significant value is 0.00, which is less than 0.01, therefore the model is statistically significant in foretelling how NP, RMS, and BDS can influence PUMA groupings. The F critical value at the 0.01 level of significant was 26.501. Given that F calculated  is greater than the F critical value of 26.501, then it means that the overall model was significant (Seber and Lee, 2012).

ANOVA(b)

a  Predictors: (Constant), RMS, NP, BDS

b  Dependent Variable: PUMA

At the same time, the beta coefficients also gives significant inferential information. According to the regression coefficients presented in table 4-23, this research found that when all independent variables (the number of persons (NP), number of rooms (RMS), and the number of bedrooms (BDS)) are kept constant at zero, the level of public use micro data area code (PUMA)  will be at 231.13. A 1% change in number of persons will lead to an 11.4% increase in PUMA, whereas a one percent change in BDS will lead to a 12.1% changes in PUMA.

Comparatively, a one percent change in  RMS will lead to a 12.8 percent change in PUMA. This leads to the conclusion that of the three variavles, RMS leads to the largest impact in PUMA when the three independent variables are pitted together. Further the statistical significance of each independent variable was tested at the 0.01 level of significance of the p-values.

Coefficients(a)

a  Dependent Variable: PUMA

In general form, it can be said that the equation used to determine the link between  Public use microdata area code, numbering of persons, rooms and bedrooms is of the form:

Y = β0+ β1X1+ β2X2+ β3X3+ ε

From the equation, β0 is a constant, whereas β1 to β3 are coefficients of the independent variables. X1 X2 and X3 are the independent variables numbering of persons, rooms and bedrooms respectively, whereas epsilon ε is an error term. Additionally, the dependent variable Y in the equation represents public use microdata area code. Pegging the present discussion in the formula above, the model would be as follows.

Y = 231.130 + .114X1 – .121X2 +.128X3

This means that the public use micordata area code = 231.130 + (0.114 x numbering of persons) – (0.121 x rooms) +(0.128 x bedrooms).

References

Kothari, C. (2004). Research methodology, methods & techniques (2nd ed.).New Delhi: Wishwa Prakashan.

Saunders, M., Lewis, P. & Thornhill, A. (2007). Research Methods for Business Students. 4th edition. England: Prentice Hall.

Seber, A. F. G. and Lee, J. A. (2012) Linear Regression Analysis. 2nd Edition. Hoboken, New Jersey: John Wiley & Sons