Survey of Employee Satisfaction Essay
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Bus409 Managing Finance and Statistical Data – Assignment 1
Introduction ………………………………………………………………………………………… …
Undertaking 1 – Sampling
Undertaking 2 – Graphs and Calculation
Undertaking 3 – ttest
Undertaking 4 – Correlation and Regression
Undertaking 5 – Written Analysis
Appendix 1 – Copy of random figure tabular arraies used
Appendix 2 – Sample of 30 records used
Introduction
The sample of informations analysed in this assignment is concerned with study consequences for 363 employees from an administration based in 8 different metropoliss. In this assignment descriptive and illative statistical analyses were practiced. The informations were collected as a footing for analysis. Descriptive statistics used for the populations, analyzing, correlativities between your graduated tables, and illative analysis utilizing appropriate ttests or analyses of discrepancy.
The first set of analyses of information is called descriptive statistics. The end of this signifier of analysis is to make quantitative sumups of the dataset. Descriptive statistics inform on the mean measurings of a graduated table or point, how much these measurings vary across observations, the scope of measurings, and other facets cognizing about informations. This signifier of analysis besides includes making ocular representations of your informations as Bar chart, histogram and cumulative frequence curve and spread secret plans. Ocular tools help the research worker identify anomalousnesss, outliers, and tendencies in informations.
In this assignment variable were used to prove hypotheses about informations utilizing illative trials. The trials used to place how categorical variables affect graduated tables or response variables.
The concluding measure in this assignment shows how the graduated tables and points of informations correlative with each other. The end of looking at correlativities is to place relationships between measurings and suggest possible causal relationships. For case, how much employee like workplace and how they rate working status runing from 1 ( low satisfaction and low importance ) to 5 ( high satisfaction and high importance ) .
Undertaking 1 – Sampling
The information was sampled utilizing a set of random figure tabular arraies bring forthing from the Excel. A random figure tabular arraies to randomly choose a sample of 30 employees from the 363 records.
A transcript of the random figure tabular arraies used and the sample of 30 records can be found in the appendix of this assignment
Undertaking 2 – Graphs and Calculation
Comparison of Employment Status and Recommend Organisation
Organization recommendation by employment type and figure of employees variables were chosen for comparing. These variables were taken from 30 employees random figure tabular array ( see appendix 2 ) . A chart was created utilizing those variables to visualize the comparing of Employment position and Recommendation.
Recommend no 
Recommend yes 

Causal 
2 
6 
Permanent wave 
4 
18 
The Charts above shows that 18 Employees who are Permanent and 6 who are Casual recommend yes. On the contrary, 4 Permanent and 2 Casual employees recommend no. There is besides a important difference can be visualised between positive and negative recommendation. The entire figure of positive recommendation is 24 ( 80 % ) out of 30. On the other manus, the combination of negative recommendation is 6 ( 20 % ) out of 30 ; which is one fifth of the entire participant.
Satisfaction with Working Conditionss
In order to cipher frequence, satisfaction of working status mark is used from a random figure tabular array which shows in appendix 2. The sores were taken into lowest to highest and bins were calculated in order. In order to cipher the frequence, a expression of frequence “=FREQUENCY ( G2: G31, G35: G40 ) ” was used. Centers were calculated by figure taken from tonss and bins by choosing their inbetween figure manually. The histogram was created utilizing frequence Numberss in excel with 2D clustered column. Then from informations series spread breadth reduced to 0 % to make a complete histogram.
Frequencies utilizing Bins
Tonss 
Bins 
Frequency 
Midpoints 
2.0 – 
2.5 
1 
2.25 
2.5 – 
3.0 
5 
2.75 
3.0 – 
3.5 
7 
3.25 
3.5 – 
4.0 
4 
3.75 
4.0 – 
4.5 
9 
4.25 
4.5 – 
5.0 
4 
4.75 
Formula
Tonss 
Bins 
Frequency 
Midpoints 
2.0 – 
2.5 
=FREQUENCY ( G2: G31, G35: G40 ) 
2.25 
2.5 – 
3.0 
=FREQUENCY ( G2: G31, G35: G40 ) 
2.75 
3.0 – 
3.5 
=FREQUENCY ( G2: G31, G35: G40 ) 
3.25 
3.5 – 
4.0 
=FREQUENCY ( G2: G31, G35: G40 ) 
3.75 
4.0 – 
4.5 
=FREQUENCY ( G2: G31, G35: G40 ) 
4.25 
4.5 – 
5.0 
=FREQUENCY ( G2: G31, G35: G40 ) 
4.75 
Histogram
The histogram above shows employee figure of frequence distribution in relation to satisfaction with working status. Tonss between 2.5 to 3.5 and 4.0 to 4.5 are rather frequent. On the other manus, tonss between 3.5 to 4.0 and 4.5 to 5.0 are somewhat lower but every bit balanced. And score 2.0 to 2.5 is a really low graduated table. Finally, due to the big value, the histogram is somewhat skewed to the left, or negative skewed. Without this value, the histogram would be moderately symmetric.
Accumulative Frequency Curve for Satisfaction with Working Conditionss
The cumulative frequence is the running sum of frequences. On a graph, it is represented by a cumulative frequence polygon, where consecutive lines connect to the cumulative frequence curve. These informations were used to pull a cumulative frequence polygon by plotting the cumulative frequences against the category. Accumulative frequence was calculated utilizing frequence Numberss which were created earlier.
Class 
Accumulative Frequency 
Bins 
up to 2.5 
1 
2.5 
up to 3.0 
6 
3.0 
up to 3.5 
13 
3.5 
up to 4.0 
17 
4.0 
up to 4.5 
26 
4.5 
up to 5.0 
30 
5.0 
Accumulative Frequency Curve
The graph above shows a smooth curve. The cumulative frequence graph above illustrates employees satisfaction with working conditions. By pulling horizontal lines it represents first quartile of the entire frequence, 2nd quartile of the entire frequence, 3rd quartile of the entire frequence, can read estimations of the lower quartile, average and upper quartile from the horizontal axis. Though here a set N value nowadays, the undermentioned computation is done in order.
Table 1. Formula
Lower quartile = 
n+1 
Thursday value 

4 

Median = 
n+1 
Thursday value 

2 

Upper quartile = 
3 ( n+1 ) 
Thursday value 

4 

Interquartile scope = 
Upper quartile – Lower quartile 

Semiinterquartile scope = 
Interquartile scope 

2 

Though there are 30 values ; n=30, is an even figure, from the undermentioned expression ;
 Lower quartile = 7.75 th value ; which would be 2.62
 Upper quartile = 23.25 th value ; which would be 3.84
 Median = 15.5 th value ; which would be 3.28 ( agencies median would be 15 Thursday and 16 Thursday values average ) .
 Interquartile scope ( 3.84 – 2.62 ) =1. 22 and
 semiinterquartile scope ( 1.22/2 ) =0.61
Statistical Calculations
The below statistical computations shows that mean 6.07 for length of services is exact reply, whereas median is 3. There is a big difference between mean and average because of utmost value e.g. 34
The mean and median for the satisfaction of working conditions are 3.79 and 3.85 severally. These are good tonss out of five. And 4.12 and 4.15 are the mean and median for the importance of working conditions at the same time. These are more good tonss out of five.
Length of service( old ages ) 
Satisfaction ofWorking conditions 
Importance of working conditions 

3 
2.8 
2.7 

1 
3.7 
3.9 

18 
2.9 
3.1 

14 
3.2 
3.9 

3 
3.2 
3.7 

13 
2.4 
4.3 

9 
3.4 
4.0 

2 
4.7 
3.9 

8 
3.3 
4.8 

5 
2.9 
3.4 

5 
3.7 
3.5 

4 
2.9 
4.2 

0 
4.1 
4.1 

14 
4.4 
4.5 

3 
4.0 
4.0 

3 
4.2 
4.2 

1 
3.5 
4.0 

34 
5.0 
5.0 

1 
4.3 
3.1 

1 
4.2 
3.9 

3 
3.2 
4.9 

0 
4.5 
4.8 

1 
4.5 
4.4 

2 
4.3 
4.2 

8 
4.8 
5.0 

1 
4.4 
4.5 

10 
4.0 
4.4 

10 
3.4 
3.9 

4 
2.9 
4.6 

1 
4.9 
4.7 

Mean 
6.07 
3.79 
4.12 
Madian 
3.00 
3.85 
4.15 
Standard Deviation 
7.16 
0.72 
0.58 
Formulas Used 

Mean 
=average ( B2: B31 ) 
=average ( C2: C31 ) 
=average ( D2: D31 ) 
Median 
=median ( B2: B31 ) 
=median ( C2: C31 ) 
=median ( D2: D31 ) 
Standard Deviation 
=stdev ( B2: B31 ) 
=stdev ( C2: C31 ) 
=stdev ( D2: D31 ) 
Remarks: consequences are dependable
Mean and Standard Deviation of a Frequency Distribution
Statisticss expressions are: mean, =and standard divergence, i?? =
Length of Service used for frequence distribution and organised lowest to highest order in order to bring forth their midpoint.
Length of 
Midpoint 
frequence 

Service 
ten 
degree Fahrenheit 
fx 
fx^{2} 
04 
2 
18 
36 
72 
59 
7 
5 
35 
245 
1014 
12 
5 
60 
720 
1519 
17 
1 
17 
289 
2024 
22 
0 
0 
0 
2529 
27 
0 
0 
0 
3034 
32 
1 
32 
1024 
amount 
30 
180 
2350 

mean 
6.00 
St.Dev. 
6.51 
Formulas used
The computation shows that the mean is 6.00 and the standard divergence is 6.51. The standard divergence measures the spread of the informations about the mean.
Drumhead Statisticss
Importance of Working Conditions allows to cipher drumhead statistics: mean, average, standard divergence, percentiles, etc.
Importanceof workingconditions 

Mean 
4.12 
Standard Error 
0.105089 
Median 
4.15 
Manner 
3.9 
Standard Deviation 
0.575596 
Sample Variance 
0.331310 
Kurtosis 
0.105686 
Lopsidedness 
0.541828 
Scope 
2.3 
Minimum 
2.7 
Maximum 
5 
Sum 
123.6 
Count 
30 
Assurance Level ( 95.0 % ) 
0.214931 
The drumhead statistics show ;
 The assurance interval ( 95.0 % ) is 4.12 ± 0.21.
 Kurtosis and Skewness less than 2*0.105 ( standard mistake ) , so the information is normal. In this table Kurtosis 0.105 and Skewness 0.541 both are less than 2*0.105.
Undertaking 3 – ttest
Ttest comparison satisfaction with working conditions for lasting and insouciant employees
Hydrogen_{0}: There is no important difference Null Hypothesis
Hydrogen_{1}: There is a important difference Alternative Hypothesis
Analysis and account
 From the findings, the average satisfaction mark of the lasting employees is 3.8 and the average satisfaction mark of the causal employees is 3.9
 The tstatistic is – 0.383 and the critical value is 2.048 for a two tailed trial.
 Ignore the subtraction mark and since 0.383 & A ; lt ; 2.048, so Accept H_{0}
 Alternatively, since the pvalue = 0.704 & A ; gt ; 0.05 so Accept H_{0}
 The sample grounds indicates that there is no important difference in the satisfaction mark of the lasting and insouciant employees at the 5 % degree of significance. The sample besides grounds indicates Null Hypothesis can non reject because P & A ; gt ; 0.05
Undertaking 4 – Correlation and Regression
Scatter graph compared tonss in correlativity of satisfaction and importance of working conditions
The spread graph shows positive correlativity at the 5 % degree, which means within these two variables there is a strong positive relationship. Arrested development used to find the quantitative relationship between two variables, and normally follows on from the constitution of important correlativity. The equation of the arrested development line is “y = a + bx” ; where a ( the intercept ) and B ( the incline ) are invariables.
Undertaking 5 – Written Analysis
Summary of findings:
 In this assignment 30 employees record was collected from 363 population as a sample of informations analysis.
 Datas used for multiple saloon chart comparing drawn between Employment position and Recommend of organisation which shows that 18 Employees who are Permanent and 6 who are Casual recommend yes. On the contrary, 4 Permanent and 2 Casual employees recommend no.
 From graphical computations shows in a histogram that frequence distribution at 2.25 represent a category 03 for length of services of 30 employees. On the other manus, frequence distribution at 4.25 represent category 4 4.5.
 The drumhead statistics show that the assurance interval ( 95.0 % ) is 4.12 ± 0.21. Kurtosis and Skewness less than 2*0.105 ( standard mistake ) , so the information is normal. In this table Kurtosis 0.105 and Skewness 0.541 both are less than 2*0.105.
 The average satisfaction mark of the lasting employees is 3.8 and the average satisfaction mark of the causal employees is 3.9. The tstatistic is – 0.383 and the critical value is 2.048 for a two tailed trial.
 Ignore the subtraction mark and since 0.383 & A ; lt ; 2.048, so Accept H0
 Alternatively, since the pvalue = 0.704 & A ; gt ; 0.05 so Accept H0
The sample grounds indicates that there is no important difference in the satisfaction mark of the lasting and insouciant employees at the 5 % degree of significance.The sample besides grounds indicates Null Hypothesis can non reject because P & A ; gt ; 0.05
 The spread graph shows positive correlativity at the 5 % degree, which means within these two variables there is a strong positive relationship. Arrested development used to find the quantitative relationship between two variables, and normally follows on from the constitution of important correlativity. The equation of the arrested development line is y = a + bx ; where a ( the intercept ) and B ( the incline ) are invariables
Appendix 1 – Copy of random figure tabular arraies used
375 
183 
106 
366 
240 
74 
62 
121 
179 
153 
174 
229 
308 
175 
38 
80 
307 
54 
46 
196 
275 
152 
388 
359 
54 
272 
267 
76 
348 
84 
234 
17 
335 
380 
266 
276 
192 
311 
180 
50 
333 
194 
149 
147 
311 
383 
25 
62 
82 
337 
284 
185 
351 
68 
328 
68 
57 
297 
223 
211 
279 
343 
383 
258 
258 
36 
359 
75 
376 
23 
83 
117 
345 
85 
276 
350 
148 
315 
395 
62 
104 
380 
238 
304 
20 
15 
240 
115 
25 
353 
229 
29 
216 
139 
153 
64 
294 
55 
192 
117 
35 
324 
374 
335 
74 
330 
195 
342 
63 
139 
378 
301 
109 
270 
184 
214 
47 
177 
124 
173 
41 
18 
145 
398 
79 
248 
333 
101 
44 
90 
38 
132 
21 
139 
371 
144 
304 
100 
247 
2 
176 
259 
138 
226 
245 
245 
292 
77 
382 
368 
372 
348 
18 
201 
394 
3 
48 
394 
306 
348 
95 
360 
312 
205 
61 
351 
130 
316 
262 
334 
396 
193 
144 
294 
57 
390 
179 
145 
35 
346 
199 
117 
16 
276 
374 
203 
188 
335 
319 
388 
76 
127 
157 
96 
57 
299 
170 
244 
157 
279 
183 
95 
87 
378 
338 
310 
154 
197 
391 
163 
37 
101 
31 
333 
358 
65 
113 
343 
31 
105 
86 
338 
21 
135 
196 
325 
298 
284 
159 
188 
387 
301 
62 
397 
320 
19 
109 
162 
376 
347 
212 
33 
46 
60 
51 
66 
26 
197 
147 
112 
156 
132 
249 
259 
62 
229 
212 
305 
293 
344 
389 
66 
370 
245 
22 
329 
92 
152 
249 
14 
RendomNumber( row ) 
Idaho 
City 
AgeGroup( old ages ) 
Length ofservice( old ages ) 
Employmentposition 
Satisfaction ofWorkingconditions 
Importanceof workingconditions 
Recommendadministration as a good topographic point to work 
183 
279 
metropolis 3 
4150 
3 
permanent 
2.8 
2.7 
yes 
106 
156 
metropolis 1 
over 50 
1 
permanent 
3.7 
3.9 
yes 
240 
362 
metropolis 3 
4150 
18 
permanent 
2.9 
3.1 
no 
74 
112 
metropolis 2 
over 50 
14 
insouciant 
3.2 
3.9 
no 
62 
92 
metropolis 2 
4150 
3 
insouciant 
3.2 
3.7 
no 
121 
176 
metropolis 1 
over 50 
13 
permanent 
2.4 
4.3 
yes 
179 
275 
metropolis 3 
over 50 
9 
permanent 
3.4 
4 
yes 
153 
227 
metropolis 2 
4150 
2 
insouciant 
4.7 
3.9 
yes 
174 
270 
metropolis 3 
2130 
8 
permanent 
3.3 
4.8 
yes 
229 
349 
metropolis 3 
over 50 
5 
permanent 
2.9 
3.4 
no 
308 
466 
metropolis 8 
3140 
5 
permanent 
3.7 
3.5 
yes 
175 
271 
metropolis 3 
3140 
4 
permanent 
2.9 
4.2 
no 
38 
63 
metropolis 7 
3140 
0 
permanent 
4.1 
4.1 
yes 
80 
118 
metropolis 2 
over 50 
14 
insouciant 
4.4 
4.5 
yes 
307 
464 
metropolis 8 
4150 
3 
permanent 
4 
4 
yes 
54 
84 
metropolis 5 
3140 
3 
permanent 
4.2 
4.2 
yes 
46 
72 
metropolis 5 