Differences in the Employment Status among Groups of Graduates - Assignment Example

Professor

QUANTITATIVE ACCOUNTING

Introduction

The report is a data analysis to determine the relationship between the employment statuses based on the degrees offered in the XYZ University. The method of data collection used was the questionnaire. The questionnaire had only three basic questions;

• Which degree did you graduate?
• What is your current employment status?
• If employed what is your annual income?

The ABC University sent a questionnaire to 1000 graduates from the XYZ University whereby a number of underlying responses were received. Of the 1000 questionnaires, only 648 respondents responded to the data. Both the quantitative data and qualitative data techniques were used. Quantitative data entails numerical data that can be either discrete or continuous. It was the case for the annual income. Qualitative data entails data that is categorical which in this report was used to get data for the degree and the current employment status of the graduates’.

However, the qualitative data was organized to be quantitative in order to get the numbers required that would help in the analysis of the report. The disciplines that were focused on include health science, commerce, law, and engineering. The data was essential for ascertaining the most preferred courses in the employment industry. This, in regard, will affect the choice of courses for the students. The research was based on four objectives that were supposed to be addressed at the end of the research. These were:

• Are there differences in the employment status among the four groups of graduates?
• Are there differences in income among the four groups of graduates?
• Is income of Commerce graduates lower than the income of Health Science graduates?
• Is the proportion of employed graduates in Law discipline different from the proportion of employed graduates in engineering discipline?

In the data analysis, the report determines any differences in the employment status among groups of graduates, income differences and the proportion of employment among the four disciplines. The research data applied the use of graphs, and techniques such as ANOVA, t-test and regression models. The report will include the definition of parameters and clear outlining of assumptions and limitations in the research data. A key assumption was made on the annual income where those who were not employed did have a $0 income.

STATISTICAL ANALYSIS

Based on the data given in the excel sheet, it was evident that majority of the students graduated with the degree of health science as illustrated in the histogram below with rank 1, who was closely followed by the commerce students with rank 2. The third were engineering students with rank 4 and lastly the law students with rank 3. However, majority of the students who did not respond to the questionnaire came from the law class.

Graph 1

Graph 2

From the data analysis shown on graph 2, the majority of graduates are employed. The unemployed graduates and the graduates who seek to complete additional education have a minor difference where the unemployed stand to be higher.

Further analysis

On the hypothesis test conducted, the following assumptions were made.

• y~ N (β0+ β1x, δ2) then β1 and δ2 have the following;
• β1~ N (β1, δ2/ 2(X- ¯X)2
• ( n-2) S2/ δ2~ chi-squared (n-2)
• β1 and S2 are independent.

On the simple regression was conducted and the following assumptions were made2

• E (€i )= 0 for all i=1, 2 …., n
• Var (€i) = δ2 for all i=1, 2 …., n
• Cov ((€I, (€j ), i≠j = 0 for all i=1, 2 …., n: where € represents the error term

Question 1

For the ANOVA test.

A further ANOVA test was conducted to establish the relationship.

H0: µ1 =µ2 =µ3= µ4; H1: the means of two populations are different.

A significance level of 5% was used

Anova: Single Factor

SUMMARY

Groups

Count

Sum

Average

Variance

Column 1

200

389

1.945

0.464296

Column 2

179

365

2.039106

0.206327

Column 3

121

248

2.049587

0.330854

Column 4

148

299

2.02027

0.496185

ANOVA

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

1.198494

3

0.399498

1.06417

0.363669

2.618736

Within Groups

241.7629

644

0.375408

Total

242.9614

647

MSR=SSR/P-1

=1.1985/3

=0.3995

MSE=SSE/N-P

= 241.7629/644

= 0.3754

F= MSR/MSE

=0.3995/0.3754

=1.0642

F0.05: 1, 644= 2.619

Since the F value of the test statistic is less than the F0.05: 1, 644 we therefore fail to reject the null hypothesis and conclude that the means were same. There was a significant difference between the graduates’ degree and the annual income of the graduates. To further understand, the relationship between the employment status and the graduates’ degree a regression analysis was done and the following was the output.

The degree was the explanatory variable while the employment status was the response variable. The parameters to be estimated were β0 which was the mean response and β1 which was the variation between thedegree and the employment status.

To get β0 we used the formula β0 = ȳ- β1 bar x

Similarly; β1= nΣxy- ΣxΣy /nΣx2- (Σx)2

n= 648; ȳ= 2.007716; Σx= 1513; Σy= 1301; Bar x= 2.334877; ΣxΣy= 1968413

(Σx)2 = 2289169; nΣxy = 1982232; nΣx2 = 2833704

Having the above data, we therefore have β1 as 0.0254 correct to four decimal places and β0 as 1.9479 correct to four decimal places. Therefore, when degree is not there the employment status chance is 1.9479 and if there is a unit of degree the employment status increases by 0.0254.

Question 2

To determine if there was a difference in the means of the degree and the employment status a hypothesis test was conducted through ANOVA. The parameters to be estimated were the mean of the annual income in the different degree categories.

Hypothesis test

H0: µ1 =µ2 =µ3= µ4 H1: at least two means differ.

The significance level used was of 5%. The test used was that of Anova and the output was as follows from the excel workbook. The population was normally distributed with an assumption that the population variances were equal.

Anova: Single Factor

SUMMARY

Groups

Count

Sum

Average

Variance

Column 1

200

4421062

22105.31

4.8E+08

Column 2

179

6148392

34348.56

3.77E+08

Column 3

121

4786532

39558.11

8.96E+08

Column 4

148

4755913

32134.54

1.12E+09

ANOVA

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

26882024121

3

8.96E+09

13.27224

2.04E-08

2.618736

Within Groups

4.34793E+11

644

6.75E+08

Total

4.61675E+11

647

MSR=SSR/P-1

=26882024121/3

= 8960674707

MSE=SSE/N-P

= 4.34793E+11/644

= 675144133.5

Test Statistic F= MSR/MSE

=8960674707/675144133.5

=13.2722

F0.05: 1, 644= 2.619

Since the F value of the test statistic is greater than the F0.05: 1, 644 we, therefore, reject the null hypothesis and conclude that one of the means was different from the others. There was a significant difference between the graduates’ degree and the annual income of the graduates. Therefore, the degree of a student affects the annual income. To further determine the relationship between the annual income and the graduates’ degree a regression analysis was calculated. The graduate’s degree was the explanatory variable while the annual income was the response variable. The parameters to be estimated were β0 which was the mean response and β1 which was the variation between thedegree and the annual income. Using the least square method; To get β0 we used the formula β0 = ȳ- β1 bar x. Nevertheless, β1 is the variation between the degree and the employment status. Therefore, we find first β1 whereby;

β1= nΣxy- ΣxΣy /nΣx2- (Σx)2

n = 648; ȳ= 31036.88062; bar x= 2.334876543; Σx= 1513; Σy= 20111898.64;

Σxy= 50101092.24; ΣxΣy=30429302640; (Σx)2=2289169; nΣxy=32465507770;

nΣx2=2833704

From the above, we apply the formula above to find β1;

32465507770 – 30429302640/ 2833704 - 2289169

Therefore; β1= 3717.229.

To get β0 we used the formula β0 = ȳ- β1 bar x

31036.88062 – 3717.229 * 2.334876543

= 22357.89

In the analysis of the data above, β1= 3717.229. This entails that when a student undertakes a degree course at the university, upon completion and gets employed, his annual income increases by $3717.229. On the other hand, β0 =22357.89. Therefore, when there is no degree, the annual income is $22357.89.

Question 3

Since we are testing the mean differences of two independent samples for the health science and that of commerce, we will test the hypothesis. The data are quantitative and the parameters to be estimated is the difference between the two means (µ1-µ2)

H0: (µ1-µ2) = 0 ; H1: (µ1-µ2) > 0 ; α= 0.05

Since the population sample is not equal we use t-test assuming equal variances.

The t test statistic is given by; t= (x1¯-x2¯) (µ1-µ2)/√δ2(1//n1 + 1/n2)

N1= 200; n2= 179; x1¯= 22106.31; x2¯= 34348.58; δ2=

= (22106.31-34348.58)/ √428685473.9(1/200 + 1/179

= -5.7279

The t value for alpha 0.05 is 1.966 therefore we reject the null hypothesis and conclude that the mean differences were not equal. Therefore, there is a difference in the income of the commerce and that of health sciences.

Null Hypothesis: there was a significant difference between the annual income of the health sciences and that of commerce.

Alternative hypothesis: there was no significant difference between the annual income of the health sciences and that of commerce.

The significance level used was 5%. The test used was that of ANOVA and the output was as follows from the excel workbook.

Anova: Single Factor

SUMMARY

Groups

Count

Sum

Average

Variance

Column 1

200

4421062

22105.31

4.8E+08

Column 2

179

6148392

34348.56

3.77E+08

ANOVA

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

1.42E+10

1

1.42E+10

32.80838

2.08E-08

3.866243

Within Groups

1.63E+11

377

4.32E+08

Total

1.77E+11

378

Since the F value is greater than the F crit we, therefore, reject the null hypothesis and conclude that there was no significant difference between the annual income of the health sciences and that of commerce. This implies that there was no mean difference between the health sciences and that of commerce in their annual income.

To find the Coefficient of determination given by r2 we use the formula; r2= SSR/SST=. This illustrates that 0.08amount of variation in degree is accounted for in the annual income of law and engineering students. The correlation coefficient is thus found by getting the square root of the coefficient of determination which is 0.28. the confidence interval at 95% for health science is 3056.51 and that of commerce is 2863.43

Question 4

Since we are testing the mean differences of two independent samples for the law and that of engineering, we will test the hypothesis. The data are quantitative and the parameters to be estimated is the difference between the two means (µ1-µ2)

H0: (µ1-µ2) = 0 ; H1: (µ1-µ2) > 0 ; α= 0.05

Since the population sample is not equal, we use t-test assuming equal variances.

The t-test statistic is given by; t= (x1¯-x2¯) (µ1-µ2)/√δ2(1//n1 + 1/n2) = 1.8974

The t value for alpha 0.05 is 1.966 therefore; we reject the null hypothesis and conclude that the mean differences were not equal. Therefore, there is a difference in the income of the law and the engineering.

ANOVA

Null Hypothesis: there was a significant difference between the annual income of the law and that of engineering. Alternative hypothesis: there was no significant difference between the annual income of the law and that of engineering. The significance level used was of 5%. The test used was that of ANOVA and the output was as follows from the excel workbook.

Anova: Single Factor

SUMMARY

Groups

Count

Sum

Average

Variance

Column 1

121

4786532

39558.11

8.96E+08

Column 2

147

4662455

31717.38

1.1E+09

ANOVA

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

4.08E+09

1

4.08E+09

4.045158

0.045307

3.876655

Within Groups

2.68E+11

266

1.01E+09

Total

2.72E+11

267

Since the F value is greater than the F crit we, therefore, reject the null hypothesis and conclude that there was no significant difference between the annual income of law and that of engineering. This implies that there was no mean difference between the law and that of engineering in their annual income.

To find the Coefficient of determination given by r2 we use the formula; r2= SSR/SST= 0.015. This illustrates that 0.015 amount of variation in degree is accounted for in the annual income of law and engineering students. The correlation coefficient is thus found by getting the square root of the coefficient of determination which is 0.12. The correlation between the degree and the employment status was found to be 0.12. This, therefore, illustrates that there was no good relationship between degree and the annual income of law and engineering students.

Summary

In summary, question one analyzed the relationship between the employment status and the graduates’ degree. The regression model revealed that when the degree is not there the employment status chance is 1.9479 and if there is a unit of degree the employment, status increases by 0.0254. In the ANOVA test, there is no clear relationship between degree and the employment status. The hypothesis testing and the t-test revealed there was no significant difference between the graduates’ degree and the employment status of the graduates. The relationship is linear. In the second research objective, when a student undertakes a degree course at the university, upon completion and is employed, his annual income increases by $3717.229. Moreover, when there is no degree, the annual income is $22357.89. The ANOVA test revealed no significant difference between the degree and the employment status.

The third objective was to determine the relationship between the annual income of the health science and that of the commerce students. The correlation revealed there was no good relationship between degree and the annual income of health science and that of commerce. In the research findings, we rejected the null hypothesis and concluded that there was no significant difference between the annual income of the health sciences and that of commerce. This implies that there was no mean difference between the health sciences and that of commerce in their annual income.

The fourth objective was to determine the relationship between the annual income of law and that of the engineering students. The hypothesis test, the null hypothesis was rejected and concluded that their mean differences are not equal to 0. This implies that there was no mean difference between the law and that of engineering in their annual income.

Conclusions

From the analysis above, it is evident that the degree of the graduates had an important role in the employment status and the annual income of the graduates. It is also clear that the annual income of the students did not differ much based on the health science and commerce and that of law and engineering graduates. All null hypotheses were rejected. Nevertheless, the study faced various limitations. The hypothesis test could not be able to calculate confidence levels of the graduates in regard to the degree they undertook. It was also highly dependent on the concentrations tested. The study used both qualitative and quantitative data that does not show the variation clearly.