UK Workers in Year 2004 - Assignment Example

The following model had been established by conducting a survey on a representative sample of UK workers in the year 2004.
(Model 1) Earn=6.047+0.043*Age+0.403*ln⁡(Size)
(0.170) (0.003) (0.023)
N = 21600, R2 = 0.025
1a.
The sample size as represented by N is 21600 while the value of R2 is equal to 0.025. The numbers in parenthesis are the standard errors of each estimated coefficient. The coefficient for the variable Age is 0.043 which indicates that keeping the size constant the earnings will change by 0.043 units with a one-unit change in Age. Since the coefficient is positive, an increase in Age will increase the Earnings while a decrease in Age will result in a decrease in Earnings.
Test for Significance of Coefficient
In order to test the significance of the coefficient, the following hypotheses have been drawn.

Ho: The coefficient is equal to zero
H1: The coefficient is greater than zero
Test Statistics
In order to test the significance of the coefficient of Age, the t statistic is needed to be calculated.
Formula
t=(estimated coefficient)/(Standard error of coefficient)
In the given case, the estimated coefficient may be referred to as βAge = 0.043 while the Standard error for βAge = 0.003.
Therefore,
t=0.043/0.003
t=14.333

Now for the given data set the t tabulated value had been found using the degree of freedom in the given case is 21600 – 3= 21597. The significance level for the test is 1%.
The t critical value for the given data is found to be 2.326 (Allison, 1991).
Decision Rule
The Null hypothesis is rejected if the t calculated is greater than the t tabulated.
Conclusion
Since the t-calculated value is greater than the t-tabulated value i.e. 14.3333>2.326, hence the criteria suggest rejecting the null hypothesis. So it can be concluded that the coefficient value is not equal to zero.
1b.
The equation required for the desired calculation is
Earn=6.047+0.043*Age+0.403*ln⁡(Size)
In the given case,
Age= 30 years
Size= 100 employees
Earn=6.047+0.043*(30)+0.403*ln⁡(100)
Earn=6.047+1.29+0.403*0
Earn=7.337 pounds per hour
Therefore, the earnings calculated for the given age and size are equal to 7.337 pounds (Belsley, Kuh & Welsch, 2005).
1c)
It depends as it sounds logical that a person with higher education would be earning greater than one who has less education. Therefore, assuming that bigger organizations hire more educated people, it may be assumed that on average there is less probability that an organization employs older people. Omitting education in case it is related to age was necessary to rule out the chances of the error caused due to the correlation existing between age and education as an increase in age is termed to be negatively associated with education. Thus the model would have implied the issue of multicollinearity. When variables are extremely correlated, the variability explained exclusively by the single variables can be minor despite the fact that the variation explicated by the variables brought together is great. (Wooldridge, 2012).
d.
Model 2 Earn=6.544+0.064*Age+0.334*ln⁡(Size)
(2.430) (0.053) (0.285)
The sample size is 83, while R2 is 0.036
1d)
By and large, with an increase in sample size, the estimated values tend to be better predictors of the population parameters. Therefore, with each additional observation that is admitted in the sample, the amount of information increases, and that additional information usually aids in providing better statistics. Thus if the model provides better-estimating results, the standard errors will be reduced. Model 1 has been framed using an extensively large size of the sample while model 2 has eliminated the general public and has enlisted only a specified fragment of the whole population resulting in a decreased size of the sample. This ultimately targets the standard errors as they are increased due to the declined size of the sample. This contemplates the information that larger samples will bring forth more accurate estimates of the coefficients in a regression analysis (Aiken & West, 1991).
e)
Hypothesis testing for Model 1
F=(R^2*(K-1))/((1-R^2 )*(N-K))
Where k=3, N= 21600 while R2=0.025 (Aiken & West, 1991)
Calculating the F statistic
F=(0.025*2)/((1-0.025)*(21600-2-1))
F=0.05/((0.975)*(21597))
F=2.3745E-06
Let the level of significance be 0.05, the F critical value, in that case, will be 2.6. Since the F calculated value is less than the F critical, hence, we may say that there exists no relationship among the variables used in the analysis (Cohen et al., 2013).
Hypothesis testing for Model 2
F=(R^2*(K-1))/((1-R^2 )*(N-K))
Where k=3, N= 83 while R2=0.036
Calculating the F statistic
F=(0.036*(2))/((1-0.036)*(83-3))
F=0.072/((0.964)*(80))
F=9.34*〖10〗^(-4)

Let the level of significance be 0.05, the F critical value, in that case, will be 2.68. Since the F critical value is greater than the F calculated, hence, we may say that there exists a relationship among the variables used in the analysis (Cohen et al., 2013).
Therefore, both models seem to be un-predictive in nature and are not representing the population in an effective manner. The second model fails to produce a significant association despite the large value of R2. Moreover, disregarding the F tests in both cases, the R2 values suggest a minimal association between the variables studied. This further adds to our finding that actually, both models need improvement as they are not able to generate an effective association between the variables employed.
Now that each model is individually tested let us compare both models considering model 2 as restricted while model 1 as unrestricted. In this case, the F statistic will be calculated using a formula.
F=(((R^2 (r)-R^2 (u)))⁄q)/((1-R^2 (u))/N-k(u))
In the given case
R2 (u) =0.05
R2(r) =0.036
K=3
N=21600
Q=number of constraints=2 as the constraints imposed are selecting a segment of society and then selecting a segment out of that segment of the society.
F=(((0.036-0.025))⁄2)/((1-0.025)/21600-3)
F=(((0.011))⁄2)/((0.975)/21597))
F=0.0055/((0.975)/21597))
F=0.0055/((4.5145*〖10〗^(-5)))
F=121.8292
Let the significance level be 0.10 at which the F critical value is found to be 2.30259. Since the F calculated is greater than the F critical value, hence we reject the null hypothesis. That suggests that the restrictions imposed are significant. Read More