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Quantitative Methods Assignment 1 Table of Contents Question 1..............................................................................................................................2 Question 2..............................................................................................................................4 Question 3..............................................................................................................................5 Question 4..............................................................................................................................6

Question 1 a) To find both the sample mean and variance for each variable, open each and select view, descriptive statistics and tests, histogram and stats. EMPL_TRAINING: (n=666)

Mean = 0.48 Variance = 0.25

EMPL_NOTRAINING: (n=1364) Mean = 0.44 Variance = 0.25

The distribution of the two variables is very similar. While the empl_notraining sample doubles that of empl_training, based on the distribution there is a slight increase in the proportion of employed people following training. b) To only consider individuals with a job, open corresponding variables (e.g. empl_training and wage_training) as a group, and then use the If condition in the sample tab.  Enter the condition empl_training=1 or empl_notraining=1.  Select view, descriptive stats, common sample. Wage_training: (n=320) Mean: 872.51 Variance: 490.052 =¿ 240144.79 Wage_notraining: (n=597) Mean: 826.60 Variance: 269.642=¿ 72707.99 c) Let wage_training = Step 1) Step 2)

X1

and wage_notraining = X 2 H 0 : σ X 2 =σ X 2 and H 1 : σ X 2 ≠ σ X 2 1

2

when H 0 : σ X =σ X 1

F=

s X 2/σ X 2 1

2

= 2

1

s X /σ X 2 2

Step 3) Step 4)

1

2

2 2

sX2 1

sX

2

is true; F 319,596

2

Significance level = 5% Reject if:  F>1.21 (found using scalar f_upper=@qfdist(0.975,319,596))

2

 Or: F 1.21, there is strong evidence at the 5% level to reject the null hypothesis and accept the alternate hypothesis; sample variances are unequal. d) Using the command: scalar pval1=2*@cfdist(1/3.3029,596,319), EViews calculates a p value of 2.78 ×10−36 (by placing X 2 on the numerator, the d.o.f. must be reversed). This p-value is incredibly small, so small it could be approximated to 0. e) As the p value is less than 0.05 (two tail test), there is very strong evidence at the 10% level to reject H 0 and accept the alternate hypothesis; the two variances are equal. f) The critical values at a 10% significance level are: 2 s1  Upper: F= 2 > F ( 0.05,319,596) =¿ 1.173 (command scalar s2 f_upper=@qfdist(0.95,319,596)) s12  Lower: F= 2 < F ( 0.95,319,596) =¿ 0.849 (command scalar s2 f_lower=@qfdist(0.05,319,596)) g) The equation for pooled variance is as follows:

¿

(320 −1 ) × 490.052 + ( 597 −1 ) × 269.642 320 + 597 −2

¿ 128777.79

Question 2 a) To test whether professional training will lead to different levels of income, we need to test the equality of means between the wage_training and wage_notraining variables.  Let X 1 = wage_training and X 2 = wage_notraining ¯ 2 and H 1 : X¯1 ≠ ¯X 2  Thus: H 0 : X¯1= X b) In (g), the pooled variance was found to be 128777.79.  The relevant test statistic (unknown population variances that are assumed to be equal) is:

¿

( 872.51−826.60 )−0

√

128777.79

( 3201 + 5971 )

¿ 1.847

c) Using command: scalar pval=1-@ctdist(1.84656,915); p value = 0.0326  ( 320 + 597 −2=915 d . o . f ¿ d) Thus, as 0.0326 < 0.05 (two tail test), there is sufficient evidence at the 10% level to reject the null hypothesis and accept the alternate hypothesis; that professional training leads to different levels of income. e) The critical values at a 10% significance level are:  Reject if t stat > t ( 0.95,915) =¿ 1.647 (using scalar tc=@qtdist(0.95,915)) −1.647 (using scalar tc=@qtdist(0.05,915))  Or: t stat t 0.95,2028=¿ 1.646 (using scalar tc=@qtdist(0.95,2028)) −1.646 (using scalar tc=@qtdist(0.05,2028))  Or: t stat 1.646, there is sufficient evidence at the 10% level to reject the null hypothesis and accept the alternate hypothesis; that professional training has an impact on employment. 2. Recent tests have concluded that professional training has impacted the employability of candidates, and their future employment. It is therefore recommended that training be conducted. Question 4 a) Let X 1 = empl_notraining and thus, empl_notraining.

μ X =¿ the mean of the population 1

H 0 : μ X =0.4125 and H 1 : μ X...