Statistic Decision Making Final Exam

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Exercise 2: Sample Design and Evaluation
The information can be summarized as follows:
N1= N2
Standard Deviation= 15
Difference in Performance= 5
Power= .8
After entering the given information, the window looks as follows, which shows us that N1= N2= 142

In the window above, change the power to .9, then N1= N2 = 190

In the window above, change the sigma1=15, sigma2=12, and don’t select Egual Sigmas checkbox, thus I get N1= N2= 156
In the window above, change the N1=200 (control group), N2=120 (testing group), and select Independent in Allocation, thus I get .9046 to be the power.

=((61-64.5)-(0))/√((16*16)/200+(13*13)/120) = (-3.5)/1.6396 = -2.1347 Critical Value: Zα/2= Z0.05/2= @qnorm(1-0.05/2)= 1.96
When comparing the test statistic to the critical value: Z=2.1347>1.96, we reject the null hypothesis. We can calculate the P-value using the EViews command:
Show @tdist (t, d.f)
In this EViews command, t stands for the appropriate test statistic and d.f are the degrees of freedom. The appropriate test statistic was calculated above, namely Z=2.1347. For the degrees of freedom, we can insert NA+NB-2. Show @tdist (2.1347, 318)= 0.03355

Since the P-value= 0.033550, and β1= 0.86361050000
ls price c assessval
Dependent Variable: PRICE
Method: Least Squares
Date: 01/21/13 Time: 16:07
Sample: 1 650 IF PRICE>50000
Included observations: 562

VariableCoefficientStd. Errort-StatisticProb.


R-squared0.701363 Mean dependent var113069.1
Adjusted R-squared0.700829 S.D. dependent var51534.97
S.E. of regression28187.83 Akaike info criterion23.33472 Sum squared resid4.45E+11 Schwarz criterion23.35013
Log likelihood-6555.056 Hannan-Quinn criter.23.34074
F-statistic1315.184 Durbin-Watson stat1.337129

Estimated intercept...
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