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Regression Analysis Assignment

A) Calculate the Least Squares (LS) estimates of α and β for the linear model.

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Solution:

A. The following is the value of β:
β=(∑(x_i-(x)) ¯*(y_i-(y)) ¯)/(∑(x_i-(x)) ¯²) = 30/47=0.64

The following is the value of α:
α=y ¯-β*x ¯=24-(0.64*11)=16.96 or 17

The following is the regression equation:
y_i=17+0.64x_i+∈_i

B) Test the hypothesis β^ =2.8.

Solution:

B. Null Hypothesis: The slope is equal to 2.8.

Alternate hypothesis: The slope is not equal to 2.8.

The calculated value of t is:

t=(β-β ^)/(S/√n)=(0.64-2.8)/(√(∑(x_i-(x)) ¯²/n?)/√n) (=-2.16)/(√(47/25)/√25)=-2.16/0.27=-8

Degree of freedom is 25-2=23.
The critical value of t is ±2.069

Since, the critical value of t = -2.069 is greater than the calculated value of t = -8, hence we reject the null hypothesis at 5% level of significance and claim that the value of β≠2.8.

C) Form a 99% CI for σ² - the variance of ε.

Solution:

C. The sample variance is as follows:
s²=(∑(y_i-(y)) ¯²-β*∑?(x_i-(x)) ¯*(y_i-(y)) ¯ )/(n-2)=(120-(0.64*30))/(25-2)=4.38

The 99% confidence interval is as follows:

((n-1) s²)/(χ_((.01/2))² ) ≤ σ² ≤ ((n-1) s²)/(χ_(1-(.01/2))² )

(24*4.38)/45.559≤ σ² ≤(24*4.38)/9.886

2.31≤ σ²≤10.63

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Question 2. Data from a different time period of the same industry are used to obtain the following estimated regression

y^ =20+0.5x, R2 =0.2; y =35; x =11;n=10; Σ_i(x_i -x‾) =86

Test the hypothesis that the slopes in this year and previous period (Question 1 above) are the same (assuming the variances of the disturbances are the same).

Solution:

Null Hypothesis: The slope of time period 1 and time period two are equal.

Alternate Hypothesis: The slope of time period 1 and time period two are not equal.

The calculated value of t is:
t=((β₁-β₂)-0)/(Pooled standard deviation√(1/n₁ +1/n₂ ))

Standard deviation of second time period is:

S₂=√(∑(x_i-(x)) ¯²/n)=√(86/10)=2.93

The pooled standard deviation is:
S_p = √(((n₁-1)S₁²+(n₂-1)S₂²)/(n₁ + n₂-2)) = √(((24*1.88)+(9*8.6))/(25+10-2))=1.93

So, t value is:

t=((0.64-0.5)-0)/(1.93√(1/25+1/10))=0.19

Degree of freedom is n1+n2-2 = 25+10-2=33

Critical value of t is ±2.03
Since, the critical value of t = 2.03 is greater than the calculated value of t = 0.19, hence we do not reject the null hypothesis at 5% level of significance and claim that the slopes of two regression time periods are equal.

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Question 3:

To test this theory, you regress the GDP per worker (relative to the United States) in 1990 (RelPersInc) on the difference between the average population growth rate of that country (n) to the U.S. average population growth rate.

Solution:

3. The following is the equation:

RelPersInc=0.518-18.831*(n-n_(U.S) ),R²=0.522,SER=0.197

a) Interpret the results carefully.

a) As the growth rate of the population of the country in comparison to the U.S. increases, the real income of the person's decreases and that too by 18.831 units.

b) Is this relationship economically important?

b) Yes, the relationship is economically important as whatever be the difference between the population growth rates of the country vis-à-vis U.S., the real income of the person will be 0.518.

c) What would happen to the slope, intercept, and regression R² if you ran another regression where the above explanatory variable was replaced by n only, i.e., the average population growth rate of the country?

c) The following will be the new equation:
RelPersInc=0.518-18.831*(n-0.009)
RelPersInc=0.518+0.169-18.831*n
RelPersInc=0.687-18.831n

The slope will be the same and the intercept would increase to 0.687 from 0.518. The R square will also be same.

There will be change in the t statistic.

d) 31 of the 104 countries have a dependent variable of less than 0.10. Does it therefore make sense to interpret the intercept?

d) 31 out of 104 countries have a dependent variable less than 0.10, it doesn't make sense to interpret the intercept as the income would be less than zero.

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