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Regression Analysis Assignment
Task B
Question 1. Calculate returns for these three series in Excel or any software of your choice using the transformation and perform the JarqueBerra test of normally distributed returns for each of Boeing and GD. What do you infer about the distribution of the two stock returns series?
Solution.
Along these lines, we see that the supposition of ordinariness isn't fulfilled
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Question 2. Test a hypothesis that the average return on GD stock is different from 2.8%. Which test statistic would you choose to perform this hypothesis test and why
Solution: So as to decide if the average GD stock is not the same as 2.8%, we perform one example t test. As per the Null and substitute theory, we can estimate that
H0: µ = 2.8% (Null Hypothesis)
That is, the average GD stock don't contrast essentially from 2.8%
Substitute Hypothesis: Ha: µ is not equal to 2.8%
That is, the average GD stock vary fundamentally from 2.8%
Dimension of Significance
Give the dimension of importance (a) = 0.05
Test Statistic
The t test measurement is
t= (x ¯μ)/(s/√n)=(5.372.8)/(39.25/√60)=1.613
Here, as per the t test, p  estimation is 0.112 which is greater than 0.05, demonstrating that there is no factual proof to dismiss the invalid speculation at 5% dimension of essentialness. In this manner, we don't have enough proof to infer that the average GD stock contrast essentially from 2.8%
Question 3. Before investing in one of the two stocks, you first want to compare risk associated with each of the two stocks. Perform an appropriate hypothesis test using 5% significance level and interpret your results.
Solution: So as to decide if there is any distinction in the average return between two stocks, we perform free example t test
H_{0}: µ_{1} = µ_{2 }(As per the Null Hypothesis)
This implies that the average return between the two stocks don't contrast fundamentally
H_{a}: µ_{1} ≠ µ_{2 }(As per the Substitute Hypothesis)
This implies that the average return between the two stocks vary appreciably
Dimension of Significance
Give the dimension of noteworthiness a chance to be α = 0.05
Test Statistic
tTest: TwoSample Assuming Equal Variances





BA

GD

Average

0.9359

5.37517

Variance

3.826107

1540.444

Observations

60

60

Pooled Variance

772.135


Hypothesized Average Difference

0


df

118


t Stat

1.243991


P(T<=t) onetail

0.107984


t Critical onetail

1.65787


P(T<=t) twotail

0.215968


t Critical twotail

1.980272


As per the above table, which gives the t test, the p  estimation value is 0.216 which is greater than 0.05, demonstrating that there is no adequate proof to dismiss the invalid speculation at 5% dimension of essentialness. In this way, we reason that the average return between the two stocks vary fundamentally
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Question 4. Perform an appropriate hypothesis test using information in your sample of 60 observations on returns. Report your findings and also mention which stock will you prefer and why?
Solution: So as to decide if there is any distinction in the average return between two stocks, we perform autonomous example t test
H_{0}: µ_{1} = µ_{2 }(As per the Null Hypothesis)
This implies that the average return between the two stocks don't contrast essentially
H_{a}: µ_{1} ≠ µ_{2 }(As per the Substitute Hypothesis)
This implies that the average return between the two stocks vary altogether
Dimension of Significance
Give the dimension of noteworthiness a chance to be α = 0.05
Test Statistic
tTest: TwoSample Assuming Equal Variances





BA

TNX

Average

0.9359

0.737684

Variance

3.826107

15.61922

Observations

60

60

Pooled Variance

9.722664


Hypothesized Average Difference

0


df

118


t Stat

0.348184


P(T<=t) onetail

0.364161


t Critical onetail

1.65787


P(T<=t) twotail

0.728323


t Critical twotail

1.980272


As per the above mentioned table, which gives the t test measurement, the pestimation value is 0.728 which is greater than 0.05, which depicts that we cannot violate the principles of Null Hypothesis (as the level of importance is nearly 5%). Along these lines, we infer that the average return between the two stocks contrast essentially
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Question 5. Compute excess return on your preferred stock as yt = rt  rf,t and excess market return as xt = rM,t  rf,t and perform the following tasks.
Solution: The CAPM model yield is given beneath
SUMMARY OUTPUT














Regression Statistics






Multiple R

0.053979






R Square

0.002914






Adjusted R Square

0.01428






Standard Error

4.324489






Observations

60













ANOVA








df

SS

MS

F

Significance F


Regression

1

3.169703

3.169703

0.169492

0.682081


Residual

58

1084.67

18.7012




Total

59

1087.839













Coefficients

Standard Error

t Stat

Pvalue

Lower 95%

Upper 95%

Intercept

0.00055

0.568604

0.00097

0.999226

1.13874

1.13763

xt

0.038207

0.092805

0.411694

0.682081

0.14756

0.223976

a. From the above relapse yield, we have
Yt =  0.00055 + 0.03821 * Xt
b. The coefficient of the free factor Xt is 0.03821, demonstrating that Yt increments by 0.03821 unit for each one unit increment in Xt
c. The coefficient of assurance is 0.0029, showing that 0.29% of the variety in the needy variable is clarified by the relapse model, while the staying 99.71% left unexplained
d. The 95% certainty interim for incline is ( 0.148, 0.224). This demonstrates, when rehashed tests are taken, at that point 95 out of multiple times the genuine slant esteem will fall inside this interim
Question 6. Using the confidence interval approach to hypothesis testing, perform the hypothesis test to determine whether your preferred stock is a neutral stock.
Solution: On observing the certainty interim esteem, we see that the esteem 0 falls inside the interim, showing that there is no valid reason to avoid the principles of Null speculation that the coefficient of the free factor don't vary essentially from zero. In this way, we have adequate proof to presume that the favored stock is an unbiased stock
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Question 7. Perform an appropriate hypothesis test to determine whether it is plausible to assume normally distributed errors.
Solution: Here, we can utilize histogram or ordinariness plot to decide if the mistakes pursue ordinary dispersion. Here, we utilize lingering plot to test the case
From the above plot, we see that the appropriation of residuals pursue ordinary conveyance around
This residual plot indicates that the error distribution do not follow any pattern indicating that the error distribution is independent of each other.
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