Finance theory suggest that αj = 0 under market efficiency. This is because αj represents excess profit or loss when the market excess return rm – rf = 0. That is, if αj > (<) 0, the stock consistently outperforms (underperforms) the market. However, if the market is efficient, this profit (loss) should disappear rather quickly by arbitrage.
1-Test whether the Microsoft stock out or under-performs the market
First: The null hypothesis is H0: αj = 0, (Tow-tailed test)
The alternative hypothesis is H1: αj ≠ 0
Second: Test statistic
t = (b – B ) / S , t = 0.006098 – 0 = 0.787109
0.007747
Third: Critical value (rejection points):
df = 132 – 2 = 130, and α/2 =0 .025
Significant level is t (0.025, 130).
As a result, t (0.025, 130) = 1.9760.
Fourth: Decision rule: we reject the null hypothesis if 1.9760 <t-statistic <-1.9760.
Since t = 0.787109 is less than 1.9760and more than -1.9760, we cannot reject the null hypothesis. This means that the αj has no significant effect on the dependent variable.
Furtherer, we can test this hypotheses by using the p-value in the E-view output.
P-value for αj is 0.4327, 0.4327 > 0.05, we cannot reject the null hypothesis.
“Tech stocks typically have higher beta. An example is the dot-com bubble. Although tech
Stocks did very well in the late 1990s, they also fell sharply in the early 2000s, with a
Much worse decline than that of the overall market.”
- The above statement implies that Microsoft (a Tech stock) is an aggressive stock.
Evaluate this claim by interpreting the relevant coefficient estimate and conducting an
Appropriate statistical test.
First: The null hypothesis is H0: β≤ 1,
The alternative hypothesis is H1: β > 1
Second: Test statistic
t = (bˆ – B) /S , t = 1.318947- 1 = 1.983625
0.160790
Third: Critical value (rejection points):
(Since N=132, df = 132 – 2 = 130 and α=0.05).
Significant level is t (0.05, 130).
So, t (0.05, 130) = 1.6555.
Forth: Decision rule: we reject the null hypothesis if t-statistic >1.6555
Since t = 1.983625 > 1.6585, we reject the null hypothesis that β≤ 1, and accept the
Alternative hypothesis that β > 1. Since a beta greater than 1 indicates an ‘‘aggressive
Stock.’’ We prove that the Microsoft (a Tech stock) is an aggressive stock.
Furthermore, we can test this hypotheses by using the p-value in the E-view output. P-value for β is ZERO, 0 < 0.05 , we reject the null hypothesis.
“Mobil-Exxon” is an oil company with stable business conditions and should be a safer
Investment than the overall market”
- The above statement implies that Mobil-Exxon is a defensive stock. Evaluate this
claim by interpreting the relevant coefficient estimate and conducting an appropriate
statistical test.
First: The null hypothesis is H0: β ≥ 1,
The alternative hypothesis is H1: β <1
Second: Test statistic
t = (bˆ – B ) /S , t = 0.413969- 1 = -6.532286
0.089713
Third: Critical value (rejection points):
(Since N=132, df = 132 – 2 = 130 and α=0.05)
Significant level is t (0.05, 130).
So, t (0.05, 130) = -1.6555.
Forth: Decision rule: we reject the null hypothesis if t-statistic < -1.6555
Since t = -6.5323 < -1.6555, we reject the null hypothesis that β ≥ 1, and accept the
Alternative hypothesis that β <1. Values of beta less than 1 indicate that the stock is
‘‘defensive’’ since its variation is less than the market’s. We prove that the Mobil-Exxon
is a defensive stock.
Furth more, we can test this hypotheses by using the p-value in the E-view output.
P-value for β is ZERO, 0 < 0.05, we reject the null hypothesis
- Based on the estimation results, fill out the following table:
| β | 95% Confidence Intervals for β | ||
| lower bound | upper bound | ||
| Microsoft | 1.3189 | 1.000843 | 1.637051 |
| GE | 0.8993 | 0.703832 | 1.094688 |
| GM | 1.2614 | 0.861336 | 1.661486 |
| IBM | 1.1882 | 0.938076 | 1.438340 |
| Disney | 0.8978 | 0.653257 | 1.142419 |
| XOM | 0.4140 | 0.236482 | 0.591456 |
At 5% significance level and df 130, the tc is t (.025.130) =1.979
- Microsoft : 1.3189 + 1.979 * 0.160790 = 1.637051
1.3189 – 1.979 * 0.160790 = 1.000843
– GE: 0.8993 + 1.979 * 0.098782 = 1.094688
0.8993 – 1.979 * 0.098782 = 0.703832
– For GM: 1.2614 + 1.979 * 0.202223 = 1.661486
1.2614 – 1.979 * 0.202223 = 0.861336
– For IBM: 1.1882 + 1.979 * 0.126433 = 1.438340
1.1882 – 1.979 * 0.126433 = 0.938076
– For Disney: 0.8978 + 1.979 * 0.123627 = 1.142419
0.8978 – 1.979 * 0.123627 = 0.653257
– For XOM: 0.4140 + 1.979 * 0.089713 = 0.591456
0.4140 – 1.979 * 0.089713 = 0.236482
- Interpret the 95% confidence intervals for GE, GM, and Disney, paying attention to
their risk profiles (aggressive, defensive, neutral).
We have obtained the estimates of b1 which is b, but we cannot say definitely that b = b.
However, it can be say that the true parameters are contained in an interval around the estimated parameters. In other words, we can say with certain probability the interval (b-tc S, b+ tc S ) has the true bi within it.
The estimated value for GE is 0.8993, GM is 1.2614 and Disney is 0.8978. As a result, GE and Disney are defensive stocks ( < 1), while GM is aggressive stock ( > 1).
The GE’s 95% confidence interval for β is from 0.7038 to 1.0947. It is believed that GE is a defensive stock in most situations, while it is a neutral or aggressive stock some times. The Disney 95% confidence interval for β is from 0.653257 to 1.142419, it is believed that Disney is a defensive stock most of times and a neutral or aggressive in some other times. In addition, The GM 95% confidence interval for β is from 0.861336 to 1.661486 which means that GM is an aggressive stock in most of times.
It is supposed that, for January 2009, the market return (rm) is expected to be either 0.05 (5%) with the probability 0.5; or -0.05 (-5%) with the probability 0.5. Assume the risk-free rate to be 0 and α value to be 0.
- An investor is seeking your recommendation to invest in three stocks among the above.
Her goal is to form a portfolio of three stocks with well-diversified risk profiles. State
your recommendations and justify your choice in relations to the properties of their β
values based on the confidence intervals obtained in the table above
We can choose three stocks of different risk (aggressive, defensive, and neutral). Based on β values for all of the stock, we can choose Microsoft as an aggressive stock with confidence interval (1.000843, 1.637051). Mobil-Exxon as a defensive stock with confidence interval (0.236482, 0.591456), and GE as natural stock with confidence interval (0.703832, 1.094688).
PART 2:
7-Based on your regression results for these five stocks, explain how their expected excess returns are related with different risk factors such as market premium, size premium and value premium. You should interpret the relevant estimated coefficients, taking account of their magnitude, signs, and statistical significance.
1/Microsoft:
Estimated Parameter (b1) for market premium is (1.031353). It is statistically significant with P-value equal zero , this means that, other thing equal, if the market return increase by 1 % Microsoft stock return will increase by 1.03 %. Therefore, there is a positive relationship between MSFT return and market return. In addition, estimated Parameter (b2) for size premium is (-0.0035). It is statistically significant at 90% with P-value equall (0.0834). This beta implies that, there is a negative relationship between the return on Microsoft stock and the company size. Furthermore, estimated Parameter (b3) for value premium is (-0.0108). It is statistically significant at 99% with P-value equal zero. This means that there is a negative relationship between the return on Microsoft stock and the value premium.
GE:
Estimated Parameter (b1) for market premium is (0.9532). It is statistically significant with P-value equall zero. It means that, other thing equal, if the market return increase by 1 % GE stock return will increase by 0.9532 %. So, there is a positive relationship between GE return and market return. In addition, Estimated Parameter (b2) for size premium is (-0.0056). It is statistically significant at 99% with P-value equal zero. It means that, there is a negative relationship between the return on GE stock and the company size. Furthermore, estimated Parameter (b3) for value premium is (-0.0021). It is not statistically significant at 90% with P-value equal (.1432). It means that here is a negative relationship between the return on GE stock and the value premium.
GM :
Estimated Parameter (b1) for market premium is (1.567) .It is statistically significant with P-value equall zero.it means that, other thing equal, if the market return increase by 1 %, GM stock return will increase by 1.567 %. So, there is a positive relationship between GM return and market return. Moreover, estimated Parameter (b2) for size premium is (-0.000788). It is not statistically significant at 90% with P-value equal (0.7651). It means that, there is a negative relationship between the return on GM stock and the company size. Furthermore, estimated Parameter (b3) for value premium is (0.008553). It is statistically significant at 99% with P-value equal (.0057).it means that here is a positive relationship between the return on GM stock and the value premium.
IBM:
Estimated Parameter (b1) for market premium is (1.1458). It is s statistically significant with P-value equall zero. It means that, other thing equal, if the market return increase by 1 %, IBM stock return will increase by 1.1458 %. So, there is a positive relationship between IBM return and market return. Moreover, estimated Parameter (b2) for size premium is (-0.002155 .It is not statistically significant at 90% with P-value equal (0.2055). it means that, there is a negative relationship between the return on IBM stock and the company size. Furthermore, , estimated Parameter (b3) for value premium is (-.002688( it is not statistically at 90% with P-value equal (.1725). it means that here is a negative relationship between the return on IBM stock and the value premium.
Disney:
Estimated Parameter (b1) for market premium is (1.004). It is statistically significant with P-value equal zero. It means that, other thing equal, if the market return increase by 1 %, DIS stock return will increase by 1.004 %. So, there is a positive relationship between DIS return and market return. . Moreover, estimated Parameter (b2) size premium is (-0.00038). It is not statistically significant at 90% with P-value equal (0.8184. it means that, there is a negative relationship between the return on DIS stock and the company size. Furthermore, , estimated Parameter (b3) for value premium is (0.002902) .It is not significant at 90% with P-value equal (0.1313). It means that here is a positive relationship between the return on DIS stock and the value premium.
Mobil-Exxon:
Estimated Parameter (b1) for market premium is (0.545523) .It is statistically significant with P-value equall zero. It means that, other thing equal, if the market return increase by 1 %, XOM stock return will increase by .0545523 %. So, there is a positive relationship between XOM return and market return. . Moreover, estimated Parameter (b2) for size premium is (-0.001713). It is not statistically significant at 90% with P-value equal (0.1446). It means that, there is a negative relationship between the return on XOM stock and the company size. Furthermore, estimated Parameter (b3) for value premium is (0.002766). It is significant at 95% with P-value equal (0.0426). It means that here is a positive relationship between the return on XOM stock and the value premium.
2. It is claimed that the stock “dis” is unrelated with the size and value premier, in the
sense that the relevant coefficients in the Fame-French model is jointly equal to 0. Test
this hypothesis using the F-test.
The null hypothesis is H0: β3 = β4 = 0
The alternative hypothesis is H1: one of the coefficient is equal to zero.
The above table shows that the P-value is 0.2129 which is higher than 1, 5, and 10 percent. Thus, we cannot reject the null hypothesis at 99, 95 and 90 percent confidence level and we conclude that DIS stock is unrelated with the size premium and value premium. Therefore, the SMB and HML coefficient is not significant
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