Stock Analysis and Risk-Return Characteristics

This assignment delves into a comparative analysis of Boeing and IBM stocks. It examines their price trends, descriptive statistics like returns and standard deviation to assess risk and return. The analysis further employs hypothesis testing to determine statistical differences in risk and return between the two stocks. A CAPM model is built using S&P 500 as the market index to understand Boeing's systematic risk and relationship with market performance. Finally, the assignment evaluates the linear assumption of the CAPM model through a normal probability plot.

STATISTICS FOR BUSINESS & FINANCE
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PART A: Computation
1. The relevant graphs highlighting the price fluctuations for the chosen stocks and index for
the period under consideration are highlighted below.
2. a) The monthly returns have been computed for the given stocks and also the index using
the given formula and following results are obtained.
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2. b) The summary statistics for the given returns data highlighted in the above part are as
highlighted below.
The requisite formula for computation of the JB statistic is as highlighted below.
Using the output of the descriptive statistics and the given formula, the JB statistic is
computed as follows.
JB Statistic (S&P Returns) = 16.36
JB Statistic (Boeing Returns) = 16.74
JB Statistic (IBM Returns) = 15.09
For all the above three cases, the JB statistic which is computed tends to be higher than the
relevant critical value. As a result, there is availability of ample statistical evidence to cause
rejection of null hypothesis and acceptance of alternative hypothesis. Hence, the non-
normality of the given distributions is confirmed.
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3. Hypothesis testing needs to be carried out so as to derive the conclusion about the population
based on given sample.
Null Hypothesis (H0): μBoeing ≤ 3%
Alternative Hypothesis (H1): μBoeing > 3%
The relevant test to be deployed here is one variable T test. A Z test is not used since the
population standard deviation is not known.
Total number of observations (n) = 65
Average returns on Boeing stock = 1.01% per month
Also, standard deviation = 5.99%
Hence, t statistic = (1.01-3)/(5.99/√65) = -2.678
Using the critical value approach, it is apparent that for the given right tailed t test, the
computed t statistic is lesser than the critical value and hence does not lead to rejection of null
hypothesis. Hence, it would be appropriate to reach the conclusion that the monthly returns
of Boeing stock are lesser than 3%.
4) The objective is to compare the associated risk of the two stock which would require the
comparison of their respective variances using the F test.
Null Hypothesis (H0): σ2Boeing = σ2IBM
Alternative Hypothesis (H1): σ2Boeing ≠ σ2IBM
The relevant excel output for the test is indicated below.
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It is apparent from the above that the two tail p value would be 0.18 (0.09*2). Assuming a
significance level of 5%, it is apparent that p value exceeds significance level. Hence, there
does not exist sufficient evidence for rejection of H0. Therefore, the logical conclusion is that
there is no significant statistical difference between the risk associated with the given two
stocks.
5) Based on the returns of the stock derived during the given period, it is essential to conduct
a hypothesis test so as to draw conclusion in relation to the population average return
comparison of Boeing stock and IBM stock.
Ho: μBoeing = μIBM
H1: μBoeing ≠ μIBM
The t test for two independent means is the appropriate test for mean comparison as the
standard deviation of the population returns is unknown. Also, equal variances option has
been chosen as the underlying standard deviation of returns for the two stocks does not show
any significant variation. The relevant excel output is highlighted below.
.
From the above, it is apparent that two tail p value comes out as 0.334 and is higher than
assumed significance level of 5%. Hence, there does not exist sufficient evidence for
rejection of H0. Therefore, the logical conclusion is that there is no significant statistical
difference between the average return associated with the given two stocks.
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6) The underlying risk of the two stocks is similar in terms of magnitude but the sample
returns tend to favour Boeing which during the given sample has delivered superior return.
Hence, despite the population returns of the two stocks not being different, it would be fair to
assume that the superior stock based on the sample performance is Boeing and hence it has
been chosen for the CAPM computations.
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7)a) Regression Output
The regression equation is highlighted below.
Excess returns on Boeing stock = 0.383 + 1.105* Market Excess Returns
c) Coefficient of Determination =0.4116
d) Lower limit of 95% confidence interval (Beta) = 0. 772
Upper limit of 95% confidence interval (Beta) = 0. 772
8) H0: βBOEING = 1
H1: βBOEING ≠ 1
The confidence interval approach is to be deployed for testing the above hypothesis. The
decision is based on whether the hypothesised value tends to be present in the 95%
confidence interval or not. In the given case, it is apparent that the hypothesised value of beta
as 1 is present in the 95% confidence interval and hence there does not exist enough
statistical evidence for rejection of null hypothesis.
9) The requisite normal probability plot for testing the assumption of the linear regression is
as highlighted below.
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0 20 40 60 80 100 120
-25
-20
-15
-10
-5
0
5
10
15
Normal Probability Plot
Sample Percentile
Boeing Excess Returns
.
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Part B: Interpretation
1) S&P 500 – Based on the given values, it is apparent S&P index has witnessed a bull run
between October 2011 and October 2014 as the index value has enhanced from 1250 to
more than 2000. As a result, it implies that the return has been in excess of 60% with no
major corrections. However, post this period, there has been a consolidation in the index
which is on expected lines considered the rally.
Boeing Stock – From December 2010 to April 2013, the stock moved sideways and did
not yield any significant returns for the investors. However, in the next one year i.e. from
April 2013 to April 2014, there was a spectacular increase in the stock price as it moved
from about $ 80 to more than $ 130. Hence, the returns during this period exceeded 60%.
However, post this move the stock has been moving sideways and consolidating in the
remaining period.
IBM Stock – The IBM stock started the period on a high note as from December 2010 to
April 2012, the stock moved from about $ 150 to more than $ 200. However, post this
move, the stock entered into a lengthy consolidation period which continued for an year.
However, during the remainder period, the stock has slowly and steadily given up all the
gains that it made during the initial period and to finally settle at around the original
starting price of $ 150.
2) b) Based on the descriptive statistics that have been derived for the two stocks in part (a),
the objective is to highlight the risk return characteristics. It is evident that Boeing tends
to exceed IBM in terms of both sample returns and sample standard deviation (risk).
However, the difference in standard deviation for the two stocks is not very significant.
But the same cannot be said about returns which are quite stellar for Boeing. Thus, the
return per unit risk is higher for Boeing as compared to IBM and hence Boeing is the
superior stock.
5) The superior stock would be one which would deliver higher returns for every unit risk
assumed. Based on the results of the hypothesis testing in relation to the risk and return, it
becomes apparent that for the two stock (i.e. Boring and IBM), there does not exist any
significant statistical difference in terms of both risk and returns. Hence, the sample data has
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to be used for the selecting the better stock which leads to the choice of Boeing as has been
explained previously.
7) b) The slope coefficient or beta comes out as 1.105. The beta of the Boeing stock is greater
than 1 which implies that the stock tends to more risky than the S &P market index. But at the
same time, an investor can expect superior returns on the stock which have been delivered for
the given period under consideration. Also, when the market changes by a percentage point,
the respective change in the Boeing stock would be 1.105% in the same direction.
c) The coefficient of determination for the given CAPM has come out as 0.4116. This implies
that the market index (S&P 500) is able to explain 41.16% of the respective changes in the
stock price of Boeing. Also, considering that majority of the changes in the Boeing stock
cannot be explained on the basis of corresponding movements in the market index which
reflects at beta being a poor indicator of systematic risk. Clearly, more predictor variables
need to be inserted in the given model to improve the coefficient of determination.
d) With an estimated confidence of 95%, it may be claimed that the Boeing stock beta would
occupy an interval between 0.772 and 1.437.
9) In order to ascertain the linear assumption with regards to the linear regression model for
the CAPM, it is imperative that the requisite normal probability plot must display a linear
pattern. Based on the requisite graph highlighted in the computations, the linear trend is
clearly visible which indicates that the normality assumption is satisfied.
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