Qualitative Research

Question 1:

Given the performance of the stock with reference to the NASDAQ stocks it is possible to construct a tree diagram that will help in the analysis of various outcomes; the tree diagram is as follows:

From the tree diagram we can determine the various probabilities, in this case we determine the probability that a stock that was a recommended buy outperformed the NASDAQ stocks, this is derived from simply multiplying the probability of outperforming the NASDQA by the probability that this stock probability of the outcome that it was recommended, this is simply multiplying 0.75 X 0.42 from the tree diagram above, the result therefore is 0.315, this is the probability that a recommended stock outperformed the NASDAQ.

B.

in this case we are given the probability that the NASDAQ will rise in any given month is 0.5, we are also provided with information that the performance of any month is independent from other outcomes, for this reason therefore we assume that the probability function assumes a binomial distribution, this distribution assumes that there exist n identical trials, there are only two possible outcomes for a trail which include success and failure, the trials are independent where the outcome of one trial does not affect the outcome of the other trial and finally success is denoted by P and failure is denoted Q.

The binomial probability function is stated as follows:

P (x) = nCx ∏x (1 – ∏) n-x

For this reason therefore we can find the various probabilities of desired outcomes, the table

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below summarises the excel output regarding the binomial distribution:

x

n

∏

P(x)

0

12

0.5

0.000244

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1

12

0.5

0.00293

2

12

0.5

0.016113

3

12

0.5

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0.053711

4

12

0.5

0.12085

5

12

0.5

0.193359

6

12

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0.5

0.225586

7

12

0.5

0.193359

8

12

0.5

0.12085

9

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12

0.5

0.053711

10

12

0.5

0.016113

11

12

0.5

0.00293

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12

12

0.5

0.000244

The probability that the rise will be observed in at least 4 months:

This probability will involve adding up the probabilities of getting 4,5,6,7,8,9,10,11 and 12

Therefore the answer is 0.927002.

The probability that the rise will be observed in at most 9 months:

The probability is derived from adding up the probability of 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9, the

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probability therefore is 0.980713

The probability that the rise will be observed in at least 1 month:

We get the probability of getting 0 months and subtract from one to get this probability:

1 – 0.000244 = 0.999756

The probability that the rise will be observed in no months:

We get probability of 0 successes which is 0.000244

C.

The average time taken for porter to walk is 45 minutes, a sample of 50 is selected and the average time is 46.8 with a standard deviation of 2 minutes, we investigate whether the average time is equal to 50, at 5% level of test we state the null and alternative hypothesis, we want to find out if the average time is greater than 45 minutes:

Null hypothesis:

X = 46.8

Alternative hypothesis

X ≠ 46.8

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Test statistic

Z = (X – U)/(s/ (n½))

Z statistics = (46.8 – 45)/ (2/ (50½))

Z statistics = 6.363961

At 5% level of test the critical value is 1.644854 a one tail test.

Due to the fact that tour t statistic exceeds the t critical value at 5% we reject the null hypothesis that the mean is equal to 46.8, therefore we reject the null hypothesis.

D.

The mean income of a population is 15,000 with a standard deviation of 5,000. A population is termed as poor if the income falls below 8000,

The proportion of the poor in the population will be calculated by the use of the Z table, we determine the Z value as follows:

Z = (8000 – 15000)/5000

Z=-1.4

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We now find the area under the Z curve that is below -1.4 critical values, the area under this point is equal to 0.4192, and this means that 41.92% of the population is poor.

When we assume that those who fall below 12% are the poor we will still use the Z table but this time we will consider the critical value that has the area under it equal to 0.12, the critical value is 0.31,

– 0.31 = X/5000 X = 1550

Therefore if we assume that those that are poor are 12% then their level of income is 1550

Question 2:

The price of NASDAQ over the period 2000 to 2007 is *summarised* by the chart below:

From the above chart it is clear that the prices have gradually been declining over time from the year 2000, the log difference on returns is summarised by the chart below:

The chart above shows that in the past years the prices have deviated by larger magnitudes, this deviation has reduced in the recent years making it less profits and less risk to invest in the NASDAQ stock.

We estimate the following model:

Ri – Rf = a + b ( Rm – Rf)

Where

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Rm is return on NASDAQ

Rf is rate of three month treasury bill

Ri is the return on stocks

We therefore assume that Ri – Rf is equal to Y and Rm – Rf is equal to X.

Y = a + b X

After estimation the model is as follows:

Y = -0.074943 + 0.85909X

This means that tan increase in X by one unit holding all the other factors constant then the level of Y will increase by 0.85909, if the level of X is equal to zero then the level of Y will be -0.074943.

The Durbin Watson coefficient for this model is equal to 0.302076 showing that there is autocorrelation, for this reason the existence of autocorrelation means that the model is not BLU (best linear unbiased estimate) this is because it violates the assumptions of OLS. Autocorrelation can be corrected through the re pecifying of the model to be estimated by adding time lagged variables into the model.

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Heteroscedasticity is the violation of the OLS estimation model that states that the variance of the error term has a constant variance across observation, it can be detected through the use of graphs where we plot the error terms versus the explanatory variable, also there are other methods such as the park test, cross sectional data is prone to this problem can be resolved by re specify our estimation method.

Estimation of Y = a + b X + y sml + δ hml

After estimation our model is as follows:

Y = -0.138609 + 0.852301X + 0.040825sml + 0.033091 hml

The model states that the autonomous value of Y is -0.1386 while if we hold all other factors constant and increase the level of x by one unit then Y will increase by 0.852301, if we hold all other factors and increase the level of SMB by one unit then the level of Y will increase by

0.040825, finally if we hold all other factors constant and increase the level of hall by one unit then the level of Y will increase by 0.033091.

The Durbin Watson test value in this case is 0.560967 meaning that there is still autocorrelation, the Durbin Watson test value equal to 2 means that there is no autocorrelation but in our case this value is less meaning that there is autocorrelation. Heteroscedasticity is the violation of the assumption that states that the variance of the error term has a constant variance across observation, it can be detected through the use of graphs where we plot the error terms versus the explanatory variable, also there are other methods such as the park test, cross sectional data is prone to this problem can be resolved by re specify our estimation method.

Multicollinearity is a situation where the explanatory variables are correlated with the explanatory variables, in this situation it can be detected by the presence of high correlation of determination and the presence of t test that show non significance estimated coefficients; it can also be detected by the presence of high pair wise correlation. To *correct* the presence of Multicollinearity we can drop the problematic variable, re specify the model, acquiring of additional data or the use of a different sample, transformation of data example from general to log form numbers.

Question 3:

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Stationary:

Stationary is the concept that applies to time series data that states that data is stationary if its mean and variance are constant over time and that the covariance between two time periods depend on the distance or gap between the two periods, in this case the variable in question is usually dependent on the previous period value Stated as follows:

Yt = PYt-1 + Ut

If the value of P is equal to one then this shows a situation of non stationary but if P is less or greater than one then the data is stationary.

The results of the model states that Yt = -0.038705Yt-1, this means that because the value of P is less than 1 then the series data is stationary.

The auto regression moving average model comprises of two parts which include the autoregressive part and the moving average part, this model is formulated for the purpose of predicting future value on the time series data, and it contains a number of autoregressive parts and the moving average parts. The autoregressive part is stated as follows:

Xt = c + b Xt-1 + ut

The moving average on the other hand is stated as follows:

Xt = et + eXt-1 where e is the error term

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For this reason therefore the autoregressive moving average model will be stated as

X t = et + b Xt-1 + eXt-1

In our case therefore we state our model as follows:

y t = et + b yt-1 + eXt-1 where Y is the stock price and is the error term

References:

Bluman A. (2000) Elementary Statistics: A Step by Step Approach, McGraw *Hill* press, New York

D. ox (2001) Applied Statistics: Principles and Examples, McGraw Hill press, New York

D. Bridge (1993) Statistics: An Introduction to Quantitative Economic Research, Rand McNally publishers, Michigan

L. Henry (1991) Statistics, Oxford University Press, Oxford

Appendixes:

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Dependent Variable: Y

Method: Least Squares

Date: 04/18/08 Time: 11:16

Sample: 1 96

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Included observations: 96

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Variable

Coefficient

Std. Error

t-Statistic

Prob.

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C

-0.138609

0.083746

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-1.655112

0.1013

X

0.852301

0.026673

31.95320

0.0000

SMB

0.040825

0.012639

3.230108

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0.0017

HML

0.033091

0.010523

3.144574

0.0022

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R-squared

0.921101

Mean dependent var

-2.357542

Adjusted R-squared

0.918528

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S.D. dependent var

1.441867

S.E. of regression

0.411557

Akaike info criterion

1.103033

Sum squared resid

15.58285

Schwarz criterion

1.209881

Log likelihood

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-48.94560

F-statistic

358.0148

Durbin-Watson stat

0.560967

Prob(F-statistic)

0.000000

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Dependent Variable: Y

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Method: Least Squares

Date: 04/18/08 Time: 11:05

Sample: 1 96

Included observations: 96

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Variable

Coefficient

Std. Error

t-Statistic

Prob.

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C

-0.074943

0.087183

-0.859610

0.3922

X

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0.859909

0.028177

30.51779

0.0000

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R-squared

0.908323

Mean dependent var

-2.357542

Adjusted R-squared

0.907347

S.D. dependent var

1.441867

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S.E. of regression

0.438888

Akaike info criterion

1.211470

Sum squared resid

18.10657

Schwarz criterion

1.264894

Log likelihood

-56.15058

F-statistic

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931.3354

Durbin-Watson stat

0.302076

Prob(F-statistic)

0.000000

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Dependent Variable: YT

Method: Least Squares

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Date: 04/18/08 Time: 12:06

Sample: 1 94

Included observations: 94

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Variable

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Coefficient

Std. Error

t-Statistic

Prob.

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YT_1

-0.038705

0.096052

-0.402959

0.6879

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R-squared

-0.004160

Mean dependent var

-0.005221

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Adjusted R-squared

-0.004160

S.D. dependent var

0.068257

S.E. of regression

0.068399

Akaike info criterion

-2.516338

Sum squared resid

0.435093

Schwarz criterion

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-2.489281

Log likelihood

119.2679

Durbin-Watson stat

1.919828

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