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Question 1: Since the performance of the stock with regard to NASDAQ stocks, it is possible to construct a tree diagram that will help in analyzing the various results, the tree diagram is as follows: From the tree diagram we can find the different probabilities, in this case we determine the probability that a stock was a recommended buy surpassed NASDAQ stocks, this comes from simply multiplying the probability of outperforming the NASDQA of probability that this stock likely outcome, it was recommended, it is simply to multiply the 0th 75 X 0th 42 from the tree diagram above, the result therefore is 0 315, this is the likelihood that a recommended stock surpassed NASDAQ. B. In this case, we get the probability that NASDAQ will increase in a given month is 0 5, we are also including information on the performance of any month is independent of other results, for this reason therefore we assume that the likelihood function assuming a binomial distribution, this distribution implies that there exist n identical trials, only two possible outcomes for a track, which includes success and failure, attempts are independent, where the outcome of a lawsuit does not affect the outcome of the second trial, and finally success is indicated by P and failure called Q. binomial probability distribution function is specified as follows: P (x) = nCx Πx (1 – Π) nx From this reason why we can find the different probabilities of desired results in the following table summarizes the excel output with respect to the binomial distribution: x n < p> Π P (x) 0 12 0th 5 0th 000 244 a 12 0th 5 0th 00293 2 12 0th 5 0th 016 113 3 12 0th 5 0th 053 711 4 12 0th 5 0th 12085 5 12 0th 5 0th 193 359 6 12 0th 5 0th 225 586 7 12 0th 5 0th 193 359 8 12 0th 5 0th 12085 9 12 0th 5 0th 053 711 10 12 0th 5 0th 016 113 11 12 0th 5 0th 00293 12 12 0th 5 0th 000 244 The probability that growth will be observed for at least 4 months: This likely will lead to aggregation of the probabilities for 4,5,6,7, 8,9,10,11 and 12 Therefore the answer is the 0th 927002nd The probability that growth will be observed in more than nine months: probability derived from the aggregation of the probability of 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9, the probability therefore is 0 980 713 The probability that growth will be observed for at least 1 month: We get the probability of getting 0 months and subtract one to get this probability 1 to 0 000244 = 0th 999 756 The probability that growth will be observed for some months: We get probability of 0 successes as “0″. 000 244 C. The average time it takes for the porter to go is 45 minutes, a sample of 50 is selected and the average time is 46th 8 with a standard deviation of 2 minutes, we investigate whether the average time is equal to 50, 5% level tests we set the null and alternative hypothesis, we want to find out whether the average time is greater than 45 minutes: < / p> Null hypothesis: X = 46th 8 Alternative hypothesis X ≠ 46th 8 Test statistics Z = (X – D) / (s / (n ½)) Z statistic = ( 46th 8-45) / (2 / (50 ½)) Z statistic = 6 363 961 At 5% level of the test the critical value is 1 644854 one one tail test. Due to the fact that the trip t statistic exceeds the t critical value at 5% we reject the null hypothesis that the average is equal to the 46th 8 reasons why we reject the null hypothesis. D. The average income for a population of 15,000 with a standard deviation of 5.000. A population identified as poor, whose income is lower than 8000, The proportion of poor population will be calculated using the Z table, we determine the Z value as follows:
Z = (8000-15000) / 5000 Z = – 1 4 We now find the area under the Z curve is -1. Four critical values, the area under this section is equal to 0 4192, and that means the 41st 92% of the population is poor. When we assume that those who fall below 12% are poor, we will still use the Z table, but this time we will consider the critical value, the area under the equal 0 12, the critical value is 0 31, – 0 31 = X/5000 X = 1550 Therefore, if we assume that those who are poor is 12%, so their income level is 1550 ;
Question 2: The price on NASDAQ for the period 2000-2007 are summarized in the table below: From the above chart it is clear that prices gradually has been declining over time from 2000, is log difference in returns summarized in the table below: chart above shows that in recent years, prices deviate from the larger sizes, this difference decreased in the recent years makes it less profits and less risk of investing in the NASDAQ stock. We estimate the following model: Ri – Rf = a + b (Rm – Rf) Where
Rm is back on NASDAQ Rf is the three-month Treasury bill Ri is the return on shares We are therefore assume that Ri – Rf equals Y and Rm – Rf is equal to X. Y = a + b X After estimating the model is as follows: Y = -0. 074,943 + 0th 85909X This means that tan increase in X with a unit holding all other factors constant then the level of Y will increase by 0 85,909 will if the level of X is equal to zero as the level of Y being -0. 074943rd Durbin Watson coefficient for this model is equal to 0 302076 shows that there is autocorrelation for this reason that there is autocorrelation means that the model is not BLU (best linear unbiased estimate) it is because it violates the assumptions of OLS. Autocorrelation can be corrected by re pecifying of the model to be estimated by adding time lagged variables in the model. Heteroscedasticity is violation of the OLS estimation model states that the variance of the error term has constant variance across observations, it can be detected by using graphs, where we plot the error terms versus the explanatory variable, there are also other methods, such as park test, cross-sectional area data are prone to this problem can be solved by re-specify our method. Estimation of Y = a + b X + Y SMG + δ HML After our estimation model is as follows: Y = -0. 138,609 + 0th 852301X + 0th 040825sml + 0th HML 033 091 The model states that the autonomous value of Y is -0. 1386, whereas if we hold all other factors constant, increasing x with a unit of Y will increase by 0 852 301 if we hold all other factors and increase the level of SMB with one unit so the level of Y will increase by 0 040 825 final, if we keep all other factors constant and increase the level of the hall with one unit so the level of Y will increase by 0 033091st Durbin Watson test value in this case is 0 560,967 which means 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 does not mean that there is autocorrelation. Heteroscedasticity is violation of the assumption that states that the variance of the error term has constant variance across observations, it can be detected by using graphs, where we plot the error terms versus the explanatory variable, there are other methods such as park test, cross-sectional area data is prone to this problem can be solved by re-specify our method. Multicollinearity is a situation where the explanatory variables are correlated with the explanatory variables in this situation can be demonstrated by the presence of high correlation between decisiveness and presence of t-tests show non-significant estimated coefficients, it can also be detected by the presence of high pair wise correlation. To correct the presence of Multicollinearity we can escape the problematic variables, re-enter model, acquisition of additional data or use of a second sample, transformation of data from such general logging form numbers. Question 3: Stationary: Stationary concept is applicable to time series data states that data are stationary if the mean value and variance are constant over time and that covariance between two periods depends on the distance or gap between the two periods, in this case the variable in question usually depends on the previous period, the value indicated by the YT = puddle-1 + Ut If the value of P is equal to one indicates that a situation is not static, but if P is less or greater than a so data is stationary. The results of the model states that yt = -0. 038705Yt-1, it means that because the value of P is less than 1 then the series data is stationary. auto regression moving average model consists of two parts, which includes autoregressive part and the moving average part, this model formulated to predict the future value of time series data, and it contains a series autoregressive parts and the moving average components. The autoregressive part is designated as follows: Xt = c + b Xt-1 + ut moving average on the other side is indicated as follows: ;
Xt = a + ext-1, where e is the error term For this reason, therefore, autoregressive moving average model will be listed as X t = a + b Xt 1 + ext-1 In our case we therefore state our model as follows: yt = a + b yt-1 + ext-1, where Y is the share price and is the error term References: Bluman A. (2000) Elementary Statistics : An incremental 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 Publishing, Michigan L. Henry (1991) Statistics, Oxford University Press, Oxford < ; strong> Appendix: dependent variable: Y Method: least squares Date: 04/18/2008 Time: 11:16 < p> test: a 96 ; Included observations: 96
Variable Coefficient ; Std. Error t-Statistic Prob.
C -0. 138 609 0th 083 746 -1. 655 112 0th 1013 X 0th 852 301 0th 026 673 31st 95320 0th 0000 SMB 0th 040 825 0th 012 639 3rd 230 108 0th 0017 HML 0th 033 091 0th 010 523 3rd 144 574 0th 0022 ;
R-squared 0th 921 101 Mean dependent was -2. 357 542 Adjusted R-squared 0th 918 528 SD-dependent were 1st 441 867 SE of regression 0th 411 557 Akaike info criterion 1st 103 033 Sum squared Resid 15th 58285 Schwarz criterion 1st 209 881 Log likelihood -48. 94560 F-statistic 358th 0148 Durbin-Watson State 0th 560 967 Prob (F-statistic) 0th 000 000 ;
dependent variable: Y Method: least squares < ; p> Date: 18/04/2008 Time: 11:05 ; test: a 96 Included observations: 96 < ; / p> < ; p> Variable Coefficient Std. Error t-Statistic Prob.
C -0. 074 943 0th 087 183 -0. 859 610 0th 3922 X 0th 859 909 0th 028 177 30th 51779 0th 0000 ;
R-squared 0th 908 323 Mean dependent was -2. 357 542 Adjusted R-squared 0th 907 347 SD-dependent were 1st 441 867 SE of regression 0th 438 888 Akaike info criterion 1st 211 470 Sum squared Resid 18th 10657 Schwarz criterion 1st 264 894 Log likelihood -56. 15058 F-statistic 931st 3354 Durbin-Watson State 0th 302 076 Prob (F-statistic) 0th 000 000 ;
dependent variable: YT Method: least squares < ; p> Date: 04/18/2008 Time: 12:06 ; sample: a 94 Included observations: 94
< ; / p> < ; p> Variable Coefficient Std. Error t-Statistic Prob.
YT_1 -0. 038 705 0th 096 052 -0. 402 959 0th 6879 ;
R-squared -0. 004 160 Mean dependent was -0. 005 221 Adjusted R-squared -0. 004 160 SD-dependent were 0th 068 257 SE of regression 0th 068 399 Akaike info criterion -2. 516 338 Sum squared Resid 0th 435 093 Schwarz criterion -2. 489 281 Log likelihood 119th 2679 Durbin-Watson State 1st 919 828 ;
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