Companies Statistics Analysis - Assignment Example

Statistics Statistics Part one Task To calculate the descriptive statistics (number of days, mean values and standard deviations) of the four companies using Excel Answers are reflected in the Excel sheet with the criteria used is indicated as well. Task 2 Company 1 Company 2 Company 3 Company 4 Total Above Average 76 75 52 80 283 Below Average 44 45 68 38 195 Total 120 120 120 118 478 The Calculations and the procedures are reflected in the excel sheet clearly Created questions regarding the Contingency table What is the probability that any of the companies selected is below the Average? What is the Probability of a random selected company is Dubai Investments and the share price is above the mean? What is the Probability of a random selected company is Air Arabia and the share price is below or above the average? What percentage contingency entails a company selected is Shuaa and its share price is below average? Answers to the questions a) Let company = x P(X) < average From the table the totals for firms below Average= 195 out of 478 P(X) < Average = 195/478 = 0.4079 Probability = 0.4079 b) P(X=Dubai & Above Average) From the Excel sheet Dubai is the first Company Therefore, P = Dubai is given as 120/478 = 0.25 Probability that it has above average = 44/478= 0.092 The joint probability becomes = 0.25 * 0.092 = 0.023 c) P(X=Air Arabia, Share price is below or Above Average) From the Excel table Air Arabia is company number 3 P(X=) = 120/478 = 0.25 Probability above average = 52/478 = 0.108 Probability below =68/ 478= 0.1422 Additive probability either the share price is above or below mean = 0.108+0.1422 = 0.2502 Probability that it is Air Arabia and is below or above takes a joint Probability given as 0.25 * 0.2502 = 0.0625 d) P(X=Shuaa, Share Price below Average) From the Excel Sheet Shuaa is Company number 4 Probability X = Shuaa = 118/478 = 0.2468 Probability that it is share price is below Average = 38/478 = 0.079 Therefore, the joint probability = 0.079 * 0.2468 = 0.01949 Percentage Contingency = 1.9% Part 2 Task 3 Normal distributions: Case study Air Arabia Using the Mean and standard deviations from part 1 (Source; Blatt et al, 2009) 4. In describing the shape of the two distributions for the Air Arabia Price and Value curves a researcher has to consider different parameters including: The standard deviations and the mean The mean, mode and median of the normal distributions should be equal The symmetry of the curves Whether the area under curves equates to 1 Skewness and kurtosis of the distribution lie -3 to +3 and -2 to +2 respectively Using these parameters the normality of the curves can be established using the Megastat software as indicated in the Excel sheet. The distribution curve for the prices for is normally distributed as indicated by Output 1 of the Excel Sheet. The Skewness and Kurtosis all lie under the recommended range of -0.0564 and – 0.86 respectively. The mean, median and the mode appears to be concentrated between 1.0 and 2.0 of the normal score. For the value distribution, the curve is skewed to the right as indicated in the Output 2 of the Excel sheet. The Skewness indicates that the distribution is skewed to the right (3.7225 > +3) While the Kurtosis (21.9142> +2) indicating that the distribution is not normal. The mean (25, 096, 397) is not equal to the median and the mode. Mean ≠ Mode ≠median 5. Calculate the Probability that Price is greater than 1.439999999 ≈ 1.44 Frequency X = 1.44 µ =1.54 normal score The above case the ð = 0.105 ≈ 0.11 µ = 1.54 X = 1.44 Z score = X- µ / ð = 1.44 – 1.54 / 0.11 = -0.9090 From the tables = P value = 0.3634 (two tailed) For Price greater than 1.44 = 0.3634 + 0.5 (0.5 indicates the second half of the curve) Therefore, Probability = 0.8413 (Shown in Output 3 in excel) 6. The probability value of the price is less than 1.439999999 will be given as Assuming area under curve has probability = I 1- Probability of P>than 1.44 = Probability P< 1.4399999 Probability = 1 – 0.8413 = 0.1587 (Output 3 of the Megastat) Comments The share price for the Air Arabia Company has a higher probability of retaining its value above its average indicating that it wise to buy shares up to unforeseeable time since the share price has a high chance of stabilizing. This indicates the share performance of the firm and it is wise to invest if the manager anticipated a share price of 1.4399999999 Dirhams. It is advisable to invest in the common stock of Air Arabia. 7. Financial manager wants to know that the probability of the firm is within 80%. Calculate the range of values; Assuming normal distribution µ = 25,096,397, ð = 24,222,101 80% of 25, 096, 397 = 20,077,117.6 (Becomes the X) Probability the value of the company= 20, 077, 117.6 Z score = 20, 077, 117.6 – 25, 096, 397 / 24,222,101 = -0.2 P Value = 0.8414 (under two-tailed curve) Probability that the value is within the 80% of the mean = 0.8414 8. Assuming normal distribution of the price, 1% of the opening prices can be calculated as; X = number that conforms to 1% Z= X – 1.54 / 0.11 = 0.01 X- 1.54 = 0.0011 The value is 1.5411 Task 4 Sampling distributions 1. Regarding the Price of the shares if the researcher was to select a sample of 64 out of the population of 120, ie N=120 & n= 64 Simple random technique is suitable using the random number generation tool. The technique offers all the subjects an equal chance to be selected. The random number generator (Megastat tool) is simple and easy to use. The sample size is reflected on Output 4 of the Excel sheet (For the Prices) 2. A sample for the number of trades is reflected on the output 5 of the Excel sheet. 3. For the two sampling distributions it is easy to describe them a) For the number of trade’s sampling distribution the calculated standard mean for the sample is given as 239.2 while the calculated standard deviation is 132.50. b) The Minimum value is -2.41 while the maximum value of the distribution is 617.70. c) The price’s sampling distribution the calculated mean and standard deviation (standard error) are 1.5498 and 0.1179 respectively. d) The maximum and minimum values for the distributions are 1.81 and 1.33 respectively. 4. Standard error of the samples (are reflected on the Outputs 4 & 5 on the upper right descriptive statistics of the sample distributions. However, they can be found by the formula = ðx = ð / (n) ½ . After substituting the values Standard errors for the two sample distributions are 132.50 and 0.1179 respectively. 5. 15% of the sample means will be above which distribution? This will be under a binomial distribution given the n=64 and the p=15% or 0.15 (As shown in the output 6 in the Excel. 6. Sample size and the standard error of the mean relationship The standard error of the distribution gives the measure of how well the sample subjects represent the underlying population. The standard error decreases with an increasing sample size. For example this can be illustrated as; Sample 1 Sample 2 Sample 3 Sample 4 9 6 5 8 2 6 3 1 1 8 6 7   8 4 1   3 7 3   8 2 3     6 4     9 7     7 1     1 8     1 9     7 9       3       1       6       8       3       4                 Mean: 4.00 6.50 4.83 4.78 Std dev, s: 4.36 1.97 2.62 2.96 Sample size, n: 3 6 12 18 sqrt(n): 1.73 2.45 3.46 4.24 Standard error, s/sqrt(n): 2.52 0.81 0.76 0.70 Part 3 Task 1, Confidence intervals (Source; Smithson, 2003) Using the Price sample distribution generated using the Megastat. Mean of 1.5498 and standard deviation of 0.1179, n=64, N= 120 1. Construct a confidence interval to 99% Confidence level to estimate the opening share price. The Confidence interval is shown in Output 7 of the Excel sheet Upper limit = 1.5878 Lower limits = 1.5118 Therefore, 1.5118≤ P≤ 1.5878 2. Width of the Confidence interval of the Mean constructed in 1 above? From the Output 7 results of the limit half width = 0.0380 Therefore, the width for the Confidence interval = 2(0.0380) = 0.076 3. Use the given sample to construct a significance level of 0.14 confidence level to estimate the opening price in your company Significance level of 0.14 indicates a Confidence level of 0.86 Therefore, using n= 64, sd= 0.1179 and mean of 1.5498, the results can be reflected in the output 8 of the excel sheet 1.5281 ≤ P ≤ 1.5715 is the Confidence interval when CL reduces to 86% 4. What is the width of the confidence interval of the mean you constructed in Q3? The width is given as 2(0.0217) since the output 8 indicates that the half width is 0.0217. Therefore, it becomes 0.0434 5. Explain the relationship between the confidence level and the width of the confidence interval of the mean? Explain to your manager which level will serve him better in estimating the population mean. The width of the confidence interval reduces as confidence level decreases as seen from the above case, When the CL is 99% the width for the interval is 0.076, however, as confidence level decreases to 86% the width reduces to 0.043. Accordingly, the Confidence level of 99% is suitable for the opening prices that they will be more than the mean or will match the estimated mean. It gives more surety regarding how confident to expect the opening prices for the shares to be estimating the population mean of 1.54 Dirhams. Task 2 Hypothesis testing, Traditional method (source; Wilcox, 2005) A) A recent article claims that the mean share Price is less than 1.35 Given mean = 1.5498, n=64, N=120, Standard error = 0.1179, At 3% significance level, can we conclude the above claim. Claim: Mean less than 1.35 F H1: Mean = 1.35 H0: Mean ≠ 1.35 P= 1.5498 From the illustration Calculate the Z score = X- mean / sd = 1.35 – 1. 5498 / 0.1179 = 1.694 Decision rule; Reject H0 if Zts > 1.694 where the ts = test statistic Zts = X- µ substituting the values = 1.35 – 1.5489 (µ2/n)1/2 (1.5498*1.5498/64)0.5 Zts = 141.903 in absolute terms Zts > 1.694 Therefore, Reject the H0; and conclude that the mean value is not less than 1.35 B. Using the P value with 1% Significance level Calculate the Z score from the above information = 1.35 – 1.5498 / 0.1179 = 1.694 P value for the above Z score = 0.0902 (Two tailed) from the Z-score table Decision Rule; Reject H0 if P< 0.05 Therefore, P=0.0902 > 0.05 hence accept the H0 and conclude that the mean value was less than 1.35 as claimed by the article. C. Types of errors that could be made Type I error- occurs when the null hypothesis is rejected when it is true Type II error – When the H0 is false but it is accepted. D. In making the decision in part A, most likely I would have made the type I error since I accepted the H0 while it might be true that the mean was less than 1.35 as per the article and as seen upon using the P-value test. However, in the part B I would have made the type II error since I accepted the H0 When it is false as per the t-test in part A. However, the results are influenced by the significance level and the test used which appear to be different in the two cases (Ruppert, 2011). References Blatt, J., Amabile, T., Mills, N., Annenberg/CPB Project., Consortium for Mathematics and Its Applications (U.S.), Chedd-Angier Production Company., Intellimation, Inc., ... American Statistical Association. (2009). Describing distributions ; Normal distributions. Santa Barbara, Calif: Intellimation [distributor. Dubai Financial Markets, PJSC. (2015). Listed securities. http://www.dfm.ae/pages/default.aspx?c=1011&smb=UPP Ruppert, D. (2011). Statistics and data analysis for financial engineering. New York: Springer. Wilcox, R. R. (2005). Introduction to robust estimation and hypothesis testing. Amsterdam: Elsevier/Academic Press. Smithson, M. (2003). Confidence intervals. Thousand Oaks, Calif: Sage Publications. Read More