Business Statistics - Assignment Example

Business statistics Table of Contents Answers to the Questions 3 Bibliography 15 Answers to the Questions In a single throw, probability of hitting bull’s eyes, p = 0.39. Therefore, probability of not hitting bull’s eyes, q = 1 – p = 0.61 Thus in five successive throws, probability of hitting bull’s eyes at least once = 1 – P(not hitting bull’s eyes once, in five throws) = 1 - 0.615 = 0.91554 (approx.) 2. To figure out the median, the given set of numbers needs to be arranged in ascending order at first, corresponded by their frequency of occurrence and less than type cumulative frequencies. Here, number of observations, N = 12; hence, (N + 1) = 13. Given that (N+1)/2 = 6.5, from the last column of the table, 6.5 is found to be greater than 5 but less than 7. Thus, the median in this case will be ‘9’. 3. Number of students appearing = 1000, Percentile score = 62. Let number of students who got less than 65 = x. Hence, (x/ 1000) x 100 = 62 x = 620 Thus, number of students who got more than 65 = 1000 – 620 = 380. 4. P (-1.52 < z < 0.29) = Φ (0.29) - Φ (-1.52) = 0.6141 – [1 - Φ (1.52)] = 0.6141 – [1 – 0.9357] = 0.5498 5. Given that there are 9 treatments in all, it could be said that the number of independent samples under consideration, k = 9. Each sample consists of 7 observations (n = 7), so that the total number of observations, N = (7 x 9) = 63 Thus, the degrees of freedom of F-distribution in One-way ANOVA Table = (N – k), (k – 1) = (63 – 9), (9 – 1) = (54, 8) 6. Number of observations, N = 83 and number of treatments or groups, k = 10. With given information about Sum of Squares, the One-way ANOVA Table could be formulated as under – Source of Variation Sum of Squares (SS) Degrees of freedom (df) Mean sum of Squares (MS) Calculated F statistic Between Groups SSB = 48.77 (k – 1) = 9 MSB = 5.4111 = MSB/MSE = 16.26899 Within Groups (Error) SSE = 24.28 (N – k) = 73 MSE = 0.332603 Total SST = 73.05 (N – 1) = 82 Hence, the calculated F-statistic is 16.26899 approximately. 7. Given that the 4-sided die is rolled 32 times, the number of times that each side is expected to show up is 8, though the observed values are quite different, which is the basis of a discrepancy in the goodness of fit of the model. The test statistic to examine the goodness of fit is chi-square, which could be calculated as follows – Observed Frequency (O) Expected Frequency (E) (O – E) (O – E)2 (O – E)2/ E Face 1 5 8 -3 9 1.125 Face 2 7 8 -1 1 0.125 Face 3 13 8 5 25 3.125 Face 4 7 8 -1 1 0.125 Total - - - - χ2 = 4.5 Hence, the goodness of fit statistic is chi-square = 4.5. 8. Confidence interval for population mean could be calculated with the help of Standard Normal Variate. 95% of the area under the standard normal curve lies within the range -1.96 ≤ z ≤ 1.96, where, z = (x - μ)/ (σ/ √n). In the present case, x = (190+125+171+98+100+58+127+135+172)/ 9 = 130.6667 σ = 42.14262 and n = 9. Thus, z = (130.6667 – μ)/ (42.14262/ 3) = (130.6667 - μ)/ 14.04754. Hence, 95% confidence interval for population mean could be represented as, -1.96 ≤ (130.6667 - μ)/ 14.04754 ≤ 1.96 => - (1.96 x 14.04754) ≤ 130.6667 - μ ≤ (1.96 x 14.04754) => - 27.53318 ≤ 130.6667 - μ ≤ 27.53318 => -158.1999 ≤ - μ ≤ -103.134 => 103.134 ≤ μ ≤ 158.1999 9. The proportion of telephone operators feeling insecure about their jobs, p = 42% = 0.42 Hence, proportion of operators feeling secure of their jobs, q = 1 – p = 58% = 0.58. So 90% confidence interval for the proportion of people in the population who feel insecure about their jobs could be represented as (p ± 1.64 S.E), since area of the standard normal curve falling between ± 1.64 is 90%. Given that, S. E. = √(pq)/ n = √(0.42 x 0.58)/ 165 = 0.038423, the 90% confidence interval about the proportion of telephone operators in UK call centres being insecure about their jobs is, [0.42 ± (1.64 x 0.038423)] = (0.356985, 0.483015). 10. Calculation of 95% confidence interval yielded the confidence limits to be (2800, 3800). Hence it could be said that, x - 1.96 (se) = 2800 and x + 1.96 (se) = 3800 Where, x = Sample estimate of mean, se = Standard error and P[-1.96 ≤ z ≤ 1.96] = 0.95, given z ~ N (0, 1) Thus, solving the above two equations, x = 3300 and se = 255.1020. Given that μ0 = 2700, in the present case, z = (3300 – 2700)/ 255.1020 = 2.35200 Hence, P (z ≤ 2.35200) = Φ (2.35) = 0.9906. For two tailed significance testing the proportion of area lying beyond ±2.35200, P (-2.35200 ≤ z ≤ 2.35200) = 2 – [2 x Φ (2.35)] = 0.0188. Thus, the sample estimated mean is found to be significant at 5% level of significance. 11. The time distribution, T, is considered to be normally distributed, such that, T ~ N (134, 625) Hence, the standard normal variable, Z = (X – 134)/ 25 ~ N (0, 1) Therefore, when X = 150, then Z = (16/ 25) = 0.64 Thus, probability that X is at least equal to 150, could be represented as, P (Z ≤ 0.64) = Φ (0.64) = 0.7389. 12. For the 1st sample comprising of 7 observations, the variance, S12 = 12345 – (291 x 291)/ 7 = 247.7143. For the 2nd sample, comprising of 9 observations, the variance, S22 = 12859 – (325 x 325)/ 9 = 1122.8889. Hence, pooled SD, So, pooled variance S122 = 747.8141. 13. The number of observations, n = 22, implying that the degrees of freedom for t-test is (22 - 1 =) 21; while the chosen confidence level for two-tailed significance testing is 0.90. Hence, the critical t22, 0.10 = 1.7207 14. Let the null hypothesis be, H0: MeanX1 – MeanX2 = 0, where 1, 2 are samples 1 and 2 respectively. The appropriate test statistic in the present case in two-sample Student’s t-statistic, defined as, Where, Here, X1 = 4.97, n1 = 6 and SX1 = 0.59, while, X2 = 4.25, n2 = 9 and SX2 = 0.51. Thus, SX1 X2 = 0.5422, implying that, estimated t = 2.5197. The estimated t will be compared with tabulated t, at 95% confidence interval and (n1 + n2 – 2 = 13) degrees of freedom, to accept or reject the null hypothesis. It is found that, t0.05, 13 = 2.1604 < 2.5197 Hence, the concerned null hypothesis has to be rejected at 5% level of significance, implying that the means of the two samples are unequal. 15. The information could be represented in the form of a joint-distribution table between variables ‘Age’ and ‘Media’. However, prior to conducting joint probability distribution, the information need to be modified as follows – Media Age Radio Television Newspaper Total Age below 25 85 22 51 = (85+22+51)=158 Age between 25 and 45 41 47 80 168 Age over 45 22 66 17 105 Total =(85+41+22)=148 135 148 431 Each of the cell values need to be divided by the total sample size to figure out the proportion of total sample that each of them represent. The resultant table will be the joint probability distribution one. Media Age Radio Television Newspaper Total Age below 25 0.197215777 0.051044084 0.118329466 0.366589327 Age between 25 and 45 0.09512761 0.109048724 0.185614849 0.389791183 Age over 45 0.051044084 0.153132251 0.039443155 0.24361949 Total 0.343387471 0.313225058 0.343387471 1 Given that each cell represents the probability of intersection between any two values of the variables, the required calculation could be made. So, the probability of a person aged above 45 years will recall the product given that it was advertised on Television could be represented as, P (Age over 45| Television) = P (Age over 45 ∩ Television)/ P (Television) = 0.15313/ 0.31323 = 0.48889 = 50% (approx.) Hence, 50% of the people aged above 45 years are expected to recall the product advertised on Television. 16. To figure out the total number of observations, the blank boxes need to be filled up at first. Procedures used to do the same, have been depicted in the table below. Source Sum of Squares Degrees of freedom Mean Sum of Squares F Treatment A 887.95 4 (= SSA/MSA) 221.9888448 (=FxMSE) 0.86976 Treatment B 1314.92709 (=MSBxdf) 3 438.3090313 (=FxMSE) 1.71731 Interaction (A*B) 5485 12 (=SSA*B/ MSA*B) 457.0811978 (FxMSE) 1.79086 Error 25522.9929* 100 (=SSE/MSE) 255.23 Total 33210.87 ‘*’ SSE = TSS – SSA – SSB – SS(A*B) It is known that the degrees of freedom for the error term = N – ab Where, N = Total Sample Size, a = dfA + 1 and b = dfB + 1 Hence, 100 = N – (5 x 13) = N – 65. Thus, N = 165 = Sample size. 17. F-statistic could be represented as, F = Regression Sum of Squares (ESS)/ σ*2 => σ* 2 = ESS/ F = 75.7/ 4.252809 = 17.8. Given that the observed values of the independent variables are 0, standard error of the estimated intercept is that of the regression model itself. Hence, se (β1*) = √ (σ* 2/ n) = 1.217922. Hence, (1 - α) = 95% confidence interval for intercept parameter, β1 is, [β1* ± tα/2 x se (β1*)] = [β1* ± tα/2 x se (β1*)] = [β1* ± 1.96 x 1.218]. Thus, the width of the confidence interval is, (2 x 1.96 x 1.218) = 4.775 (approx.). 18. The formula to estimate slope coefficient is, β* = Cov (x, y)/ Var (x) n2. Cov (x, y) = n. ∑xy - ∑x. ∑y = (30 x 734) – (98.7 x 74.9) = 14627.37 n2. Var (x) = n. ∑x2 – (∑x)2 = (30 x 799.1) – (98.7)2 = 14231.31. Hence, β* = 14627.37/ 14231.31 = 1.02783019 (approx.) 19. The formula for exponential smoothing is Y (t)* = Y (t-1)* + w [Y (t) – Y (t-1)*], where w = 0.8. For simplicity, it is assumed that Y (t-1)* is the observed value for the previous year, for t = 1. Thus, the predicted value for t = 6 could be yielded as follows, Years (t) Y(t) Y(t-1)* Y (t) – Y (t-1)* w [Y (t) – Y (t-1)*] Y (t+1)* 1 97         2 84 97 -13 -10.4 86.6 3 98 86.6 11.4 9.12 95.72 4 93 95.72 -2.72 -2.176 93.544 5 89 93.544 -4.544 -3.6352 89.9088 6 82 89.9088 -7.9088 -6.32704 Thus, the predicted value for Y (7) is 83.5818. 20. The information being provided could be used to figure out the correlation coefficient, r2 = Sxy2/ (Sxx. Syy) = 0.60714 (approx.) Now, r2 = 1 – (RSS/ TSS); where, RSS = Residual sum of squares, ∑ (ui – ui*)2 and TSS = Total Sum of Squares = ∑ (yi – yi*)2 = n. Syy = (14 x 40) = 560. Hence, RSS = (1 – r2) x TSS = (1 – 0.60714) x 560 = 220.0016. So, error variance of the regression = 1/n x RSS = 220.0016/ 14 = 15.7144. 21. Here, n = 10, ∑x = 72.6, ∑x2 = 676, ∑y = 60, ∑y2 = 606 and ∑xy = 578.8624. Hence, correlation coefficient, r2 = [(578.8624/ 10) – (72.6 x 60)/ 100]2/ [(676/ 10) – (72.6/ 10)2] x [(606/ 10) – (60/ 10)2] = 0.560228. Thus, r = √0.560228 = 0.748484 The hypothesis to be tested is, H0: Covariance between x and y = 0, which indicates that the correlation coefficient between the x and y is zero. To carry on with hypothesis testing, the correlation coefficient has to be portrayed to Fisher’s transformation, as follows, F (r) = ½ x ln [(1 + r)/ (1 – r)] = 0.969498. Convert it into z-score as, z = √(10 – 3 x 0.287806) = 2.66298. To figure out the p-value for two-tailed test, the area above 2.66298 and below -2.66298 has to be calculated. Given that the total area under the standard normal curve is equal to 1, the percentage falling under the required area is, [Φ (2.66298) + Φ (-2.66298)] = 0.992 = 99.2% Hence, area falling outside this = 1 – 0.992 = 0.008 = 0.8%, which is the estimated p-value of the predicted Fisher’s statistic. Assuming the level of significance to be α = 0.05, the p-value is found to be lower than the former, suggesting that the statistic falls in the rejection region. Hence, the corresponding null hypothesis has to be rejected at 5% level of significance. 22. To figure out the year in which the Index adjusted price is minimum the average price needs to be divided with the Retail Price Index (RPI) and the answer has to be multiplied with 100. Year Price RPI Index Adjusted Price 1987 61107 100.0 61107 1988 64728 103.3 62660.21297 1989 67440 110.0 61309.09091 1990 74200 119.5 62092.05021 1991 79636 130.2 61164.36252 1992 83306 135.6 61435.10324 1993 83945 137.9 60873.82161 1994 87256 141.3 61752.30007 1995 89062 146.0 61001.36986 1996 91783 150.2 61107.19041 1997 94827 154.4 61416.45078 1998 98187 159.5 61559.24765 1999 99766 163.4 61056.30355 2000 102271 166.6 61387.15486 2001 104267 171.1 60939.21683 Hence, the index adjusted price is found to be lowest for the year 1993. 23. Adjusted multiple correlation coefficient (R2) could be calculated in a similar way as R2, with the only difference being that the numerator and denominator have to be divided by their respective degrees of freedom. Thus, in the present case, R2 = 1 – [(86.71/ 14)/ (150.18/ 19)] = 1 – 0.78358 = 0.216421. RSS (n-k) TSS (n-1) 24. Let the 4 continuous covariates be represented as X1, X2, X3 and X4. The three binary covariates on the other hand, assume two values each, so that each of them could be represented by two dummies, with one representing ‘0’, while the other representing ‘1’. Hence, assuming the parent variables to be X5, X6 and X7, the dummies could be considered as (X50, X51), (X60, X61) and (X70, X71) respectively. Following a similar logic, the two discrete covariates, say, X8 and X9 could each be represented by 5 dummies as (X80, X81, X82, X83, X84) and (X90, X91, X92, X93, X94). Given that in multiple regression, each of these dummies are represented by a separate explanatory variable, there will be [4 + (3 x 2) + (2 x 5)] = 20 explanatory variables. Thus, the degrees of freedom associated with the regression sum of squares = 20 – 1 = 19. 25. According to the formula for calculating the confidence interval based on Student’s t-distribution, the upper limit of the 95% confidence interval for the coefficient variable X2 at (21-1 =) 20 degrees of freedom is, [-1.6900 + (2.0860 x 0.7260)] = -0.1756 Sample Tabulated Standard Estimate Student’s t Error Bibliography Gujarati, D. N. (2004). Basic Econometrics (4th Edition). New York, USA: McGraw-Hill. Read More