We can conclude that the data are Poisson distributed. Chi-Square test of independence Problem 12.12 Use the following contingency table to determine whether variable 1 is independent of variable 2. Let α = .01 | Variable 2 | Variable1 | 24 | 13 | 47 | 58 | | 93 | 59 | 187 | 244 | Step 1 Ho: the two classifications are independent Ha: the two classifications are dependent Step 2 d.f = (r – 1) (c – 1) Step 3 α = 0.01 x 2 0.01, 3df = 11.3449 Step 4 Reject Ho if x 2 > 11.3449 | Variable 2 | Total | Variable1 | 24 (22.92) | 13 (14.10) | 47 (45.83) | 58 (59.15) | 142 | | 93 (94.08) | 59 (57.90) | 187 (188.17) | 244 (242.85) | 583
This can be displayed on a time line: | | | | | $P | $400,000 | | | | | | 0 | 90 | 180 | 270 | | | | | | | | | P | = | price | = | unknown | | S | = | Maturity value | = | $400,000 | | r | = | Simple interest rate (decimal) | = | 9.16 | 100 | | = | 0.0916 | | t | = | Time period (years) | = | 90 | 365 | | = | 0.24657534... years. | | The step-by-step calculation is: P | = | S(1 + rt)-1 | | = | 400,000(1 + 0.0916 x 0.24657534...)-1 | | = | 400,000 x 0.97791257...
Chapter 6 8. 1. The parameters of the opportunity set are: E(rS) = 15%, E(rB) = 9%, sS = 32%, sB = 23%, r = 0.15, rf = 5.5% From the standard deviations and the correlation coefficient we generate the covariance matrix [note that Cov(rS, rB) = rsSsB]: Bonds Stocks Bonds 529.0 110.4 Stocks 110.4 1024.0 The minimum-variance portfolio proportions are: wMin(B) = 0.6858 The mean and standard deviation of the minimum variance portfolio are: E(rMin) = (0.3142 × 15%) + (0.6858 × 9%) = 10.89% = [(0.31422 × 1024) + (0.68582 × 529) + (2 × 0.3142 × 0.6858 × 110.4)]1/2 = 19.94% % in stocks % in bonds Exp. return Std dev. 00.00 100.00 9.00 23.00 20.00 80.00 10.20 20.37 31.42 68.58 10.89 19.94 Minimum variance
1,054,848 c. 1,405,888 d. 1,045,828 5. Evaluate: 12xy, when x = 8 and y = 11. a. 31 b. 228 c. 1056 d. 188 6. Write
The standard deviation is 932 and minimum and maximum credit balances are 1864 and 5678 respectively. Relationships III The four more significant relationships from this sample are: Income and Credit Balance, Income and Household Size along with Years and Credit Balance. The bivariate relationship between income and credit balance is reflected in the below scatter gram /scatterplot that reflects a clear and definite linear positive relationship/correlation between income and credit balance, indicating that the higher the income, the higher the credit balance will be and vice versa. The below scatterplot indicates that there is no definite relationship between Income and household size. The points on the scatter plot seems random and do not reflect a clear pattern indicating that there is no correlation among the variables Income and Size The below scatter plot indicates that there is no relationship between years and credit balance, the scatter plot does not reflect a clear pattern indicating that there is there is correlation among the variables Years and Credit
5. a) Summary Statistics: | | | | | | | | Variable | Mean | Median | StDev | Max | Min | Q1 | Q3 | IQR | n | | 1039.60 | 1076.00 | 367.8835 | 1499.00 | 570.00 | 794.00 | 1259.00 | 465.00 | 5.00 | (Those random data is different from the other questions) Mean=1039.60 Standard Deviation=367.8835 Standard error= StDev/ n=73.58 t= (1039.6-5)/73.58=14.1 b) t= (1039.60-1000)/(367.8835/1000)=107.6 :=1000 A: A<1000 P-value<0.05 then reject, otherwise accept. According calculation, I get P-value=2.56835. So, in conclusion, the government’s declare could be accept because P-value is bigger than 0.05 6. 7. Positive relationships extend from bottom left to top right – this happens when both variables increase together.
Yes because the 190 falls under class b. Exercise 8.1.5 What is the slash notation equivalent of subnet mask 255.255.255.0? 11111111.11111111.11111111.00000000 What is the
Problems Answer Grade Problem-1 a x+y=56 /3 b x+y=56 x+3x=56 56/4= 14 x=14 56-14=42 y=42 check 14+42=56 or 14+(14*3)=56 /3 c before we use elimination, we simplify the second equation by dividing by 25,000: new equation for b): 7x + 8y = 288 in order to eliminate a variable, multiply the first equation by -7: new equation for a): -7x + -7y = -266 Elimination: add the two equations and the x's cancel out: y = 22 x = 38-22 = 16 /3 d For the first equation, the intercepts are (56, 0) and (0,56). The intercept for the second equation is (0, 0). The lines would intersect at (14, 42) /3 Problem-2 a x+y=38 /3 b $175,000x+$200,000y=$7,200,000 /3 c Before we use elimination, we simplify the second equation
160.131 Mathematics for Business (1) Test 1 4th of April, 2012 Formulas you may need Slopes (Gradients) The slope (or gradient) m of a curve measures the rate of change in the dependent variable (y), i.e. m = [pic] = [pic] = [pic] = [pic] Determining the Equation of a Line: If we know 2 distinct points (x1, y1) and (x2, y2) on the line: y – y1 = [pic](x – x1); if we know a point (x0, y0) and the slope m of the line: y – y0 = m(x – x0); if we know the slope m and the y-intercept (0, c) of the line: y = mx + c. multiplier = - [pic] Inverse Matrices for 2 × 2 matrices: If A = [pic] then A-1 = [pic][pic], provided ad – bc [pic] 0. Best Fit Curves and
Management 371,101.00 total variable cost 691,691.00 Other + 79,888.00 Total fixed cost 624,989.00 Breakeven point for 2003 PX=A+BX P=1,244,261/6,821 =182.42 X= breakeven point A=624,989.00 B=542,105/6,821= 101.41 182.42X= 624,989+101.41X (subtract 101.41 from both sides) 81.01X=642,989 (divide both sides by 81.01) X= 7937 BEP= 7937 Fixed cost 2004 variable cost 2004 Rent and utilities 150,000.00 payroll and benefits 915,787.00 Telephone 24,000.00 supplies + 320,525.52 Management 445,819.00 total variable cost 1,236,312.52 Other + 115,999.00 Total fixed cost 735,818.00 Breakeven point 2004 PX=A+BX P= 2,191,243/11,822= 185.35 X=breakeven point A=735,818 B=1,236,312.52/11,822=104.57 185.35X=735,818+104.57 (subtract from both sides by 104.57) 80.78X=735,818 (divide both sides by 80.78) X= 9109 BEP= 9109 Purpose, Advantages, disadvantages, and