Statistics For Business And Economics Solution Manuals
- Statistics For Business And Economics
- Statistics For Business And Economics 12th Edition Solution Manual
This manual contains completely worked-out solutions for all the odd-numbered exercises in the text. This product accompanies Statistics for Business and Economics Plus NEW MyLab Statistics with Pearson eText -- Access Card Package, 12th Edition.
Solution Exercise 5.1 a. Covariance and correlation coefficient always have the same sign.
A correlation coefficient can never be larger than 1. A consequence would be that the correlation coefficient is −1.1259, but this is impossible. The covariance of uncorrelated variables is always 0. Solution Exercise 5.2 a. I xi yi 1 2 3 4 5 total 14 18 13 21 18 84 90 88 91 64 80 413 xi − x yi − y ( xi − x )2 ( yi − y ) 2 ( xi − x )( yi − y ) -2.8 1.2 -3.8 4.2 1.2 0 7.4 5.4 8.4 -18.6 -2.6 0 7.84 1.44 14.44 17.64 1.44 42.80 54.76 29.16 70.56 345.96 6.76 507.20 -20.720 6.480 -31.920 -78.120 -3.120 -127.4 x = 16.8; y = 82.6; s X2 = 42.80/4 = 10.70; sY2 = 507.20/4 = 126.80; s X = 3.2711; sY = 11.2606; s X,Y = -127.4/4 = -31.85; r = -31.85 / (3.27) = -0.8647 b. The x-data and the y-data are strongly negatively linearly related.
B1 = X2,Y = -2.9766 and b0 = y − b1 x = 132.6075; yˆ = 132.6075 – 2.9766x sX d. If x increases by 1 unit, then the sample regression line decreases by 2.9766 units. The intercept cannot be interpreted since 0 is not part of the range of the x-data. # hrs worked Solution Exercise 5.3 a.
Statistics For Business And Economics
8 7 6 5 4 3 2 1 0 0 2 4 6 8 10 # hrs studied It looks that y is negatively linearly dependent of x. Yi2 = 133 and xi2 = 179, i =1 7 7 7 b. Since x = 4.7143, y = 3.8571, i =1 xi yi = 104, it i =1 follows that: 1 s X2 = ( s X,Y = ( xi2 − 7 × x 2 ) / 6 = 3.9048; sY2 = ( yi2 − 7 × y 2 ) / 6 = 4.8095; xi yi − 7 x y ) / 6 = −3.8810 c. R = −0.8956; the x- and the y-data are strongly linearly related. Solution Exercise 5.4 a. −20.4, 170.2, 12.8 and 97.2 b.
−0.722 and −25.4669 Solution Exercise 5.5 a. V = 0.3 × (X – 3000) = 0.3X – 900 and W = 0.25 × (Y – 3000) = 0.25Y – 750. Hence: µV = 0.30×20000 – 900 = 5100; µW = 0.25×15000 – 750 = 3000; σ V = 0.30×6000 = 1800 and σ W = 0.25×5000 = 1250 b.
Σ V,W = 0.3×0.25× σ X,Y Since σ X,Y = 0.75×6000×5000 = 22500000, it follows that σ V,W = 1687500. ΡV,W = ρ X,Y = 0.75 c. 5100 = 0.3 × µX − 900, so µX = 20000 d. Men, after taxes: X – 0.3X + 900 = 0.7X + 900; mean = 14900 and variance = (0.7)2×(6000)2 = 17640000 women, after taxes: Y – 0.25Y + 750 = 0.75Y + 750; mean = 12000 and variance = 14062500 Solution Exercise 5.6 a. X = 4.8571 and y = 1.5714; s X2 = 2.4762 and sY2 = 1.9524; s X,Y = -1.9048 r = -0.8663 b. Yˆ = 5.3077 − 0.7692 x; if you watch television one more hour, then it is estimated that the number of hours that you will study decreases by 0.7692.
Prediction = 5.3077 – 0.7692×1.5 = 4.1539 hrs. Yˆ3 = 5.3077 – 0.7692×6 = 0.6925, so e3 = y3 – yˆ3 = −0.6925. The actual observation of Y at day 3 is 0.6925 units less than the prediction that follows from the sample regression line. SSE = = 2.9234.
This is the sum of the squared residuals; it measures the variation of the dots in the sample plot around the sample regression line. Solution Exercise 5.7 a. S X,Y = rX,Y × s X × sY = 7.4001 s X,Y = 2.114. If in a certain week the costs of advertising are 10000 dollars s X2 more, then the sales in that week are estimated to increase by 211400 dollars. W = 0.7692Y and V = 0.7692 X; v = 2.6922; w = 6.4100; sV2 = 2.0708; sW2 = 18.8545; sV,W = 4.3784; rV,W = 0.7007 d.
Slope = 2.114 b. Slope = 2 s X,Y 10.8 = = 1.2857, intercept = y − b1 x = 7.2 − 1.2857 × 1.8 = 4.8857, so: s X2 8.4 yˆ = 4.8857 + 1.2857 x c. V = 4 + 3 x = 9.4 and w = 5 − 2 y = −9.4; sV2 = 9 s X2 = 9 × 8.4 = 75.6 and sW2 = (−2) 2 × 16.8 = 67.2; b. Slope = sV,W = 3 × (−2) × s X,Y = −64.8 − 64.8 = −0.8571 75.6 wˆ = −1.3433 − 0.8571v rV,W = − rX,Y = −0.9091; and and intercept = w − (−0.8571)v = −1.3433, so: slope = Solution Exercise 5.11 a. 1 2 3 4 5 6 7 8 9 10 total x 269.89 177.44 210.33 108.35 171.75 109.75 255.05 105.65 180.36 3.17 x2 7284.95 4429.72 2945.06 65050.5 1119.73 90349.6 y 700 1020 300 199967 y2 00016400 01009010036090 xy 2064818055024 x = 188.917; y = 7429996.7; 1 s X2 = × (400949.6 − 10 × (188.917) 2 ) = 4894.8079; s X = 69.9629; 9 1 sY2 = × (511090 − 10 × (7429996.7) 2 ) = 1.; 9 s X = 113; 1 s X,Y = × ( − 10 × 188.917 × 7429996.7) = −460052426.856; 9 s rX,Y = X,Y = −0.5772 s X sY b.
Yˆ = -0.000003544x + 292 If the number of inhabitants of a country is 1000000 more, then the number of PC’s per 1000 people is on average 3.544 less. “A country without inhabitants has 215.2497 PC’s per 1000 people”. But the intercept of the regression line cannot be interpreted like this: 0 is not in the range of the x-data.
The prediction is: yˆ = 179.1009. So: e = y − yˆ = 177.44 – 179.1009 = −1.6609 Solution Exercise 5.12 Since ei = yi − yˆ i = yi − b0 − b1 xi, it follows that e= 1 n n ( yi − yˆ i ) = i =1 1 n n yi + i =1 1 n n ( −b0 ) + i =1 1 n n ( −b1 xi ) = y − b0 − b1 x, i =1 4 which equals y − ( y − b1 x ) − b1 x = y − y + b1 x − b1 x = 0. Solution Exercise 5.13 a. Such variables would have: ρ = 7 = 1.1667. But this number is larger than 1, 2×3 which is not possible.
Such variables would have: ρ = −7 = −1.1667. But this number is smaller than −1, 2×3 which is not possible.
By Tables 5.17 and 5.18 it follows: σ Y = b σ X ρ X,Y = and σ X,Y = b × 1 × σ X, X = bσ X2; σ X,Y bσ X2 b b = = ×1 = = ±1 b σ Xσ Y σ X b σ X b Since Y is strictly linearly related to X, all dots in the population cloud fall precisely on one (increasing or decreasing) straight line. The linear transformations Y = a + bX has to satisfy: 9 = σ Y2 = b 2σ X2 = 4b 2, so b = ± 9 / 4 = ±1.5 If ρ = 1, the constant b has to be positive and all linear transformations Y = a + 1.5 X satisfy the requirements. If ρ = −1, then b has to be negative and all linear transformations Y = a − 1.5 X satisfy the requirements. Solution Exercise 5.14 a. It is expected that inflation rate is more or less depending on the GDP growth. Yˆ = 4.5242 + 0.3489 x.
If the GDP growth is 1% more, then the percentage inflation is on average 0.3489% more. If the growth is 0%, then the inflation is on average 4.5242. Since 0 falls in the range of the growth data, this interpretation is valid.
Yˆ Zimb = 4.5242 + 0.3489×(−7.1) = 2.047. This is extrapolation, since –7.1 lies far below the minimum value –3.6 of the growth data. GDP growth x Inflation y Prediction yˆ Residual e = y − yˆ Belgium Denmark France Germany Norway Sweden 1.2 3.2 1.2 0.9 2.3 2.7 2.5 1.7 1.9 1.9 1.6 0.8 4.94 5.64 4.94 4.84 5.33 5.47 -2.44 -3.94 -3.04 -2.94 -3.73 -4.67 e. R = Multiple R = 0.2185. There is a weak positive linear relationship between the GDP-growth data and the corresponding inflation data. 5 mileage 0 0 0 0 10000 0 y = 10242x + 3222.4 0 1 2 3 4 5 6 7 age b. Covariance s X,Y = 20422.62 (if you use the Excel-command covar, don’t forget to multiply by 22/21); correlation coefficient rX,Y = 0.925247.
The positive linear relationship is strong. 100000 y = 12071x - 1306.5 mileage 80000 Petrol 60000 Diesel 40000 Linear (Diesel) Linear (Petrol) 20000 y = 8123.3x + 8711.5 0 0 2 4 6 8 age (years) d.
Statistics For Business And Economics 12th Edition Solution Manual
For each extra year, the diesel cars on average drive 12071 miles more and the petrol cars only 8123.3 miles. The two lines are deviating.
Apparently, diesel cars drive more miles than petrol cars. Petrol: 0.9412; diesel: 0.9765. For both type of cars, the linear relationship between age and mileage is positive and strong. Solution Exercise 5.19 a. The first variable will somehow depend on the second, so take the first variable as dependent variable (Y). The two variables have a rather strong linear relationship.
Since ρ = σ X,Y, it follows that σ XσY σ X,Y = ρσ X σ Y = 0.749367× 333.68640 × 176.82560 = 182.0272 β1 = σ X,Y = 0.5455 σ X2 d. Β 0 = µY − β1µ X = 18.12 – 0.5455×34.44 = −0.6672; y = −0.6672 + 0.5455 x If the percentage of the households with broadband connection increases by 1, then the percentage of individuals buying over the Internet increases on average by 0.5455. Since 0 is not in the range of the x-data, we cannot give a valid interpretation of the intercept. 7 Solution Exercise 5.20 a. Yˆ = −0.667 + 0.545 x, in accordance with d. Of Exercise 5.19. SSE = 1938.227 measures the variation around the regression line.
Germany is medium as far as ‘% of households with broadband connection’ is concerned. However, as far as ‘% of individuals buying over the Internet’ is concerned, Germany is very progressive.
GDPpc Solution Exercise 5.21 a. 0 GDPpc Neth 20000 GDPpc USA 10000 0 1950 1970 1990 2010 time Netherlands and USA are running up jointly over time. GDPpc Neth 0 y = 0.8553x - 1053.5 0 0 0 0 0 40000 GDPpc USA This picture suggests that GDPpc in the Netherlands is strongly linearly dependent on GDPpc in USA. However, this relationship is – at least partially – dependent on developments that are included in time. Note that wt = 100( yt − yt −1 ) / yt −1 and vt = 100( xt − xt −1 ) / xt −1. SUMMARY OUTPUT Regression Statistics Multiple R 0.322658 R Square 0.104108 Adjusted R Square 0.08688 Standard Error 2.119477 Observations 54 ANOVA df 1 52 53 SS 27.14505 233.5936 260.7386 Coefficients 1.735601 0.31978 Standard Error 0.404451 0.130087 Regression Residual Total Intercept X Variable 1 MS 27.14505 4.492184 F 6.04273 t Stat 4.291255 2.458196 P-value 7.75E-05 0.017332 wˆ = 1.735601 + 0.31978v - If the growth of GDPpc in the USA is 1 percentage point more, then the growth in the Netherlands will on average be 0.3198 percentage point more.
Since 0 is in.