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INDIVIDUAL FINAL ASSESSMENT DICISION SCIENCE & STATISTICS FOR MANAGERS Name: Le Thi Bich Diep Date of birth: 09/10/1984 Class: GaMBA01.N01 Question 1: A True or False questions and explain: False because relate autocorrelation is awake lysiss tight middle inchoate nexus True because big claw pattern then accuracy claw returns of a census False because average development speed is geometric mean of uninterrupted development speeds True because can have arithmetic mean and number harmonic mean False because still to need to based on squads middle distance B Choose the correct answer e e 3.a 4.e 5.e Question 2: n = 50, X = 30, S = 5, α = 5% a Find about estimation to worker's a hour average yield: We estimate the median, and we know σ, n>30 Use Z test,α =0,05: - α/2 = 1-0.05/2 = 0.975, Find in Statistical tables: Zα/2 = 1.960 30 −1,960 × 5 ≤ µ ≤30 +1.960 × 50 50 28,614 ≤ µ ≤ 31,386 So with 95% confidence we state that worker's a man hour average yield is between 28.614 products to 31.386 products b there shall not happen lay off worker because reference level average of taller worker layed off yield (25p/h) Question 3: We canculate median, variation, standard deviation for each plan: (xi1- X ̅ STT xi1 ̅X ̅ ̅)2 xi2 24 28.083 16.674 26 27 28.083 1.174 32 25 28.083 9.507 35 29 28.083 0.840 38 23 28.083 25.840 35 26 28.083 4.340 26 28 28.083 0.007 30 30 28.083 3.674 28 32 28.083 15.340 24 10 34 28.083 35.007 26 11 33 28.083 24.174 12 26 28.083 4.340 Tổng 337 140.917 300 S = 140.917/(12-1) = 12,811 X ̅ ̅2 30 30 30 30 30 30 30 30 30 30 (xi2- X ̅ ̅)2 16 25 64 25 16 36 16 206 S22 = 206/(10-1)= 22,889 This is the case of not know σ12,σ22 (n1, n2 < 30) Assumption: µ1: Plan 1, µ2: plan H0 : µ1 = µ2 (the average cost of Plan is same as the the average H1 : average cost of plan 2) µ1 ≠ µ2 (the average cost of Plan is not same as the the cost of plan 2) Use T test t= X1 − X S2 S2 + n1 n2 (n1 − 1) S12 + (n2 − 1) S 22 S = n1 + n2 − 2 Replace S21= 12,811, S22 =22,889, n1 = 12, n2 = 10, X ̅1= 28.083, X ̅2=30 to this equation, we canculate t= -1.075 Find in Statistical tables: tα / 2;( n1 +n2 − ) With 95% confidence => α = 5%;α / = 2,5% tα / 2;( n1 +n2 −2 ) = 2,086 => t < tα / 2;( n1 + n2 −2 ) t is not in rejection, We conclude that with 95% confidence, two plans have the same average cost Question 4: Draw a stem and leaf display the data Steam Leaf Total leaf 0 7 0 1 4 0 Total 6 30 With aforementioned stipes can break tuple into class With distance evenly is tonnes Classes Mid points tons but under tons 3,5 13% 13% tons but under tons 4,5 20% 10 33% tons but under tons 5,5 17% 15 50% tons but under tons 6,5 27% 23 77% 7tons but under tons 7,5 23% 30 100% 30 100% 82 100% Total Percentage Commulative Frequency (%) frequency Commulative (%) frequency Draw Histograms: Remark about steel product mass in 30 months: The histograms show that intersection steely quantitative production in the past 30 months are dynamic In level from tonnes to tonnes, among them centralize much In through tonne level 4 Canculate median of steel output: - From the data set: X = ∑x n i = 168.3 = 5,61 30 - From frequency table: ∑x f ∑f i X = i i = 173 = 5,76667 30 Compare the two canculation results show that calculate average from frequency distribution table though simple but result accounts for low-built pecision such as case calculated direct from tuple Question 5: Definitely linear regression equation signifies nexus between checkpoint and proceeds of a sale Estimate the simple linear regression relationship between points (x) and sales (y) - The simple linear regression : = + x From the data set, we canculate: số TT y x 20 xy 160 64 400 15 90 36 225 28 252 81 784 10 50 25 100 12 72 36 144 16 112 49 256 15 105 49 225 13 78 36 169 27 243 81 729 10 25 200 64 625 Total 181 71 1.362 521 3.657 average 18,1 7,1 136,2 52,1 365,7 = - = 52,1 – (7,1)2 = 1,69 = - = 365,7 – (18,1)2 = 38,09 =( - )/ = - = (136,2 – 7,1 x 18,1) / 1,69 = 4,55 = 18,1 – 4,55 x 7,1 = -14,20 The simple linear regression: = -14,2 + 4,55x Equation is told when the checkpoint of the day proceeds of a sale, boost sales representative candidate point the those candidate increases 4.55 million dong Measure the strength of relationship between x and y by the coefficient of determination r: r = ( - ) / ( x y) = (136,2 – 7,1 * 18,1) / ( * ) = 0,958 Because r is near by 1, so x and y have strong linear relationship r2 = 0,918 or 91,8% show that 91.8% variation of y can be explained by the variability of x Assumption: Use t test: t = (b1 - H0 : H1: = (no linear relationship ) (x and y have linear relationship) ) / Sb1 Sb1 = = We canculate: Sb1 = 1,97 / = = 0,48 = 1,97 t = 4,55 / 0,48 = 9,48 With 95% confidence ( /2 = 0,025), Find in A2 we have t Because /2;n-2 = t0,025; = 2,306 = 9,48 > t0,025; we reject H0 With 95% confidence, we conclude that the seller’s points and the sales realy have linear relationship Confidence interval estimation for yx: t t /2;n-2 /2;n-2 = t0,025; = 2,306, = 1,97, = 10, = 6, = 7,1 = -14,2 + 4,55 x = 13,09 = 16,9 Replace these values into the formula, we have: 13,09 – 2,306 x 1,97 x 0,41 13,09 + 2,306 x 1,97 x 0,41 yx 11,21 yx 14,97 With 95% confidence, person whose point is has the maximum sale of 14.97 million dong, this level is less than the level sales efficiency of recruiment, so the manager will not recruise him ... quantitative production in the past 30 months are dynamic In level from tonnes to tonnes, among them centralize much In through tonne level 4 Canculate median of steel output: - From the data set: X... points (x) and sales (y) - The simple linear regression : = + x From the data set, we canculate: số TT y x 20 xy 160 64 400 15 90 36 225 28 252 81 784 10 50 25 100 12 72 36 144 16 112 49 256 15... ( - ) / ( x y) = (136,2 – 7,1 * 18,1) / ( * ) = 0,958 Because r is near by 1, so x and y have strong linear relationship r2 = 0,918 or 91,8% show that 91.8% variation of y can be explained by