Solutions Key for Assignment 2 - Regression Analysis | BA 303, Assignments of Introduction to Business Management

Download Solutions Key for Assignment 2 - Regression Analysis | BA 303 and more Assignments Introduction to Business Management in PDF only on Docsity! 1 BA 303 Assignment #2 – Solution Key Chapter 10 1. ASW - page 409 - Question # 17 µ1 for 2001 season µ2 for 1992 season H0: 1 2 0µ µ− ≤ Ha: 1 2 0µ µ− > b. 1 2x x− = 60 - 51 = 9 days 9/51(100) = 17.6% increase in number of days. c. ( )1 2 2 2 2 2 1 2 1 2 0 (60 51) 0 2.48 18 15 45 38 x x t s s n n − − − − = = = ++ 2 22 2 2 2 1 2 1 2 2 2 2 22 2 2 2 1 2 1 1 2 2 18 15 45 38 81 1 18 1 151 1 44 45 37 381 1 s s n n df s s n n n n ⎛ ⎞ ⎛ ⎞ + +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠= = = ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ++⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− − ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ Using t table, p-value is between .005 and .01. Actual p-value = .0075 p-value≤ .01, reject H0. There is a greater mean number of days on the disabled list in 2001. d. Management should be concerned. Players on the disabled list have increased 32% and time on the list has increased by 17.6%. Both the increase in inquiries to players and the cost of lost playing time need to be addressed. 2. ASW - page 416 - Question # 27 Using matched samples, the differences are as follows: 4, -2, 8, 8, 5, 6, -4, -2, -3, 0, 11, -5, 5, 9, 5 H0: µd ≤ 0 Ha: µd > 0 2 d = 3 and sd = 5.21 t d s n d d = − = − = µ / . / .3 0 5 21 15 2 23 df = n - 1 = 14 Using t table, p-value is between .01 and .025 Actual p-value = .0213 Reject H0. Conclude that the population of readers spends more time, on average, watching television than reading. Chapter 14 3. ASW - page 562 - Question # 1 a. b. There appears to be a linear relationship between x and y. c. Many different straight lines can be drawn to provide a linear approximation of the relationship between x and y; in part d we will determine the equation of a straight line that “best” represents the relationship according to the least squares criterion. d. Summations needed to compute the slope and y-intercept are: 215 40 ( )( ) 26 ( ) 10i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − = 1 2 ( )( ) 26 2.6 10( ) i i i x x y yb x x Σ − − = = = Σ − b y b x0 1 8 2 6 3 0 2= − = − =( . )( ) . ˆ 0.2 2.6y x= + e. ˆ 0.2 2.6(4) 10.6y = + =