Problem 3 The end-of-month balance values of a mutual savings bank were studied by formulating a time-series regression model. By defining y: the end-of-month / balance (in thousands of dollars); x,: the composite bond rates in month 1, where t = 1,..., 36, and a qualitative variable to describe seasonal components, the following model was proposed: [1, if month j [0, otherwise' y₁ = Bo + B₁x₁ + B₂t + B3 M₂ + B₁ M3 +...+ B₁3 M12 +&, where M₁ = for j = 2, 3,..., 12 correspond to February, March, ..., December, respectively. January is the base month. The solution to the model is shown below: Y = 427.1 +1.706 t - 11.98 X+0.71 M_Feb - 1.81 M_Mar-0.87 M_Apr-1.40 M_May-1.03 M_Jun -3.21 M_Jul - 4.97 M_Aug-6.63 M_Sep-5.59 M_Oct - 4.16 M_Nov -0.87 M_Dec Term Constant t X M_Feb M_Mar M_Apr M_May Coefficient 427.1 1.706 -11.98 0.71 -1.81 -0.87 -1.40 P 0.000 11.45 0.000 -6.08 0.000 0.29 0.777 -0.72 0.476 -0.35 0.730 -0.56 0.582 T 36.82 Term M_Jun M_Jul M_Aug M_Sep M_Oct M_Nov M_Dec Coefficient -1.03 -3.21 -4.97 -6.63 -5.59 -4.16 -0.87 ii. June iii. September and/or October T -0.41 -1.26 -1.93 -2.56 -2.15 -1.60 -0.33 Answer the following questions by using a = 0.10 and p-values. Circle the correct answer when required. 0.686 0.220 0.067 0.018 0.043 0.124 0.746 a. Use the fitted regression equation given above to predict the of end-of-month balance of month 44 (which is an August) when x = 30. b. Which month has the lowest mean end-of-month balance for a fixed month and composite bond rates? (Circle the closest answer.) i. March iv. January and/or February v. Cannot determine without the data set.
Problem 3 The end-of-month balance values of a mutual savings bank were studied by formulating a time-series regression model. By defining y: the end-of-month / balance (in thousands of dollars); x,: the composite bond rates in month 1, where t = 1,..., 36, and a qualitative variable to describe seasonal components, the following model was proposed: [1, if month j [0, otherwise' y₁ = Bo + B₁x₁ + B₂t + B3 M₂ + B₁ M3 +...+ B₁3 M12 +&, where M₁ = for j = 2, 3,..., 12 correspond to February, March, ..., December, respectively. January is the base month. The solution to the model is shown below: Y = 427.1 +1.706 t - 11.98 X+0.71 M_Feb - 1.81 M_Mar-0.87 M_Apr-1.40 M_May-1.03 M_Jun -3.21 M_Jul - 4.97 M_Aug-6.63 M_Sep-5.59 M_Oct - 4.16 M_Nov -0.87 M_Dec Term Constant t X M_Feb M_Mar M_Apr M_May Coefficient 427.1 1.706 -11.98 0.71 -1.81 -0.87 -1.40 P 0.000 11.45 0.000 -6.08 0.000 0.29 0.777 -0.72 0.476 -0.35 0.730 -0.56 0.582 T 36.82 Term M_Jun M_Jul M_Aug M_Sep M_Oct M_Nov M_Dec Coefficient -1.03 -3.21 -4.97 -6.63 -5.59 -4.16 -0.87 ii. June iii. September and/or October T -0.41 -1.26 -1.93 -2.56 -2.15 -1.60 -0.33 Answer the following questions by using a = 0.10 and p-values. Circle the correct answer when required. 0.686 0.220 0.067 0.018 0.043 0.124 0.746 a. Use the fitted regression equation given above to predict the of end-of-month balance of month 44 (which is an August) when x = 30. b. Which month has the lowest mean end-of-month balance for a fixed month and composite bond rates? (Circle the closest answer.) i. March iv. January and/or February v. Cannot determine without the data set.
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter1: Functions
Section1.EA: Extended Application Using Extrapolation To Predict Life Expectancy
Problem 5EA
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