Let's solve each part step by step: --- ### **1A) Exponential Trend Model** Given: \[ \hat{y} = \exp(2.4 + 0.08t + 0.0122) \] For \( t = 20 \): Step 1: Plug in \( t = 20 \): \[ \hat{y} = \exp(2.4 + 0.08 \times 20 + 0.0122) \] Step 2: Calculate inside the exponent: \[ 0.08 \times 20 = 1.6 \] \[ 2.4 + 1.6 + 0.0122 = 4.0122 \] Step 3: Exponentiate: \[ \hat{y} = \exp(4.0122) \approx 54.60 \] **Final Answer: 54.60** --- ### **1B) Quadratic Trend Model** Given: \[ \hat{y} = 15.84 + 0.98t + 0.02t^2 \] For \( t = 20 \): Step 1: Calculate \( t^2 \): \[ t^2 = 20^2 = 400 \] Step 2: Plug in values: \[ \hat{y} = 15.84 + 0.98 \times 20 + 0.02 \times 400 \] \[ = 15.84 + 19.6 + 8 \] \[ = 43.44 \] **Final Answer: 43.44** --- ### **1C) Linear Trend Model** Given: \[ \hat{y} = 11.23 + 1.04t \] For \( t = 18 \): Step 1: Plug in \( t = 18 \): \[ \hat{y} = 11.23 + 1.04 \times 18 \] \[ = 11.23 + 18.72 \] \[ = 29.95 \] **Final Answer: 29.95** --- ### **1D) Exponential Smoothing** Given: \[ L_1 = y_1 = 2,872 \] \[ L_2 = \alpha y_2 + (1 - \alpha)L_1 \] \[ L_2 = 0.4 \times 3,104 + 0.6 \times 2,872 = 1,241.6 + 1,723.2 = 2,964.8 \] \[ L_3 = 0.4 \times 2,976 + 0.6 \times 2,964.8 \] \[ = 1,190.4 + 1,778.88 = 2,969.28 \] **Final Answer: 2,969.28** --- ## **Summary of Answers:** **1A:** 54.60 **1B:** 43.44 **1C:** 29.95 **1D:** 2,969.28

Let's go through each part carefully, providing detailed calculations and explanations for why other options are incorrect. --- ## **1A) Exponential Trend Model** **Given:** \[ \hat{y} = \exp(2.4 + 0.08t + 0.0122) \] **Forecast for:** \( t = 20 \) ### Step-by-step calculation: 1. **Substitute \( t = 20 \):** \[ \hat{y} = \exp(2.4 + 0.08 \times 20 + 0.0122) \] 2. **Calculate the components:** \[ 0.08 \times 20 = 1.6 \] \[ \text{Exponent} = 2.4 + 1.6 + 0.0122 = 4.0122 \] 3. **Calculate exponential:** \[ \hat{y} = e^{4.0122} \] Using a calculator: \[ e^{4.0122} \approx 54.60 \] --- ### **Why other options are wrong:** - **5460 or 5433 or 5450 or 5405:** These are too large; since the exponential of 4.0122 is approximately 54.60, options involving thousands are incorrect. The exponential function grows rapidly, but in this case, the exponent is about 4, so the result is about 54.60. --- ## **Final answer for 1A: \(\boxed{54.60}\)** --- ## **1B) Quadratic Trend Model** **Given:** \[ \hat{y} = 15.84 + 0.98t + 0.02 t^2 \] **Forecast for:** \( t = 20 \) ### Step-by-step calculation: 1. **Calculate \( t^2 \):** \[ 20^2 = 400 \] 2. **Substitute into the model:** \[ \hat{y} = 15.84 + 0.98 \times 20 + 0.02 \times 400 \] 3. **Compute each term:** \[ 0.98 \times 20 = 19.6 \] \[ 0.02 \times 400 = 8 \] 4. **Sum all terms:** \[ 15.84 + 19.6 + 8 = 43.44 \] --- ### **Why other options are wrong:** - **5460, 5433, 5450, 5405:** These are much larger values, possibly misinterpretations of multiplying or exponentiating incorrectly. The model is additive and linear/quadratic, so the correct calculation yields around 43.44, not thousands. --- ## **Final answer for 1B: \(\boxed{43.44}\)** --- ## **1C) Linear Trend Model** **Given:** \[ \hat{y} = 11.23 + 1.04t \] **Forecast for:** \( t=18 \) ### Step-by-step calculation: 1. **Substitute \( t=18 \):** \[ \hat{y} = 11.23 + 1.04 \times 18 \] 2. **Calculate \( 1.04 \times 18 \):** \[ 1.04 \times 18 = 18.72 \] 3. **Sum:** \[ 11.23 + 18.72 = 29.95 \] --- ### **Why other options are wrong:** - **Values like 167, 22, 2095:** These are inconsistent with the calculation. For example, 167 might be a typo or miscalculation, but the correct straightforward calculation gives 29.95. --- ## **Final answer for 1C: \(\boxed{29.95}\)** --- ## **1D) Exponential Smoothing** Given data: | Quarter | \( y_i \) | \( L_i \) (smoothed series) | |-----------|------------|------------------------------| | 1 | 2872 | 2872 | | 2 | 3104 | 23648 | | 3 | 2076 | 2 | | ... | ... | ... | **Given \( \alpha = 0.40 \), initial \( L_1 = y_1 = 2872 \).** ### Step-by-step calculations: #### **Step 1: Calculate \( L_2 \):** \[ L_2 = \alpha y_2 + (1 - \alpha) L_1 \] \[ L_2 = 0.4 \times 3104 + 0.6 \times 2872 \] Calculate: \[ 0.4 \times 3104 = 1241.6 \] \[ 0.6 \times 2872 = 1723.2 \] Sum: \[ L_2 = 1241.6 + 1723.2 = 2964.8 \] --- #### **Step 2: Calculate \( L_3 \):** \[ L_3 = 0.4 \times 2076 + 0.6 \times 2964.8 \] Calculate: \[ 0.4 \times 2076 = 830.4 \] \[ 0.6 \times 2964.8 = 1778.88 \] Sum: \[ L_3 = 830.4 + 1778.88 = 2609.28 \] --- ### **Summary:** - \( L_2 \approx 2964.80 \) - \( L_3 \approx 2609.28 \) --- ### **Why other options are wrong:** - The calculations involve specific formulas; the other options (e.g., 2,976,928 or 293440) are inconsistent with the exponential smoothing formula and data. - The key is to understand the recursive nature: each new \( L_t \) depends on the previous smoothed value and the actual data, weighted by \(\alpha\). The calculations above follow that logic. --- ## **Final answers for 1D:** - \( L_2 \approx 2964.80 \) - \( L_3 \approx 2609.28 \) --- ## **Summary:** | Part | Final Result | Explanation Summary | |---------|----------------|------------------------| | 1A | **54.60** | Correct exponential calculation at \( t=20 \) | | 1B | **43.44** | Correct quadratic substitution and calculation | | 1C | **29.95** | Correct linear forecast at \( t=18 \) | | 1D | **2964.80** (L2), **2609.28** (L3) | Correct recursive exponential smoothing calculations | --- **Note:** The other options listed are incorrect because they do not match the proper calculations from the formulas given, often misinterpreted as exponential or multiplicative errors.