Answer: 30

The model is given as: $$ y_t = 0.2 \times \text{Time}_t + 10.5 + 3.0 \times D_{2t} + 2.1 \times D_{3t} + 2.8 \times D_{4t} $$ Since we have three dummies and an intercept, quarterly seasonality is reflected by the intercept (10.5) plus the three seasonal dummy variables ($D_2, D_3$ and $D_4$). If $y_{(T+1)} = \text{May 2019}$, then April 2020 = $y_{(T+12)}$ and July 2020 = $y_{(T+15)}$. Finally, note that April falls under $D_{(2t)}$, while July falls under $D_{(3t)}$. **Step-by-step calculation:** 1. **April 2020 (y_{T+12}):** - Time_t = 12 (since it's the 12th month from the start) - April is in Spring season, so D_{2t} = 1, D_{3t} = 0, D_{4t} = 0 - y_{April2020} = 0.2 × 12 + 10.5 + 3.0 × 1 + 2.1 × 0 + 2.8 × 0 - y_{April2020} = 2.4 + 10.5 + 3.0 = 15.9 (in hundreds) 2. **July 2020 (y_{T+15}):** - Time_t = 15 (since it's the 15th month from the start) - July is in Summer season, so D_{2t} = 0, D_{3t} = 1, D_{4t} = 0 - y_{July2020} = 0.2 × 15 + 10.5 + 3.0 × 0 + 2.1 × 1 + 2.8 × 0 - y_{July2020} = 3.0 + 10.5 + 2.1 = 15.6 (in hundreds) 3. **Difference:** - Difference = y_{April2020} - y_{July2020} = 15.9 - 15.6 = 0.3 (in hundreds) - Since the answer is in actual number of guests (not hundreds), multiply by 100: 0.3 × 100 = 30 guests Therefore, April 2020 is expected to have 30 more guests than July 2020.

Author: Nikitesh Somanthe