campaign_users +---------+-----------+ | user_id | treatment | +---------+-----------+ | 1001 | control | | 1002 | test | | 1003 | control | | 1004 | test | | 1005 | control | +---------+-----------+ ​ sessions +---------+-----------+--------+---------------------+ | user_id | session_id| region | timestamp | +---------+-----------+--------+---------------------+ | 1001 | s1 | SF | 2024-01-05 10:00:00 | | 1002 | s2 | NY | 2024-01-06 11:00:00 | | 1002 | s3 | SF | 2024-01-10 12:00:00 | | 1004 | s4 | SF | 2024-01-12 09:30:00 | | 1005 | s5 | SF | 2024-02-01 08:20:00 | +---------+-----------+--------+---------------------+ ​ orders +---------+-----------+----------+---------+--------+ | user_id | session_id| order_id | revenue | profit | +---------+-----------+----------+---------+--------+ | 1001 | s1 | o101 | 50.00 | 10.00 | | 1002 | s2 | o102 | 80.00 | 15.00 | | 1002 | s3 | o103 | 45.00 | 12.00 | | 1004 | s4 | o104 | 60.00 | 20.00 | | 1004 | s4 | o105 | 20.00 | 5.00 | +---------+-----------+----------+---------+--------+ ##### Scenario Using three relational tables that record campaign assignment, user sessions, and orders, evaluate the January-2024 SF-only promotion experiment and perform follow-up analyses. ##### Question Write a SQL query that joins the three tables to compute average revenue per user for treatment vs. control, restricting data to sessions in January-2024 and region = 'SF'. For each treatment arm, return the three user_id values with the highest number of orders. For each treatment arm, calculate the percentage of users who have placed more than five orders. ##### Hints Use proper JOINs, WHERE filters on date and region, GROUP BY treatment, COUNT orders, window functions or conditional aggregation as needed.