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Snowflake SnowPro Advanced: Data Analyst Certification Exam Sample Questions (Q47-Q52):

NEW QUESTION # 47
You have a Snowflake table 'EMPLOYEES' with columns 'EMPLOYEE ID' ONT, PRIMARY KEY), 'SALARY' and 'DEPARTMENT' (VARCHAR). You need to enforce the following business rules: 1. 'SALARY' must be a positive value. 2. 'DEPARTMENT' must be one of the following values: 'SALES', 'MARKETING', 'ENGINEERING'. 3. If the Employee is in 'SALES' Department, Salary should be between 50000 and 100000. Which of the following is the most appropriate and efficient approach using Snowflake constraints and features?

A. Use a CHECK constraint for 'SALARY > , a CHECK constraint 'DEPARTMENT IN ('SALES', 'MARKETING', 'ENGINEERING') , and a third CHECK constraint 'CASE WHEN DEPARTMENT = 'SALES' THEN SALARY BETWEEN 50000 AND 100000 ELSE TRUE END'.
B. Use CHECK constraints for both rules and a third CHECK constraint to combine rules 2 and 3 to apply on each record.
C. Use CHECK constraint for SALARY, Create a Lookup table for Departments, and apply a Foreign key relationship for DEPARTMENT field in EMPLOYEE table.
D. Use a CHECK constraint for 'SALARY > , an ENUM type for 'DEPARTMENT' , and a TRIGGER to enforce the salary range rule for the 'SALES' department.
E. Enforce rules using stored procedure at the time of insertion and updation.

Answer: A

Explanation:
Option D is the most appropriate and efficient. CHECK constraints are designed for these types of validations. The 'CASE' statement within the third CHECK constraint allows conditional validation based on the 'DEPARTMENT value. CHECK constriants are enforced at the time of record insert or update. Stored procedures could be an option but are not the most appropriate. Snowflake does not directly support ENUM types for column definitions. Creating Lookup table with Foreign key is another option.

NEW QUESTION # 48
A financial institution uses Snowflake to store transaction data'. They observe an unexpected spike in fraudulent transactions during a specific weekend. The 'TRANSACTIONS table contains columns like 'transaction id', 'transaction date', 'amount', 'merchant_category', 'user_id', and 'is_fraudulent. Which of the following approaches, leveraging Snowflake's capabilities, would be MOST effective in diagnosing the potential causes of this fraud spike?

A. Analyze transaction patterns by 'merchant_category' and 'user_id' before, during, and after the spike to identify any unusual concentrations or emerging fraudulent schemes. Consider using 'QUALIFY' clause to filter top N merchant_category by transaction amount.
B. Join the 'TRANSACTIONS table with a table containing external threat intelligence data (e.g., IP address reputation, known fraudulent merchants) to identify transactions originating from suspicious sources.
C. Create cohorts of users based on transaction behavior before the spike and compare their transaction patterns during the spike to identify anomalies within specific user groups. Use window functions like 'FIRST_VALUE to define cohorts.
D. Use a machine learning model (deployed as a Snowflake user-defined function) to re-score all transactions from the affected weekend based on historical patterns and identify potentially misclassified transactions.
E. Only check the transaction details, like amount and the users involved, and assume one of the users triggered those Fraud Trasanctions.

Answer: A,B,C,D

Explanation:
Options A, B, C, and D provide a robust diagnostic approach. A uses machine learning for anomaly detection, B analyzes patterns by merchant category and user, C incorporates external threat intelligence, and D examines cohort behavior. Only checking specific transaction is not effective.

NEW QUESTION # 49
You are tasked with analyzing website traffic patterns using Snowflake. The data is stored in a table 'WEB TRAFFIC' with columns 'VISIT DATE (DATE) and 'PAGE_VIEWS (NUMBER). You need to identify anomalies (unusually high or low traffic days) using the 'STDDEV POP function to calculate the standard deviation and flag days that fall outside a certain number of standard deviations from the mean. Which of the following SQL queries BEST implements this anomaly detection logic, flagging days with page views more than 2 standard deviations above the mean?

Answer: E

Explanation:
Option D correctly uses window functions to calculate the average and standard deviation across the entire dataset without grouping, then filters the rows where 'PAGE_VIEWS' are more than 2 standard deviations above the mean. Window functions (OVER ()) allow aggregate calculations without collapsing the rows, making it suitable for anomaly detection on a row-by-row basis. Option A and C incorrectly tries to use aggregate function without group by condition, and option B without CTE its creating ambiguity. Option E has logical issue.

NEW QUESTION # 50
You are preparing data for analysis in Snowflake. You have two tables: 'Products' (ProductlD, ProductName, CategorylD) and 'Categories' (CategorylD, CategoryName). You need to create a view that combines product and category information, but also needs to efficiently handle situations where a 'ProductlD' might be missing a 'CategorylD' assignment in the 'Products' table. Furthermore, you need to ensure the view performs optimally for ad-hoc querying by analysts. Which of the following SQL statements creates the MOST performant and robust view, ensuring minimal impact from missing 'CategorylD' values?

A. CREATE VIEW ProductCategories AS SELECT p.ProductlD, p.ProductName, c.categoryName FROM Products p FULL OUTER JOIN categories c ON p.CategorylD = c.CategorylD;
B. CREATE VIEW ProductCategories AS SELECT p.ProductlD, p.ProductName, c.CategoryName FROM Products p INNER JOIN Categories c ONp.CategorylD = c.CategorylD;
C. CREATE VIEW ProductCategories AS SELECT ProductlD, ProductName, (SELECT CategoryName FROM Categories WHERE CategorylD =Products.CategorylD) AS CategoryName FROM Products;
D. CREATE VIEW ProductCategories AS SELECT p.ProductlD, p.ProductName, NVL(c.CategoryName, 'Unknown') AS CategoryName FROM Products p LEFT JOIN Categories c ON p.CategorylD = c.CategorylD;
E. CREATE OR REPLACE VIEW ProductCategories AS SELECT p.ProductlD, p.ProductName, CASE WHEN c.CategoryName IS NULL THEN 'Unknown' ELSE c.CategoryName END AS CategoryName FROM Products p LEFT JOIN Categories c ON p.CategorylD = c.CategorylD;

Answer: D

Explanation:
The best option is B. It uses a 'LEFT JOIN' to ensure all products are included, even those without a 'CategorylD' in the 'Categories' table. ' NVL' efficiently handles 'NULL' values in 'CategoryName' by replacing them with 'Unknown'. A 'FULL OUTER JOIN' (option A) is not appropriate as we specifically need all Products. Using 'CASE WHEN' (option C) is functionally equivalent to 'NVL' but often less efficient. 'INNER JOIN' (option D) will exclude products without a matching 'Categoryl[Y. The subquery approach (option E) is highly inefficient and will not perform well, especially with large tables.

NEW QUESTION # 51
You're building a dashboard to monitor the performance of various marketing campaigns. The data resides in Snowflake, and you're using a BI tool that supports direct query The table has columns: 'CAMPAIGN ID, 'DATE, 'IMPRESSIONS', 'CLICKS, 'SPEND , and You need to create a calculated field in the BI tool representing the Cost Per Conversion (CPC), but you want to optimize query performance and avoid division by zero errors. Assume 'SPEND' and 'CONVERSIONS' are both numeric columns. Which SQL expression, suitable for use in a direct query BI tool, is the MOST performant and robust way to calculate CPC, avoiding zero conversion issues?

Answer: E

Explanation:
The ' DIV0' function is specifically designed by Snowflake to handle division by zero gracefully, returning NULL. It's the most concise and performant way to achieve the desired result. 'NULLIF(CONVERSIONS, 0)' is also correct way, but DIVO is more accurate for Snowflake environment. 'CASE WHEN' and 'IFF are functionally equivalent in this scenario, but is shorter to write. ZEROIFNULL' will return 0 when input is null which won't solve the zero conversion issues. Furthermore ZEROIFNULL' is not a valid Snowflake function.

NEW QUESTION # 52
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