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Snowflake DAA-C01 Exam Syllabus Topics:

Section	Weight	Objectives
Topic 1: Data Presentation and Data Visualization	28%-29%	- Integrate with BI tools 1. Tableau integration 2. Other partner visualization tools 3. Power BI integration - Create dashboards 1. Present analytical results 2. Snowsight dashboards
Topic 2: Data Ingestion and Data Preparation	15%-20%	- Prepare data and load into Snowflake 1. Load files using Snowsight 2. Load data from external/internal stages into a table - Implement data processing solutions 1. Automate and implement data pipelines (scheduling) 2. Respond to processing failures 3. Use logging and monitoring solutions (auditing, data lineage) 4. Cleanse, conform, and enrich data - Use a collection system to retrieve data 1. Retrieve data from semi-structured sources (Parquet, Avro, ORC, JSON, XML) 2. Retrieve data from structured sources (CSV) 3. Retrieve data from unstructured sources 4. Synthetic Data Generation - Perform data discovery to identify what is needed from available datasets 1. Determine the level of data granularity required 2. Query tables to assess data elements and statistics maintained by Snowflake 3. Identify elements required for business goals using BI reports or SQL analysis 4. Use commands to read metadata or alter context (DESCRIBE, SHOW, USE) 5. Evaluate required transformations (table joins, set operations, ASOF JOINS) - Enrich data by identifying and accessing relevant data from the Snowflake Marketplace 1. Use Secure Data Sharing (Marketplace, Internal Marketplace, Private Listings, Listings) 2. Find external data sets that correlate with available data 3. Create tables and views - Use best practice considerations relating to data integrity structures 1. Perform table joins between parent/child tables 2. Implement constraints 3. Define primary keys for tables
Topic 3: Data Transformation and Data Modeling	22%-30%	- Transform data using SQL 1. PIVOT/UNPIVOT operations 2. Window functions 3. Common Table Expressions (CTEs) 4. QUALIFY clauses - Design data models 1. Data vault models 2. Snowflake schema design 3. Star schema design
Topic 4: Data Analysis	30%-32%	- Perform advanced analytics using SQL 1. Aggregate functions 2. Time-series analysis 3. Snowflake-specific analytical features

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Snowflake SnowPro Advanced: Data Analyst Certification Exam Sample Questions (Q54-Q59):

NEW QUESTION # 54
How does the utilization of Regular views differ from Materialized views in data analysis?

A. Materialized views offer limited data accessibility compared to Regular views.
B. Regular views are exclusively used for exploratory analyses.
C. Regular views offer better query performance compared to Materialized views.
D. Materialized views provide a persisted snapshot, unlike Regular views.

Answer: D

Explanation:
Materialized views provide a persisted snapshot of data, differentiating them from Regular views.

NEW QUESTION # 55
A marketing company is analyzing customer purchase data stored in Snowflake to understand which customer demographics are most likely to purchase a newly launched product. The 'CUSTOMERS table has columns: 'customer_id', 'age' , 'gender' , 'location' , and 'household income'. The 'PURCHASES' table has columns: 'customer_id', 'purchase_date', and 'product id'. Which SQL query would most effectively identify the top three age groups with the highest purchase rate for the new product (product_id = 'NEW PRODUCT')?

A. Option C
B. Option E
C. Option A
D. Option D
E. Option B

Answer: D

Explanation:
Option D is the most effective because it calculates the purchase rate by dividing the count of distinct customers who purchased the new product by the total number of customers in each age group. This provides a normalized purchase rate for each age group, allowing for a fair comparison regardless of the size of each group. Options A and C only provide the raw count of purchases or unique purchasers, which doesn't account for the size of each age group. Option B uses 'SUM(CASE...V , which can work, but is less efficient than using 'COUNT(DISTINCTV in combination with a left join to account for all customers, even those who didn't make any purchases. Option E would produce an incorrect rate as it is dividing a customer's count by the distinct count of each customer, resulting in a rate close to 1.0 for each customer that made a purchase.

NEW QUESTION # 56
In Snowsight, why is it significant to create various chart types (e.g., bar charts, scatter plots, heat grids) for data visualization?

A. It limits data presentation options for complex datasets.
B. Different chart types offer varied data representation for better analysis.
C. Snowsight doesn't support multiple chart types for visualization.
D. It restricts users to specific chart types for simplicity.

Answer: B

Explanation:
Diverse chart types offer varied data representation, facilitating better analysis in Snowsight.

NEW QUESTION # 57
How do diverse chart types (e.g., bar charts, scatter plots, heat grids) contribute to effective data presentation and visualization in reports and dashboards?

A. Different chart types offer varied data representation for better analysis.
B. Diverse chart types restrict data exploration in reports and dashboards.
C. Charts don't impact data visualization in reports or dashboards.
D. They limit data representation options for simplicity.

Answer: A

Explanation:
Different chart types offer varied data representation, aiding better analysis in reports and dashboards.

NEW QUESTION # 58
You are tasked with enriching your company's customer transaction data with external economic indicators (e.g., unemployment rate, GDP) obtained from a Snowflake Marketplace data provider. The transaction data resides in a table 'TRANSACTIONS' with columns 'TRANSACTION (INT), 'TRANSACTION DATE (DATE), and (VARCHAR). The economic indicators data, obtained from the Marketplace, is available in a table 'ECONOMIC DATA' with columns 'DATE (DATE), ZIP_CODE (VARCHAR), 'UNEMPLOYMENT RATE (NUMBER), and 'GDP' (NUMBER). Due to data quality issues, some zip codes in both tables are missing or malformed. You need to create a view that efficiently joins these two tables, handles missing or malformed zip codes, and provides the transaction data enriched with the economic indicators. Which of the following approaches is the MOST robust and efficient way to create this enriched view, minimizing data loss and maximizing data quality?

A. Create a view that first filters out all rows with missing or malformed zip codes from both 'TRANSACTIONS' and 'ECONOMIC DATA' using 'WHERE clauses and regular expressions to validate the zip code format. Then, perform an ' INNER JOIN' between the filtered datasets on 'TRANSACTIONS.TRANSACTION DATE = ECONOMIC DATA.DATE-' and 'TRANSACTIONS.CUSTOMER ZIP = ECONOMIC DATA.ZIP CODE.
B. Create a view using a 'LEFT OUTER JOIN' between 'TRANSACTIONS and ECONOMIC_DATX on 'TRANSACTIONS.TRANSACTION_DATE =ECONOMIC_DATA.DATE' and 'TRANSACTIONS.CUSTOMER_ZIP = ECONOMIC_DATA.ZIP_CODE'. Additionally, use the function to handle malformed zip codes and the 'NVL' function to replace missing or malformed zip codes with a default zip code (e.g., '00000') for joining purposes. Also include a new column "ENRICHMENT SUCCESS' that flag indicates that the join was successful or whether data was enriched using the default zip code.
C. Create a stored procedure that iterates through each transaction in 'TRANSACTIONS' , attempts to find a matching economic data record in ECONOMIC_DATA' based on date and zip code, and updates a new 'TRANSACTIONS_ENRICHED table with the economic indicators. Handles missing zipcodes by setting the 'UNEMPLOYMENT RATE' and 'GDP ' to 0 for any record in transaction which zip code is missing.
D. Create a Snowflake Task that runs daily to update a materialized view that joins 'TRANSACTIONS' and 'ECONOMIC_DATX on 'TRANSACTIONS.TRANSACTION DATE = ECONOMIC DATA.DATE-' and 'TRANSACTIONS.CUSTOMER ZIP = ECONOMIC DATA.ZIP CODE , handling missing zip codes by skipping those records entirely.
E. Create a view that performs a simple 'JOIN' between 'TRANSACTIONS' and 'ECONOMIC DATA' on 'TRANSACTIONS.TRANSACTION DATE = ECONOMIC_DATDATE and 'TRANSACTIONS.CUSTOMER_ZIP = ECONOMIC_DATA.ZIP_CODE. This approach ignores missing or malformed zip codes.

Answer: B

Explanation:
Option C provides the most robust and efficient solution. Using 'LEFT OUTER JOIN' ensures that all transactions are included in the view, even if there is no matching economic data. 'TRY TO NUMBER handles malformed zip codes gracefully by converting valid zip codes to numbers and returning NULL for invalid ones, preventing errors. 'NVL' replaces NULL zip codes (either originally missing or resulting from TRY_TO_NUMBER) with a default value, allowing the join to proceed using a fallback. Adding the 'ENRICHMENT_SUCCESS' flag provides transparency about which records were enriched using the default zip code, enabling users to assess the reliability of the enriched data. Option A is inadequate because it ignores missing or malformed zip codes, leading to data loss. Option B is inefficient and not scalable due to row-by-row processing. Option D discards records with missing or malformed zip codes, resulting in significant data loss. Option E does not specifically handle data quality issues related to missing or malformed zip codes. Further the use of Tasks and materialized views, while increasing performance, doesn't necessarily address the issue of data quality.

NEW QUESTION # 59
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