To solve this problem, we need to ensure that the data satisfies the condition of statistical independence. This means that the probability of a customer rating the product as "Good" or "Bad" is independent of whether they are a new or old customer.

For two variables to be independent, the probability of a customer being from one group and giving a particular rating is the product of the individual probabilities. In other words, the expected count for each combination of categories (New/Old and Good/Bad) can be found using:

$\text{Expected count} = \frac{\text{(row total)} \times \text{(column total)}}{\text{(grand total)}}$

Step-by-step process:

1. Calculate the row and column totals:

   • New customers (Good + Bad): $100 + 200 = 300$
   • Old customers (Good + Bad): $175 + x$ (where $x$ is the missing number, which we need to find)
   • Good ratings (New + Old): $100 + 175 = 275$
   • Bad ratings (New + Old): $200 + x$

   The grand total (sum of all the cells) is:

   $300 + (175 + x) = 475 + x$

2. Set up the equation for statistical independence:

   For statistical independence, the expected frequency of old customers rating the product as "Bad" should equal the observed frequency of old customers giving a "Bad" rating. The expected count is:

   $E_{\text{Old, Bad}} = \frac{\text{(Old Total)} \times \text{(Bad Total)}}{\text{Grand Total}} = \frac{(175 + x) \times (200 + x)}{475 + x}$

   But we know that the observed count for Old, Bad is $x$. So, set the equation as:

   $\frac{(175 + x) \times (200 + x)}{475 + x} = x$

3. Solve for $x$:

   Expanding both sides:

   $(175 + x)(200 + x) = x(475 + x)$ $35000 + 375x + 200x + x^2 = 475x + x^2$ $35000 + 575x = 475x$ $35000 = 475x - 575x$ $35000 = -100x$ $x = -\frac{35000}{100} = -350$

   So, the missing value in the "Old, Bad" cell is 350.

Final Answer:

The value that must have been in the empty cell in order for statistical independence to hold is 350.

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Would you like further details on the solution, or do you have any questions about the process?

Here are some follow-up questions you might find interesting:

1. How do you calculate expected frequencies in a contingency table?
2. What is the difference between independent and dependent variables in a contingency table?
3. Can you apply this method to more complex datasets?
4. How does statistical independence relate to chi-square tests?
5. What is the chi-square test formula for testing independence?

Tip: Always check that the row and column totals are correctly calculated when solving for missing values in contingency tables.