To find the expected frequencies in the blanks of a contingency table under the assumption that the two variables (color vision and trial outcome) are independent, we’ll use the formula for expected frequency in each cell:

$E_{ij} = \frac{( \text{Row Total for Row i}) \times (\text{Column Total for Column j})}{\text{Grand Total}}$

The given table is as follows, with blanks indicated:

Trial Outcome	Hit	Miss	False Alarm	Correct Rejection	Total
Colorblind	77	639.13	22	23	263.36
Not Colorblind	630	269	271	638	1808
Total	707	291	294	708	2000

Step 1: Calculate Missing Expected Frequencies

Using the formula, let’s calculate each expected frequency for the blank cells.

1. Expected number of Colorblind participants with outcome "Miss"
   $E = \frac{(\text{Row Total for Colorblind}) \times (\text{Column Total for Miss})}{\text{Grand Total}} = \frac{263.36 \times 291}{2000} = 38.22$

2. Expected number of Colorblind participants with outcome "Correct Rejection"
   $E = \frac{(263.36) \times (708)}{2000} = 93.01$

3. Expected number of Colorblind participants with outcome "False Alarm"
   $E = \frac{(263.36) \times (294)}{2000} = 38.67$

Summary of Expected Frequencies in Blank Cells

• Miss (Colorblind): $38.22$
• Correct Rejection (Colorblind): $93.01$
• False Alarm (Colorblind): $38.67$

Let me know if you'd like further clarification on any of these steps or if you have any other questions!

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Related Questions

1. How would you calculate the probability of a hit outcome regardless of color vision?
2. If these values were actual outcomes, what would be the next step to analyze independence between variables?
3. What are common reasons to test for independence in contingency tables?
4. How does sample size affect the expected frequencies in contingency tables?
5. What are the limitations of using expected frequencies in assessing independence?

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Tip

When dealing with large contingency tables, breaking down each calculation step can help reduce mistakes and provide a clear path to the final answer.