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The problem in the image involves a chi-square test for goodness of fit to determine if the observed frequencies of different genetic traits in fruit flies match the expected frequencies based on genetic principles. Here’s a step-by-step breakdown of how to proceed:

### 1. **Null and Alternative Hypotheses**
   - **Null Hypothesis (H₀)**: The observed frequencies fit the expected proportions according to genetic principles.
   - **Alternative Hypothesis (H₁)**: The observed frequencies do not fit the expected proportions.

### 2. **Chi-Square Test Statistic (χ²)**
   - Formula: 
     \[
     \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}
     \]
     Where \( O_i \) is the observed frequency, and \( E_i \) is the expected frequency.

   Based on the image:
   - Red Eye/Normal Wing: Observed = 86, Expected proportion = \( \frac{9}{16} \)
   - Sepia Eye/Normal Wing: Observed = 30, Expected proportion = \( \frac{3}{16} \)
   - Red Eye/Vestigial Wing: Observed = 36, Expected proportion = \( \frac{3}{16} \)
   - Sepia Eye/Vestigial Wing: Observed = 8, Expected proportion = \( \frac{1}{16} \)

   Total = \( 86 + 30 + 36 + 8 = 160 \)

   **Expected Frequencies**:
   - Red Eye/Normal Wing: \( \frac{9}{16} \times 160 = 90 \)
   - Sepia Eye/Normal Wing: \( \frac{3}{16} \times 160 = 30 \)
   - Red Eye/Vestigial Wing: \( \frac{3}{16} \times 160 = 30 \)
   - Sepia Eye/Vestigial Wing: \( \frac{1}{16} \times 160 = 10 \)

   **Now calculate each term for the chi-square formula:**
   \[
   \chi^2 = \frac{(86 - 90)^2}{90} + \frac{(30 - 30)^2}{30} + \frac{(36 - 30)^2}{30} + \frac{(8 - 10)^2}{10}
   \]
   Simplifying:
   \[
   \chi^2 = \frac{16}{90} + 0 + \frac{36}{30} + \frac{4}{10} = 0.1778 + 1.2 + 0.4 = 1.7778
   \]

   The calculated chi-square value \( \chi^2 \approx 1.778 \).

### 3. **P-value**
   Use the degrees of freedom (df). Here, \( df = k - 1 = 4 - 1 = 3 \), where \( k \) is the number of categories.
   
   Based on the chi-square table or a calculator for \( \chi^2 = 1.778 \) and \( df = 3 \), we find the p-value is approximately 0.619.

### 4. **Conclusion**
   - Since the p-value (0.619) is greater than the significance level (0.05), we **fail to reject the null hypothesis**.
   - This means there is **not sufficient evidence** to reject the claim that the observed frequencies match the expected proportions.

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Would you like a more detailed explanation of any part? Here are 5 questions to think further:

1. How do you calculate expected frequencies in a chi-square test?
2. What do the null and alternative hypotheses represent in this context?
3. How does changing the significance level affect the outcome?
4. Why do we use degrees of freedom in the chi-square test?
5. What are the limitations of the chi-square goodness-of-fit test?

**Tip:** In hypothesis testing, always compare the p-value to the significance level to make conclusions about the null hypothesis.

A high school biology class conducted genetics experiments with randomly selected fruit flies, and the results in the table are based on the frequencies they observed. Use a 0.05 significance level to test the claim that the observed frequencies agree with the proportions expected based on genetic principles.

Test if observed genetic frequencies in fruit flies match the expected proportions using a chi-square goodness-of-fit test. A step-by-step breakdown with a significance level of 0.05, suitable for high school students in grades 11-12.

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