SOLUTION: A genetic experiment involving peas yielded one sample of offspring consisting of 441 green peas and 168 yellow peas. Use a 0.01 significance level to test the claim that under the
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Question 1208909: A genetic experiment involving peas yielded one sample of offspring consisting of 441 green peas and 168 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 26% of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution.
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to conduct a hypothesis test for the proportion of yellow peas:
**1. Hypotheses:**
* **Null Hypothesis (H0):** p = 0.26 (The proportion of yellow peas is 26%.)
* **Alternative Hypothesis (H1):** p ≠ 0.26 (The proportion of yellow peas is *not* 26%.) This is a two-tailed test.
**2. Sample Proportion:**
* Total peas = 441 green + 168 yellow = 609
* Sample proportion of yellow peas (p̂) = 168/609 ≈ 0.2759
**3. Test Statistic (z-score):**
We'll use the normal distribution as an approximation to the binomial. The formula for the test statistic is:
z = (p̂ - p) / sqrt[(p(1-p))/n]
Where:
* p̂ is the sample proportion
* p is the hypothesized proportion under H0
* n is the sample size
z = (0.2759 - 0.26) / sqrt[(0.26 * 0.74)/609]
z = 0.0159 / sqrt(0.000317)
z ≈ 2.82
**4. P-value:**
Since this is a two-tailed test, we need to find the area in *both* tails of the standard normal distribution that is beyond our z-score of 2.82. Using a z-table or calculator:
P(z > 2.82) ≈ 0.0024
P(z < -2.82) ≈ 0.0024
P-value = 2 * 0.0024 = 0.0048
**5. Conclusion about the Null Hypothesis:**
* Significance level (α) = 0.01
Since the P-value (0.0048) is *less than* the significance level (0.01), we *reject* the null hypothesis.
**6. Final Conclusion:**
There is sufficient evidence at the 0.01 significance level to reject the claim that 26% of offspring peas will be yellow. The sample data suggests that the true proportion of yellow peas is different from 26%.
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