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Sagot :
First, let's identify the hypotheses for this test.
### Hypotheses
- Null Hypothesis ([tex]\( H_0 \)[/tex]): [tex]\( p = 0.25 \)[/tex]
- Alternative Hypothesis ([tex]\( H_1 \)[/tex]): [tex]\( p > 0.25 \)[/tex]
This indicates we are conducting a one-tailed test.
### Sample Data
- Total sample size ([tex]\( n \)[/tex]) = 557
- Number of yellow peas ([tex]\( k \)[/tex]) = 153
### Claimed Proportion
- Claimed proportion ([tex]\( p \)[/tex]) = 0.25
### Observed Proportion
- Observed proportion ([tex]\( \hat{p} \)[/tex]) = [tex]\( \frac{153}{557} \approx 0.2747 \)[/tex]
### Significance Level
- Significance level ([tex]\( \alpha \)[/tex]) = 0.05
### Standard Error Calculation
The standard error ([tex]\( SE \)[/tex]) is calculated using the formula:
[tex]\[ SE = \sqrt{\frac{p (1 - p)}{n}} \][/tex]
Given [tex]\( p = 0.25 \)[/tex] and [tex]\( n = 557 \)[/tex]:
[tex]\[ SE = \sqrt{\frac{0.25 \cdot (1 - 0.25)}{557}} \approx 0.0183 \][/tex]
### Test Statistic (z)
The test statistic (z-score) is calculated by:
[tex]\[ z = \frac{\hat{p} - p}{SE} \][/tex]
Where [tex]\( \hat{p} = 0.2747 \)[/tex], [tex]\( p = 0.25 \)[/tex], and [tex]\( SE \approx 0.0183 \)[/tex]:
[tex]\[ z = \frac{0.2747 - 0.25}{0.0183} \approx 1.35 \][/tex]
### P-value
Using the z-score, we can find the P-value for the one-tailed test:
[tex]\[ \text{P-value} \approx 0.0892 \][/tex]
### Conclusion About the Null Hypothesis
To determine whether to reject the null hypothesis, compare the P-value to the significance level:
- If P-value < [tex]\(\alpha\)[/tex], reject [tex]\(H_0\)[/tex].
- If P-value [tex]\(\geq\)[/tex] [tex]\(\alpha\)[/tex], do not reject [tex]\(H_0\)[/tex].
In this case:
[tex]\[ 0.0892 > 0.05 \][/tex]
We do not reject the null hypothesis.
### Final Conclusion
Given the evidence, we do not have sufficient evidence to reject the claim that 25% of the offspring peas will be yellow. Thus, there isn't enough statistical support to conclude that the proportion of yellow peas is greater than 25%.
### Answer Summary
- Hypotheses:
[tex]\[ H_0: p = 0.25 \][/tex]
[tex]\[ H_1: p > 0.25 \][/tex]
- Test Statistic:
[tex]\[ z = 1.35 \][/tex]
- P-value:
[tex]\[ \approx 0.0892 \][/tex]
- Conclusion about the null hypothesis:
Do not reject [tex]\( H_0 \)[/tex].
- Final Conclusion:
There is not enough evidence to support the claim that more than 25% of the offspring peas will be yellow.
### Hypotheses
- Null Hypothesis ([tex]\( H_0 \)[/tex]): [tex]\( p = 0.25 \)[/tex]
- Alternative Hypothesis ([tex]\( H_1 \)[/tex]): [tex]\( p > 0.25 \)[/tex]
This indicates we are conducting a one-tailed test.
### Sample Data
- Total sample size ([tex]\( n \)[/tex]) = 557
- Number of yellow peas ([tex]\( k \)[/tex]) = 153
### Claimed Proportion
- Claimed proportion ([tex]\( p \)[/tex]) = 0.25
### Observed Proportion
- Observed proportion ([tex]\( \hat{p} \)[/tex]) = [tex]\( \frac{153}{557} \approx 0.2747 \)[/tex]
### Significance Level
- Significance level ([tex]\( \alpha \)[/tex]) = 0.05
### Standard Error Calculation
The standard error ([tex]\( SE \)[/tex]) is calculated using the formula:
[tex]\[ SE = \sqrt{\frac{p (1 - p)}{n}} \][/tex]
Given [tex]\( p = 0.25 \)[/tex] and [tex]\( n = 557 \)[/tex]:
[tex]\[ SE = \sqrt{\frac{0.25 \cdot (1 - 0.25)}{557}} \approx 0.0183 \][/tex]
### Test Statistic (z)
The test statistic (z-score) is calculated by:
[tex]\[ z = \frac{\hat{p} - p}{SE} \][/tex]
Where [tex]\( \hat{p} = 0.2747 \)[/tex], [tex]\( p = 0.25 \)[/tex], and [tex]\( SE \approx 0.0183 \)[/tex]:
[tex]\[ z = \frac{0.2747 - 0.25}{0.0183} \approx 1.35 \][/tex]
### P-value
Using the z-score, we can find the P-value for the one-tailed test:
[tex]\[ \text{P-value} \approx 0.0892 \][/tex]
### Conclusion About the Null Hypothesis
To determine whether to reject the null hypothesis, compare the P-value to the significance level:
- If P-value < [tex]\(\alpha\)[/tex], reject [tex]\(H_0\)[/tex].
- If P-value [tex]\(\geq\)[/tex] [tex]\(\alpha\)[/tex], do not reject [tex]\(H_0\)[/tex].
In this case:
[tex]\[ 0.0892 > 0.05 \][/tex]
We do not reject the null hypothesis.
### Final Conclusion
Given the evidence, we do not have sufficient evidence to reject the claim that 25% of the offspring peas will be yellow. Thus, there isn't enough statistical support to conclude that the proportion of yellow peas is greater than 25%.
### Answer Summary
- Hypotheses:
[tex]\[ H_0: p = 0.25 \][/tex]
[tex]\[ H_1: p > 0.25 \][/tex]
- Test Statistic:
[tex]\[ z = 1.35 \][/tex]
- P-value:
[tex]\[ \approx 0.0892 \][/tex]
- Conclusion about the null hypothesis:
Do not reject [tex]\( H_0 \)[/tex].
- Final Conclusion:
There is not enough evidence to support the claim that more than 25% of the offspring peas will be yellow.
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