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A genetic experiment involving peas yielded one sample of offspring consisting of 404 green peas and 153 yellow peas. Use a 0.05 significance level to test the claim that under the same circumstances, [tex]25\%[/tex] 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.

E.
[tex]\[
\begin{array}{l}
H_0: p = 0.25 \\
H_1: p \neq 0.25
\end{array}
\][/tex]

F.
[tex]\[
H_0: p = 0.25 \\
H_1: p \ \textgreater \ 0.25
\][/tex]

What is the test statistic?

[tex]\[
z = \square
\][/tex]

(Round to two decimal places as needed.)



Answer :

GinnyAnswer
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.