A genetic experiment involving peas yielded one sample of offspring consisting of 450 green peas and 175 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 25% of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesi: 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. What are the null and alternative hypotheses? O A. Ho: p= 0.25 H,:p>0.25 о В. Но: р3D0.25 H,:p#0.25 Ос. Но: р#0.25 O D. Ho: p+0.25 H1: p<0.25 H1:p>0.25 ОЕ. Но: р#0.25 H1:p=0.25 O F. Ho:p= 0.25 H1: p<0.25 What is the test statistic? (Round to two decimal places as needed.) What is the P-value? P-value = (Round to four decimal places as needed.) What is the conclusion about the null hypothesis? O A. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a. B. Reject the null hypothesis because the P-value is greater than the significance level, a. OC. Fail to reject the null hypothesis because the P-value is greater than the significance level, a. O D. Reject the null hypothesis because the P-value is less than or equal to the significance level, a. What is the final conclusion? O A. There is sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow. O B. There is sufficient evidence to support the claim that less than 25% of offspring peas will be yellow. OC. There is not sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow. D. There is not sufficient evidence to support the claim that less than 25% of offspring peas will be yellow.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

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A genetic experiment involving peas yielded one sample of offspring consisting of 450 green peas and 175 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 25% of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesi:
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.
What are the null and alternative hypotheses?
O A. Ho: p= 0.25
H,:p>0.25
о В. Но: р3D0.25
H,:p#0.25
Ос. Но: р#0.25
O D. Ho: p+0.25
H1: p<0.25
H1:p>0.25
ОЕ. Но: р#0.25
H1:p=0.25
O F. Ho:p= 0.25
H1: p<0.25
What is the test statistic?
(Round to two decimal places as needed.)
What is the P-value?
P-value =
(Round to four decimal places as needed.)
What is the conclusion about the null hypothesis?
O A. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a.
B. Reject the null hypothesis because the P-value is greater than the significance level, a.
OC. Fail to reject the null hypothesis because the P-value is greater than the significance level, a.
O D. Reject the null hypothesis because the P-value is less than or equal to the significance level, a.
What is the final conclusion?
O A. There is sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow.
O B. There is sufficient evidence to support the claim that less than 25% of offspring peas will be yellow.
OC. There is not sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow.
D. There is not sufficient evidence to support the claim that less than 25% of offspring peas will be yellow.

Expert Solution

Step 1

Let p is defined as the proportion of offspring peas will be yellow.

The claim of the test is 25% of offspring peas will be yellow.

The hypothesis is,

Null hypothesis:

$H_{0} : p = 0.25$

Alternative hypothesis:

$H_{1} : p \neq 0.25$

The proportion of offspring peas will be yellow is,

$\begin{array}{rcl} \hat{p} & = & \frac{175}{450 + 175} \\ = & \frac{175}{625} \\ = & 0.28 \end{array}$

The test statistic is,

$\begin{array}{rcl} z & = & \frac{\hat{p} - p}{\sqrt{\frac{p q}{n}}} \\ = & \frac{0.28 - 0.25}{\sqrt{\frac{0.25 (1 - 0.25)}{625}}} \\ = & \frac{0.03}{0.0173} \\ = & 1.73 \end{array}$

Thus, the test statistic is 1.73.

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