Question

A genetic experiment involving peas yielded one sample of offspring consisting of 444 green peas and 161 yellow peas. Use a 0.05 significance level to test the claim that under the same circumstances, 25% 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. What are the null and alternative hypotheses? O A. Ho: p= 0.25 H:p#0.25 | 0 C. Ho: p= 0.25 He: p=0.25 O E. Ho: p=0.25 Hy: p<0.25 OB. Ho: p=0.25 Hy:p>0.25 OD. Ho:p#0.25 Hy:p<0.25 OF. Ho:p#0.25 H:p>0.25 What is the test statistic? za (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, c. O B . Fail to reject the null hypothesis because the P-value is greater than the significance level, a. O C. Reject the null hypothesis because the P-value is less than or equal to the significance level, a. O D. Reject the null hypothesis because the P-value is greater than the significance level, a.

Answer 1

Answer)

Null hypothesis Ho : p = 0.25

alternate hypothesis Ha : p not equal to 0.25

N = 444 + 161 = 605

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 151.25

N*(1-p) = 453.75

Both the conditions are met so we can use standard normal z table to estimate the P-Value

Test statistics z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

Observed P = 161/605

Claimed P = 0.25

N = 605

Test statistics z = 0.92

From z table, P(z>0.92) = 0.1788

But our test is two tailed

So, p-value is = 2*.1788 = 0.3576

As the obtained p-value is greater than 0.05 (significance level)

We fail to reject the null hypothesis Ho

Option B

There is sufficient evidence to support the claim that 25% offsprings are yellow