A scientist conducted a hybridization experiment using peas with green pods and

Question

A scientist conducted a hybridization experiment using peas with green pods and yellow pods. He crossed peas in such a way that 25% (or 140) of the 560 offspring peas were expected to have yellow pods. Instead of getting 140 peas with yellow pods, he obtained 146. Assume that the rate of 25% is correct. Find the probability that among the 560 offspring peas, exactly 146 have yellow pods. Find the probability that among the 560 offspring peas, at least 146 have yellow pods. Which result is useful for determining whether the claimed rate of 25% is incorrect? (Part (a) or part (b)?) Is there strong evidence to suggest that the rate of 25% is incorrect? The probability that exactly 146 have yellow pods is (Round to four decimal places as needed.) The probability that at least 146 have yellow pods is. (Round to four decimal places as needed.) is useful for determining whether the claimed rate is incorrect. Is there strong evidence to suggest that the rate of 25% is incorrect? No Yes

Explanation / Answer

here n=560 and p=0.25

here we use binomial distribution with n=560 and p=0.25

and Binomial distribution ,P(X=r)=nCrpr(1-p)n-r  

(a)P(X=146)=0.0324 ( using ms-excel command =BINOMDIST(146,560,0.25,0))

(b)P(X at least 146)=P(X>146)=1-P(X<=146)=1-0.7386=0.2614

(c) part(b)

(d) here we use z-test and z=(0.2614-0.25)/sqrt(0.25*(1-0.75)/560)=1.079

the critical z(0.05)=1.96 is less than the calculated z=1.079, which means p=0.25 at 5% level of significance

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