Question

A genetic experiment involving peas yielded one sample of offspring consisting of 412 green peas and 151 yellow peas. Use a 0.01 significance level to test the claim that under the same​circumstances, 27​% 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 is the test statistic?

What is the P- value?

What is the conclusion about the null hypotheses? Fail to reject or reject and why?

What is the final conclusion?

Answer 1

Let n = number of offspring = 412 + 151 = 563

X : Number of offspring consisting Yellow peas = 151

p = propertion of offspring consisting yellow peas = X / n = 151 / 563 = 0.2682

Let P be the Population propertion of offspring consisting yellow peas.

X ~ B ( n, P)

E(X) =nP and V(X) = nPQ where Q =1-P

For large n binomial distribution tends to Normal distribution

Sample propertion is an unbiased estimator for population propertion.

E(p) = P and V(p) = PQ /n.

Since X and p is asymptotically normal for large n. The normal test for the propertion of success becomes

We have to testing the hypothesis that

whether or not propertion of offspring consisting yellow peas is 27%

i.e. Null Hypothesis : Ho : P =0.27

against

Alternative Hypothesis : H1 : P # 0.27 ( Two-tailed test)

we use one sample Z-test for population propertion.

Under Ho the value of test statistic is

E(p) = P = 0.27 and V(p) = PQ /n = 0.27 *0.73 /663 = 0.0003500 S.D.(p) = 0.01871

Value of test statistic Z = -0.0962

Since the test is two-tailed, p-value is obtained by

p-value = 2 * P ( Z > 0.0962)

= 2 * 0.4617

= 0.9234

p-value= 0.9234

alpha = level of significance = 0.01

Decision : Since p-value > level of significance , the test is not significant. We fail to reject the null hypothesis.

Conclusion : There is no significant evidence to reject the claim that 27% offspring will be Yellow.