A genetic experiment involving peas yielded one sample of offspring consisting of 438 green peas and 173 yellow peas. Use a 0.05 significance level to test the claim that under the same circumstances, 26% 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 hypothesis?
What is the test statistic?
What is the P-value? What is the conclusion about the null hypothesis?
What is the final conclusion?
Answer:
Given,
sample proportion p^ = x/n = 173/438 = 0.395
Ho : p = 0.26
Ha : p != 0.26
test statistic z = (p^ - p)/sqrt(pq/n)
substitute values
= (0.395 - 0.26)/sqrt(0.26(1-0.26)/438)
= 6.44
P value = 0
Here we observe that, p value < alpha, so we reject Ho.
So there is sufficient evidence to support the claim.
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