Question

# A genetic experiment involving peas yielded one sample of offspring consisting of 442 green peas and...

A genetic experiment involving peas yielded one sample of offspring consisting of 442 green peas and 123 yellow peas. Use a 0.01 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

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H0 : p = 0.26

Ha : p 0.26 = x / n = 123 / 442 = 0.2783

P0 = 0.26

1 - P0 = 0.74

Test statistic = z

= - P0 / [ P0 * (1 - P0 ) / n]

= 0.2783 - 0.26 / [ (0.26 * 0.74) / 442]

= 0.876

P(z > 0.876) = 1 - P(z < 0.876) = 0.1905

P-value = 0.381 = 0.01

P-value > Fail to reject the null hypothesis .

There is not sufficient evidence that the  0.01 significance level to test the claim that under the same​ circumstances, 26​% of offspring peas will be yellow .

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