Question

# a genetic experience involving peas yielded one sample of Offspring consisting of 408 green peas and...

a genetic experience involving peas yielded one sample of Offspring consisting of 408 green peas and 162 yellow pill use a .05 significance level to test the claim that under the same circumstances 27% of Offspring peas will be yellow identify the null hypothesis and alternative hypothesis test statistics 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

Solution:

A genetic experiment involving peas yielded one sample of offspring consisting of 408 green peas and 162 yellow peas.
The claim is that 27% of offspring peas will be yellow.
The null and alternative hypotheses are,
H0: p = 0.27
H1: p ≠ 0.27

The proportion of yellow peas in sample is,
p̂ = X/n = 162/408+ 162 = 162/570 = 0.284

The value of z-statistic is obtained as shown below:

z = (p̂ - p)/ sqrt(p(1-p)/n)
= (0.284 - 0.27)/sqrt(0.27(1-0.27)/570)
= 0.014/0.019 = 0.737
The value of test statistic is 0.737.

The P-value is obtained as shown below:
The value of z statistic is 0.737
P-value = P(Z < 0.737)
= 0.4611
as p value is greater than 0.05 level we can not eject null hypothesis

we do not have sufficient evidence to reject the claim that under the same​ circumstances, 24​% of offspring peas will be yellow.

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