A genetic experiment involving peas yielded one sample of of

A genetic experiment involving peas yielded one sample of offspring consisting of 414 green peas and 166 yellow peas. Use a 0.05 significance level to test the claim that under the samecircumstances, 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.

A.

Upper H 0 : p equals 0.27

Upper H 1 : p less than 0.27

B.

Upper H 0 : p equals 0.27

Upper H 1 : p not equals 0.27

C.

Upper H 0 : p not equals 0.27

Upper H 1 : p equals 0.27

D.

Upper H 0 : p not equals 0.27

Upper H 1 : p less than 0.27

E.

Upper H 0 : p equals 0.27

Upper H 1 : p greater than 0.27

F.

Upper H 0 : p not equals 0.27

Upper H 1 : p greater than 0.27

Solution

Formulating the null and alternatuve hypotheses,          
          
Ho:   p   =   0.27
Ha:   p   =/=   0.27 [ANSWER, B]

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Here,

x = 166

n = 166 + 414 = 580.

As we see, the hypothesized po =   0.27      
Getting the point estimate of p, p^,          
          
p^ = x / n =    0.286206897      
          
Getting the standard error of p^, sp,          
          
sp = sqrt[po (1 - po)/n] =    0.018434413      
          
Getting the z statistic,          
          
z = (p^ - po)/sp =    0.879165313      
          
As this is a    2   tailed test, then, getting the p value,  
          
p =    0.379311649      
significance level =    0.05      

Comparing p > 0.05, we   FAIL TO REJECT THE NULL HYPOTHESIS.      

Thus, there is no significant evidence that the proportion of yellow peas is not 0.27. [CONCLUSION]