A genetic experiment involving peas yielded one sample of of

A genetic experiment involving peas yielded one sample of offspring consisting of 428 green peas and 133 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.

A. Upper H 0 : p equals 0.26 Upper H 1 : p less than 0.26

B. Upper H 0 : p not equals 0.26 Upper H 1 : p equals 0.26

C. Upper H 0 : p equals 0.26 Upper H 1 : p not equals 0.26

D. Upper H 0 : p equals 0.26 Upper H 1 : p greater than 0.26

E. Upper H 0 : p not equals 0.26 Upper H 1 : p less than 0.26

F. Upper H 0 : p not equals 0.26 Upper H 1 : p greater than 0.26

Solution

Formulating the null and alternatuve hypotheses,          
          
Ho:   p   =   0.26
Ha:   p   =/=   0.26 [OPTION C]

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As we see, the hypothesized po =   0.26      
Getting the point estimate of p, p^,          
          
p^ = x / n =    0.310747664      
          
Getting the standard error of p^, sp,          
          
sp = sqrt[po (1 - po)/n] =    0.021202186      
          
Getting the z statistic,          
          
z = (p^ - po)/sp =    2.393510861      
          
As this is a    2   tailed test, then, getting the p value,  
          
p =    0.016687987      

significance level =    0.01      

Comparing p < 0.01, we   FAIL TO REJECT THE NULL HYPOTHESIS.      

Thus, there is no significant evidence that the proportion of yellow peas is not 0.26. [CONCLUSION]