A genetic experiment involving peas yielded one sample of offspring consisting o

Question

A genetic experiment involving peas yielded one sample of offspring consisting of 407 green peas and 130 yellow peas. Use a 0.05 significance level to test the claim that under the same circumstances, 23%

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 is the test statistic?

Z= ­­__ Round to two decimal places as needed

What is the P-value?

­­­___ Round to four decimal places as needed

What is the conclusion about the null hypothesis?

What is the final conclusion?

Explanation / Answer

Formulating the null and alternatuve hypotheses,          
          
Ho:   p   =   0.23
Ha:   p   =/=   0.23
As we see, the hypothesized po =   0.23      
Getting the point estimate of p, p^,          
          
p^ = x / n = 130/(130+407) =    0.242085661      
          
Getting the standard error of p^, sp,          
          
sp = sqrt[po (1 - po)/n] =    0.018160263      
          
Getting the z statistic,          
          
z = (p^ - po)/sp =    0.665500327   [ANSWER, TEST STATISTIC]

************************************  
          
As this is a    2   tailed test, then, getting the p value,  
          
p =    0.505730534   [ANSWER, P VALUE]

****************************************

As this is a large P value,   FAIL TO REJECT THE NULL HYPOTHESIS.   [DECISION]

Hence, there is no significant evidence that the true population proportion of yellow peas is not 0.23. [CONCLUSION]  

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