Question

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

404404

green peas and

158158

yellow peas. Use a

0.050.05

significance level to test the claim that under the same? circumstances,

2424?%

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.

?P-valueequals=

Round 4 decimal places

Answer 1

total peas=404+158=562
p(hat)=   158/562=0.2811      
n=   562              
          
            
.......      two tail test      
alpha=   0.05              
          
Test Statistics                  
Z=   (p(hat)-P)/sqrt(p*q/n)
=(0.2811-0.24)/sqrt(0.24*0.76/562)
=2.2835  
P-value=1-P(Z<2.2835)=   0.011199   (this p-value is for one tail to get p-value for two tail we need to multiply it by 2.)          
rounded p-value=2*0.011199=0.0224
              
Decision making                   
P-value ? Alpha   ......... Reject H0              
0.0224<0.05                  
              
there is enough evidence to support the claim that offspring peas will be different from yellow peas.

here alternate hypothesis become true as we rejected null hypothesis.