A genetic experiment involving peas yielded one sample of of

A genetic experiment involving peas yielded one sample of offspring consisting of 402 green peas and 145 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 25% 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.

Solution

Set Up Hypothesis
Null, H0:P=0.25
Alternate, H1: P!=0.25
Test Statistic
No. Of Success chances Observed (x)=145
Number of objects in a sample provided(n)=547
No. Of Success Rate ( P )= x/n = 0.2651
Success Probability ( Po )=0.25
Failure Probability ( Qo) = 0.75
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.26508-0.25/(Sqrt(0.1875)/547)
Zo =0.8146
| Zo | =0.8146
Critical Value
The Value of |Z | at LOS 0.01% is 2.58
We got |Zo| =0.815 & | Z | =2.58
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != 0.81463 ) = 0.41528
Hence Value of P0.01 < 0.4153,Here We Do not Reject Ho

we have evidence to indicate that under the same circumstances, 25% of offspring peas will be yellow