According to the Washington Post nearly 45 of all Americans

According to the Washington Post, nearly 45% of all Americans are born with brown eyes, although their eyes don’t necessarily stay brown. A random sample of 80 adults found 32 with brown eyes. Is there sufficient evidence at the 0.01 level to indicate that the proportion of brown-eyed adults differs from the proportion of Americans who are born with brown eyes

Solution

Set Up Hypothesis
Null, brown-eyed adults differs from Americans H0:P=0.45
Alternate, brown-eyed adults same from Americans H1: P!=0.45
Test Statistic
No. Of Success chances Observed (x)=32
Number of objects in a sample provided(n)=80
No. Of Success Rate ( P )= x/n = 0.4
Success Probability ( Po )=0.45
Failure Probability ( Qo) = 0.55
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.4-0.45/(Sqrt(0.2475)/80)
Zo =-0.9
| Zo | =0.9
Critical Value
The Value of |Z | at LOS 0.01% is 2.58
We got |Zo| =0.899 & | 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 != -0.89893 ) = 0.36869
Hence Value of P0.01 < 0.3687,Here We Do not Reject Ho

brown-eyed adults differs from Americans