In a sample of 47 adults selected randomly from 1 town it is

In a sample of 47 adults selected randomly from 1 town, it is found that 9 of them have been exposed to a particular strain of the flu. Use a 0.05 confidence level to test the claim that the proportion of all adults in the town that have been exposed to this strain of the flu is 10%. Use the p-value method.
A) Null hypothesis: ?
B) Alternative hypothesis: ?
C) p hat: ?
D) test statistic: ?
E) p-value: ?
F) initial conclusion: ?
G) conlsuion: ?

Solution

Set Up Hypothesis
Null, H0:P=0.1
Alternate, H1: P!=0.1
Test Statistic
No. Of Success chances Observed (x)=9
Number of objects in a sample provided(n)=47
No. Of Success Rate ( P )= x/n = 0.1915
Success Probability ( Po )=0.1
Failure Probability ( Qo) = 0.9
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.19149-0.1/(Sqrt(0.09)/47)
Zo =2.0907
| Zo | =2.0907
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =2.091 & | Z | =1.96
Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != 2.09073 ) = 0.03655
Hence Value of P0.05 > 0.0366,Here we Reject Ho

We don\'t have evidence to indicate that the proportion of all adults in the town that have been exposed to this strain of the flu is 10%