A local newspaper recendy published an article stating that

A local newspaper recendy published an article stating that 20% of students at UNL are from out of state. All of your friends live in N ebraska, so you think the percentage is much lower than 20%. You randomly sample 200 students and only 32 of them are from out of state. Perform a hypothesis test to test your claim. Use a significance level of a = 0.10. You must show all relevant steps and all of your work. Suppose the alternative hypothesis was two-sided. What is the result of the test now? Fully interpret your results.

Solution

a)
Set Up Hypothesis
Under The Null Hypothesis H0:P=0.2
Under The Alternate Hypothesis H1: P<0.2
Test Statistic
No. Of Success chances Observed (x)=32
Number of objects in a sample provided(n)=200
No. Of Success Rate ( P )= x/n = 0.16
Success Probability   ( Po )=0.2
Failure Probability ( Qo) = 0.8
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.16-0.2/(Sqrt(0.16)/200)
Zo =-1.41
| Zo | =1.41
Critical Value
The Value of |Z ?| at LOS 0.1% is 1.28
We got |Zo| =1.414 & | Z ? | =1.28
Make Decision
Hence Value of | Zo | > | Z ?| and Here we Reject Ho
P-Value: Left Tail -Ha : ( P < -1.41421 ) = 0.07865
Hence Value of P0.1 > 0.07865,Here we Reject Ho

We have the evidence of less than 0.20

b)
Set Up Hypothesis
Under The Null Hypothesis H0:P=0.2
Under The Alternate Hypothesis H1: P!=0.2
Test Statistic
No. Of Success chances Observed (x)=32
Number of objects in a sample provided(n)=200
No. Of Success Rate ( P )= x/n = 0.16
Success Probability   ( Po )=0.2
Failure Probability ( Qo) = 0.8
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.16-0.2/(Sqrt(0.16)/200)
Zo =-1.41
| Zo | =1.41
Critical Value
The Value of |Z ?| at LOS 0.1% is 1.64
We got |Zo| =1.414 & | Z ? | =1.64
Make Decision
Hence Value of |Zo | < | Z ? | and Here we Do not Reject Ho