The US Department of Education reports that 46 of fulltime c

The US Department of Education reports that 46% of full-time college students are also employed. A recent survey of 60 full time students at a university found that 29 of those students were employed.

a. Use the 5-step p-value approach to hypothesis testing, and a 0.05 level of significance to determine whether the proportion of full-time students is different from the national norm of 0.46

b. Assume they found 36 of the 60 were employed and repeat (a). Are the conclusions the same?

Solution

a)
Set Up Hypothesis
Null, both are same H0:P=0.46
Alternate, It is diffrent H1: P!=0.46
Test Statistic
No. Of Success chances Observed (x)=29
Number of objects in a sample provided(n)=60
No. Of Success Rate ( P )= x/n = 0.4833
Success Probability ( Po )=0.46
Failure Probability ( Qo) = 0.54
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.48333-0.46/(Sqrt(0.2484)/60)
Zo =0.3626
| Zo | =0.3626
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =0.363 & | Z | =1.96
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != 0.36264 ) = 0.71687
Hence Value of P0.05 < 0.7169,Here We Do not Reject Ho

we conclude that both are same

b)
( P )= x/n = 36/60 = 6/10 = 0.6
( Po )= 0.46
( Qo) = 0.54
Zo=0.6-0.46/(Sqrt(0.2484)/60)
Zo =2.1758
| Zo | =2.1758
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =2.176 & | Z | =1.96
Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho

we conclude that both are diffrent