The US bureau of Labor reported that a person between the ag

The US bureau of Labor reported that a person between the ages 18 -34 ha an average of 9.2 jobs. A random sample of 10 workers were asked how many places they work. The results were :   

5, 6, 12, 8, 1, 3, 3, 3, 7, 11 . At a= 0.10 can it be concluded that the mean is 9.2. Use the P- value method. Assume the population is normally distributed

Solution

Set Up Hypothesis
Null Hypothesis H0: U=9.2
Alternate Hypothesis H1: U!=9.2
Test Statistic
Population Mean(U)=9.2
Sample X(Mean)=5.9
Standard Deviation(S.D)=3.6347
Number (n)=10
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =5.9-9.2/(3.6347/Sqrt(9))
to =-2.871
| to | =2.871
Critical Value
The Value of |t a| with n-1 = 9 d.f is 1.833
We got |to| =2.871 & | t a | =1.833
Make Decision
Hence Value of | to | > | t a| and Here we Reject Ho
P-Value :Two Tailed ( double the one tail ) -Ha : ( P != -2.8711 ) = 0.0184
Hence Value of P0.1 > 0.0184,Here we Reject Ho

We have evidence to claim Mean is diffrent for 9.2