Determine the probability of t in A B C and decide for each

Determine the probability of t in A., B. & C and decide for each whether you would accept or reject H0

a.H0: = 11, H1: > 11,T = 12.2, standard deviation = 8.9, n = 40 Thus the t value is: 12.2-11 = 0.82 89-40 b. H0: -11, H1: > 11,T = 14.0, standard deviation = 8.9, n = 40 Thus the t value is: 14.0-11 8.9/40 89 / 40 =2.13 c. H0: = 11, H1: > 11,T= 13.7, standard deviations 8.9, n = 40 Thus the t value is: 13.7-11 =1.91 8.9/40

Solution

a)
to =0.853
| to | =0.853
Critical Value
The Value of |t | with n-1 = 39 d.f is 1.304
We got |to| =0.853 & | t | =1.304
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho

b)
to =2.132
| to | =2.132
Critical Value
The Value of |t | with n-1 = 39 d.f is 1.304
We got |to| =2.132 & | t | =1.304
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho

c)
to =1.919
| to | =1.919
Critical Value
The Value of |t | with n-1 = 39 d.f is 1.304
We got |to| =1.919 & | t | =1.304
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho