8. Ho
There is no sig diff
between the sample
Critical value of t
(from table)
lower
Don’t reject
H0 Reject H0
higher
9. • Degree of freedom = n1+n2-2
• 16+16-2
• = 30
• For p= 0.05
• t = 2.0523
Calculated value of t = 6.5
Which is higher than table value
HENCE NULL HYPOTHESIS IS REJECTED
There is difference between yields of two fields