In: Statistics and Probability
Ho : µ = 10
Ha : µ < 10
(Left tail test)
Level of Significance , α =
0.01
sample std dev , s = 2.2000
Sample Size , n = 70
Sample Mean, x̅ = 9.5000
degree of freedom= DF=n-1= 69
Standard Error , SE = s/√n = 2.2000 / √
70 = 0.2630
t-test statistic= (x̅ - µ )/SE = ( 9.500
- 10 ) / 0.2630
= -1.90
p-Value = 0.0307 [Excel formula
=t.dist(t-stat,df) ]
Decision: p-value>α, Do not reject null
hypothesis
There is not sufficient evidence to prove that new system indeed reduced the waiting time .
Please revert back in case of any doubt.
Please upvote. Thanks in advance.