Consider the Minitab output shown below.
Two sample T-Test and CI
|
Sample
|
N
|
Mean
|
StDev
|
SE Mean
|
|
1
|
15
|
50.20
|
1.75
|
0.45
|
|
2
|
15
|
51.98
|
2.15
|
0.56
|
Difference = nu (1) - mu (2)
Estimate for difference: 7
958 CI for difference: (-3.246, -0.314)
T-Test of difference = 0 (vs not =): T-Value =-2.49 P-Value = 0.019 DF = 7 Both use Pooled StDev = 1.9602
(a) Fill in the missing values.
(b) Can the null hypothesis be rejected at a the 0.05 level? Why?
(c) Use the output and the t-table to find a 99% CI on the difference in means.
(d) Suppose that the alternative hypothesis was H1: μ1 = μ2 versus H1: μ1 > μ2. What is the P-value? What conclusions would you draw?