In this file, I have listed some examples of output from Minitab. You should go
through and make sure you can understand what is being tested. I may not cover
this in class but if you have questions, please don’t hesitate to ask!
1. This is a 95% confidence interval for µ with σ known.
The assumed sigma = 2.00
Variable
C1
N
6
Mean
1.538
StDev SE Mean
1.914
0.816 (
95.0 % CI
-0.062, 3.139)
2. This is a one sample z-test. Notice that σ is known to be 2.
Test of mu = 1.500 vs mu not = 1.500
The assumed sigma = 2.00
Variable
C1
N
6
Mean
1.538
StDev
1.914
SE Mean
0.816
Z
0.05
P
0.96
(a) H : 0 : µ = 1.5 vs. H1 : µ 6= 1.5
√
(b) n = 6, X̄ = 1.538, σ = 2, σ/ n = 0.816
X̄ −µ
√ 0 = 0.0465. They rounded up to 0.05.
(c) Test Statistic (TS) =
σ/ n
(d) The p-value is 0.96.
3. This is a one sample t-test.
Test of mu = 1.500 vs mu not = 1.500
Variable
C1
N
6
Mean
1.538
StDev
1.914
(a) H : 0 : µ = 1.5 vs. H1 : µ 6= 1.5
SE Mean
0.781
T
0.05
√
(b) n = 6, X̄ = 1.538, s = 1.914, s/ n = 0.781
X̄ −µ
√ 0 = 0.0486. They rounded up to 0.05.
(c) Test Statistic (TS) =
s/ n
(d) The p-value is 0.96.
1
P
0.96
4. This is a one sample t-test. Notice how they don’t make H0 and H1 opposites.
H1 determines what you’re testing!
Test of mu = 1.500 vs mu < 1.500
Variable
C1
N
6
Mean
1.538
StDev
1.914
SE Mean
0.781
T
0.05
P
0.52
(a) H : 0 : µ = 1.5(≥) vs. H1 : µ < 1.5
√
(b) n = 6, X̄ = 1.538, s = 1.914, s/ n = 0.781
X̄ −µ
√ 0 = 0.0486. They rounded up to 0.05.
(c) Test Statistic (TS) =
s/ n
(d) The p-value is 0.52.
5. This is a one sample test of proportion.
Test of p = 0.2 vs p > 0.2
Sample
1
X
1024
N
4860
Sample p
95.0 % CI
Z-Value P-Value
0.210700 (0.199234, 0.222165)
1.86
0.031
(a) H : 0 : p = 0.2(≤) vs. H1 : p > 0.2. p0 = 0.2
(b) n = 4660, p̂ = 1024/4860 = 0.2107. A 95% confidence interval for p is given.
p̂−p0
(c) Test Statistic (TS) = r
= 1.86.
p0 (1−p0 )
n
(d) The p-value is 0.031.
6. This is a test of 2 proportions.
Sample
1
2
X
101
56
N Sample p
10239 0.009864
9877 0.005670
Estimate for p(1) - p(2): 0.00419451
95% CI for p(1) - p(2): (0.00177439, 0.00661463)
Test for p(1) - p(2) = 0 (vs not = 0): Z = 3.40 P-Value = 0.001
(a) H : 0 : p1 = p2 (≤) vs. H1 : p1 6= p2 .
(b) Test Statistic (TS) = 3.40
(c) The p-value is 0.001.
7. This is a two sample test of means. A two-sample t-test was used.
Two sample T for C1 vs C2
C1
C2
N
21
8
Mean
13.29
24.00
StDev
3.74
1.69
SE Mean
0.82
0.60
95% CI for mu C1 - mu C2: ( -12.80, -8.63)
T-Test mu C1 = mu C2 (vs not =): T = -10.59 P = 0.0000 DF = 25
(a) H : 0 : µ1 = µ2 vs. H1 : µ1 6= µ2 .
(b) They give the summary information: sample sizes, sample means, sample
standard deviations, sample standard errors.
X̄1 −X̄2 √ 13.29−24
(c) Test Statistic (TS) = v
= -10.59. (I get -10.54
=
u
2
u s2
0.822 +0.602
s
t 1
2
n1 + n2
when I use this formula. I suspect they rounded the standard errors above.)
(d) The p-value is reported as 0.0000.