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Q1
One-Sample Z: C1, C2, C3, C4, C5, C6, C7, C8, ...
Test of μ = 50 vs ≠ 50
The assumed standard deviation = 10
Variable
N
Mean
StDev
SE Mean
95% CI
Z
P
C1
16
47.21
11.67
2.50
(42.31, 52.11)
-1.12
0.264
C2
16
51.49
10.92
2.50
(46.59, 56.39)
0.59
0.552
C3
16
51.15
13.13
2.50
(46.25, 56.05)
0.46
0.646
C4
16
49.82
8.29
2.50
(44.92, 54.72)
-0.07
0.942
C5
16
52.19
9.63
2.50
(47.29, 57.09)
0.88
0.381
C6
16
46.03
8.69
2.50
(41.13, 50.93)
-1.59
0.112
C7
16
45.10
9.84
2.50
(40.20, 50.00)
-1.96
0.050
C8
16
50.79
10.69
2.50
(45.89, 55.69)
0.32
0.751
C9
16
51.48
8.93
2.50
(46.58, 56.38)
0.59
0.554
C10
16
53.09
7.37
2.50
(48.19, 57.99)
1.23
0.217
C11
16
50.13
8.12
2.50
(45.23, 55.03)
0.05
0.960
C12
16
50.08
9.33
2.50
(45.18, 54.98)
0.03
0.974
C13
16
50.96
9.31
2.50
(46.06, 55.86)
0.38
0.701
C14
16
53.57
12.94
2.50
(48.67, 58.47)
1.43
0.153
C15
16
50.71
11.22
2.50
(45.81, 55.61)
0.29
0.775
C16
16
49.10
9.04
2.50
(44.20, 54.00)
-0.36
0.719
C17
16
51.37
8.67
2.50
(46.47, 56.27)
0.55
0.584
C18
16
48.85
9.25
2.50
(43.95, 53.75)
-0.46
0.645
C19
16
48.81
9.80
2.50
(43.91, 53.71)
-0.48
0.634
C20
16
48.42
5.25
2.50
(43.52, 53.32)
-0.63
0.528
C21
16
49.96
11.65
2.50
(45.06, 54.86)
-0.01
0.989
C22
16
51.06
11.79
2.50
(46.16, 55.96)
0.43
0.670
C23
16
55.63
9.12
2.50
(50.73, 60.53)
2.25
0.024
C24
16
51.01
8.65
2.50
(46.11, 55.91)
0.40
0.687
C25
16
50.01
9.96
2.50
(45.11, 54.91)
0.00
0.996
C26
16
51.13
8.77
2.50
(46.23, 56.03)
0.45
0.651
C27
16
53.32
10.58
2.50
(48.42, 58.22)
1.33
0.184
C28
16
48.86
10.71
2.50
(43.96, 53.76)
-0.46
0.649
C29
16
46.95
7.84
2.50
(42.05, 51.85)
-1.22
0.223
C30
16
51.81
8.49
2.50
(46.91, 56.71)
0.72
0.470
Now based on the values of the 30 test statistics, answer these
questions.
(A) How many researchers would reject H
0
. That is, how many of them made an “incorrect decision”?
........
(B) If we change the level of the test from α = 0.05 to α = 0.001
then, does this change any of your decisions to reject or not
reject H
0
? ........... (C) In general, should the number of rejections
increase or de-crease if α = 0.001 is used instead of α = 0.05?
.............
Results for: Worksheet 2
One-Sample Z: C1, C2, C3, C4, C5, C6, C7, C8, ...
Test of μ = 50 vs ≠ 50
The assumed standard deviation = 10
Variable
N
Mean
StDev
SE Mean
95% CI
Z
P
C1
16
53.04
11.53
2.50
(48.14, 57.94)
1.22
0.224
C2
16
50.53
12.80
2.50
(45.63, 55.43)
0.21
0.832
C3
16
52.28
14.36
2.50
(47.38, 57.18)
0.91
0.362
C4
16
50.85
7.81
2.50
(45.95, 55.75)
0.34
0.734
C5
16
50.24
11.92
2.50
(45.34, 55.14)
0.10
0.923
C6
16
48.41
10.56
2.50
(43.51, 53.31)
-0.64
0.525
C7
16
51.80
11.90
2.50
(46.90, 56.70)
0.72
0.471
C8
16
57.36
9.19
2.50
(52.46, 62.26)
2.94
0.003
C9
16
47.54
10.93
2.50
(42.64, 52.44)
-0.98
0.325
C10
16
54.86
9.12
2.50
(49.96, 59.76)
1.94
0.052
C11
16
54.77
9.27
2.50
(49.87, 59.67)
1.91
0.057
C12
16
50.39
11.51
2.50
(45.49, 55.29)
0.16
0.876
C13
16
50.67
7.94
2.50
(45.77, 55.57)
0.27
0.789
C14
16
55.83
10.80
2.50
(50.93, 60.73)
2.33
0.020
C15
16
54.01
10.05
2.50
(49.11, 58.91)
1.61
0.108
C16
16
55.27
12.24
2.50
(50.37, 60.16)
2.11
0.035
C17
16
49.03
12.04
2.50
(44.13, 53.93)
-0.39
0.698
C18
16
50.78
9.75
2.50
(45.88, 55.68)
0.31
0.756
C19
16
52.70
8.58
2.50
(47.80, 57.60)
1.08
0.281
C20
16
53.36
9.61
2.50
(48.46, 58.26)
1.35
0.179
C21
16
49.12
7.53
2.50
(44.22, 54.02)
-0.35
0.726
C22
16
53.62
9.01
2.50
(48.72, 58.52)
1.45
0.147
C23
16
52.85
9.69
2.50
(47.95, 57.75)
1.14
0.255
C24
16
48.39
13.29
2.50
(43.49, 53.29)
-0.64
0.521
C25
16
52.49
9.25
2.50
(47.59, 57.39)
0.99
0.320
C26
16
49.86
12.00
2.50
(44.96, 54.76)
-0.06
0.956
C27
16
51.55
11.96
2.50
(46.65, 56.45)
0.62
0.536
C28
16
54.48
7.26
2.50
(49.58, 59.38)
1.79
0.073
C29
16
51.82
8.13
2.50
(46.92, 56.72)
0.73
0.468
C30
16
51.89
11.28
2.50
(46.99, 56.79)
0.76
0.449
(D) If after a while we realized that the actual mean of the
population is currently μ = 52 dollars per hour and it is no
longer 50. Once again, using α = 0.05 and assuming σ is still 10
dollars per hour, in how many tests did you reject H
0
? .............
(E) A rejection of H
0
in part (A) is a “correct decision”. True or False? ......... (F) A
rejection of H
0
in part (D) is a “correct decision”. True or False? ..........
One-Sample T: C1, C2, C3, C4, C5, C6, C7, C8, ...
Test of μ = 50 vs ≠ 50
Variable
N
Mean
StDev
SE Mean
95% CI
T
P
C1
16
49.78
8.31
2.08
(45.35, 54.20)
-0.11
0.916
C2
16
48.78
8.36
2.09
(44.33, 53.24)
-0.58
0.569
C3
16
46.76
9.94
2.48
(41.46, 52.05)
-1.30
0.212
C4
16
48.45
10.02
2.51
(43.11, 53.79)
-0.62
0.547
C5
16
49.66
7.42
1.86
(45.70, 53.62)
-0.18
0.857
C6
16
48.97
7.95
1.99
(44.73, 53.20)
-0.52
0.611
C7
16
47.90
10.00
2.50
(42.57, 53.23)
-0.84
0.413
C8
16
47.31
9.69
2.42
(42.15, 52.48)
-1.11
0.285
C9
16
49.30
12.22
3.06
(42.79, 55.81)
-0.23
0.822
C10
16
47.73
9.62
2.41
(42.60, 52.86)
-0.94
0.360
C11
16
52.47
11.26
2.82
(46.47, 58.47)
0.88
0.394
C12
16
44.25
13.11
3.28
(37.26, 51.23)
-1.75
0.100
C13
16
47.21
10.86
2.72
(41.43, 53.00)
-1.03
0.321
C14
16
48.91
10.92
2.73
(43.09, 54.73)
-0.40
0.695
C15
16
47.33
12.47
3.12
(40.68, 53.97)
-0.86
0.405
C16
16
47.83
8.09
2.02
(43.52, 52.14)
-1.08
0.299
C17
16
51.65
8.38
2.09
(47.19, 56.12)
0.79
0.442
C18
16
49.15
7.86
1.97
(44.96, 53.35)
-0.43
0.673
C19
16
51.86
9.92
2.48
(46.57, 57.14)
0.75
0.466
C20
16
47.20
10.72
2.68
(41.49, 52.91)
-1.04
0.313
C21
16
46.87
8.91
2.23
(42.12, 51.62)
-1.40
0.180
C22
16
50.88
6.68
1.67
(47.32, 54.44)
0.53
0.606
C23
16
48.66
12.16
3.04
(42.18, 55.14)
-0.44
0.666
C24
16
55.54
8.67
2.17
(50.92, 60.17)
2.56
0.022
C25
16
51.28
12.36
3.09
(44.69, 57.87)
0.41
0.685
C26
16
52.08
7.44
1.86
(48.12, 56.04)
1.12
0.281
C27
16
49.84
6.21
1.55
(46.53, 53.15)
-0.10
0.919
C28
16
53.80
9.26
2.31
(48.86, 58.73)
1.64
0.122
C29
16
47.33
10.44
2.61
(41.77, 52.90)
-1.02
0.323
C30
16
52.64
8.92
2.23
(47.89, 57.39)
1.18
0.255
Repeat parts (A), (B), and (C) of Question 1, using ttest in-
stead of ztest, and answer (Thus ‘ztest 50 10 c1-c30’ changes to
‘ttest 50 c1-c30’) (A) In how many tests did you reject H
0
. That is, how many times did you make an “incorrect
decision”? .......... (B) Suppose you used α = 0.00008 instead
of α = 0.05. Does this change any of your decisions to reject or
not? ........... (C) In general, should the number of rejections
increase or decrease if α = 0.00008 is used instead of α = 0.05?
............
1. I am deciding on whether to invest 400,000 CAD to open a
convenience store in particular spot in Ottawa. I know that the
business will be profitable, with income of 70,000 CAD per
year, if the store will have in average μ = 60 or more customers
per day. If I am convinced that the business will be profitable
then I will go ahead and open the store otherwise I won’t.
Hence I am dealing with this hypothesis testing problem.
H
0
: μ = 60 vs H
a
: μ < 60
(A) Explain what Type I error means in the context of this
problem. (B) What is the consequence of Type I error. (C)
Explain what Type II error means in this context.
(D) What is the consequence of Type II error in the context of
this problem. (E) Which error is more expensive in your
opinion. Explain.
2.
A recent study conducted by the government attempts to
determine proportion of people who support further increase in
cigarette taxes. In this study, 2500 voting age citizens were
sampled, and it was found that 1900 of them were in favor of an
increase in cigarette taxes. At level α = 0.05, do you believe
that 78% of all citizens are in favor of an increase in cigarette
taxes. (A) State the null and alternative hypotheses and
compute the p-value. (B) What is the smallest α you need to
reject the null hypothesis.
3.
To study the average amount of debts in Canada, 100 Canadians
were surveyed. The amount of debt of each one was recorded.
The sample gave an average of 28110 CAD with standard
deviation of 3500 CAD. It is believed that the actual mean value
of the debt in Canada is 27500. Given our data, we would like
to test if the actual mean of debt is higher than 27500. (A)
State the null and alternative hypotheses and compute the value
of the test statistic. (B) Compute the p-value for this test (C)
Draw a conclusion by comparing the p-value you obtained in
part (B) to α = 0.05.
4
-A caviar producing company packs its product in containers.
They claim that the average weight of the caviar in the
containers is 1 kilogram. They sell each container for 8,000
CAD. We had a budget of approximately 90,000 CAD. So, we
could buy 11 of the containers and weighted the caviar in each.
Here are the results in Kilograms: 0.89, 1.01, 1.00, 0.90,
0.90, 0.91, 0.91, 0.89, 0.95, 1.00, 0.96. If we assume that the
wights are normally distributed is there sufficient evidence for
us to believe that the actual average weight of caviar in the
containers is less that what the producer claims it is. Use the
critical value approach with α = 0.05
5. The mid-distance running coach, Zdravko Popovich, for the
Olympic team of an eastern European country claims that his
six-month training program significantly reduces the average
time to complete a 1500-meter run. Four mid-distance runners
were randomly selected before they were trained with coach
Popovich’s six-month training program and their completion
time of 1500-meter run was recorded (in minutes). After six
months of training under coach Popovich, the same four
runners’ 1500 meter run time was recorded again. The results
are given below.
Runner
1
2
3
4
Completion time before training
Completion time after training
6.0
5.5
7.5
7.1
6.2
6.2
6.8
6.4
Assume that the times before training and after training are
normally distributed.
(A) Are the two samples independent? Explain.
(B) Construct 95% confidence interval for the difference
between the actual mean of completion time before training and
the actual mean of completion time after training.
(C) Based on the confidence interval in part (B) can we
conclude that there is no significant difference between the
between the actual mean of completion time before training and
the actual mean of completion time after training. Hint: Check
and see if your confidence interval in part (B) includes 0.

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