Midterm 2 Version B - 2023
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MAT 2379C
Midterm Examination
November 15, 2023
Instructor: Xiao Liang
Time: 80 minutes
Student Number:
Family Name:
First Name:
•
This is a closed book examination.
•
You can bring your own formula sheet (one page, one-sided).
•
Some statistical tables are included in the last pages of the booklet.
•
Only Faculty standard calculators are permitted: TI30, TI34, Casio fx-260, Casio fx-300.
•
You are not allowed to use any electronic device during the exam. Cell phones should be put away.
•
The exam consists of 6 multiple choice questions and 4 long answer questions.
•
Each multiple choice question is worth 5 marks and each long answer question is worth 10 marks. The
total number of marks is 70.
NOTE: At the end of the examination, hand in the entire booklet.
*****************************************************
For professor’s use:
Number of marks
Total for all MC Questions
Long Answer Question 1
Long Answer Question 2
Long Answer Question 3
Long Answer Question 4
Total
1
Part 1: Multiple Choice Questions
Record your answer to the multiple choice questions in the table below:
Question
Answer
1
E
2
C
3
C
4
E
5
A
6
C
1. Our scientists study polar bears in the Beaufort Sea. Their research indicates that bears, while fasting,
give birth to cubs of lower weight.
We measured the birth weight (in grams) for a sample of 52 cubs, obtaining a sample mean
¯
x
= 715
g
and a standard deviation
s
= 123
g. Calculate a confidence interval of
99%
for the mean birth weight
μ
of the cubs, and the estimated standard error of the mean.
A)
715
±
1
.
96; 17
.
057
B)
715
±
43
.
92; 2
.
46
C)
715
±
43
.
92; 2
.
46
D)
715
±
34
.
09; 17
.
057
E)
715
±
43
.
92; 17
.
057
Solution
This is an interval for a large sample. A confidence interval of
99%
for
μ
is given by :
715
±
2
.
575
123
√
52
= 715
±
43
.
9218
..
The estimated standard error of the mean is
s/
√
n
= 123
/
√
52 = 17
.
057
. Here, since
P
(
Z
≤
2
.
57) =
0
.
9949
and
P
(
Z
≤
2
.
58) = 0
.
9951
, we choose
z
α/
2
= 2
.
575
. The answer is E.
2. We aim to estimate the mean
daily
temperature on San Crist´obal Island during the dry season. Assuming
that daily temperatures (during the dry season from
June to December
) follow a normal distribution,
we wish to determine a 90% confidence interval for this mean temperature during
three weeks
of the
dry season, given that the daily measured mean was 21.5
°
C with a variance of 2.083.
A)
21
.
5
±
0
.
327
B)
21
.
5
±
0
.
418
C)
21
.
5
±
0
.
543
D)
21
.
5
±
0
.
682
E)
21
.
5
±
0
.
701
Solution (Section 8.1) We have
¯
x
= 21
.
5
and
s
2
= 2
.
083
. The confidence interval is based on a random
variable
T
which follows a
t
(20)
distribution. To construct this interval, we first need to find the value of
t
such that
P
(
-
t
≤
T
≤
t
) = 0
.
90
, where
T
is a random variable following a
t
(20)
distribution. Therefore,
P
(
T > t
) = (1
-
0
.
90)
/
2 = 0
.
05
and, consequently,
P
(
T
≤
t
) = 1
-
0
.
05 = 0
.
95
. Using Table 18.4, we
find
t
0
.
05
,
20
= 1
.
725
.
A 90% confidence interval for
μ
is
21
.
5
±
1
.
725
√
2
.
083
√
21
!
= 21
.
5
±
0
.
543
.
The answer is C.
3. Glaucoma is an eye disease characterized by elevated intraocular pressure. Let’s assume that in the general
population, intraocular pressure follows an approximately normal distribution, with a mean of 15 mm Hg
and a standard deviation of 4 mm Hg. The usual range of intraocular pressure is considered to be between
10 mm Hg and 19 mm Hg. What proportion of the general population has an intraocular pressure within
2
the usual range?
A)
0
.
7164
B)
0
.
0718
C)
0
.
7357
D)
0
.
7426
E)
0
.
8012
Solution
(Section 5.2) We want to calculate
P
(10
≤
X
≤
19)
, where
X
follows a normal distribution
with mean
μ
= 15
and standard deviation
σ
= 4
. Using standardization, we have :
P
(10
≤
X
≤
19)
=
P
10
-
15
4
≤
X
-
15
4
≤
19
-
15
4
=
P
(
-
1
.
25
≤
Z
≤
1)
=
P
(
Z
≤
1)
-
P
(
Z <
-
1
.
25)
=
0
.
8413
-
0
.
1056 = 0
.
7357
.
The answer is C.
4. We’re looking for information on humidity levels in three cities: Gatineau, Montreal and Quebec City.
For each city, we selected a sample of 100 houses and recorded their humidity levels. We then created
3 data sets of 100 observations each, called ”Gatineau”, ”Montreal” and ”Quebec City”.
Below are
box-and-whisker plots and histograms for these data sets. Which of the following is
correct
?
(a) Gatineau
(b) Qu´
ebec
(c) Montr´
eal
(d)
(e)
(f)
A) Histogram (d) is for Gatineau, (e) Montr´
eal, and (f) Qu´
ebec.
B) Histogram (f) is for Gatineau, (e) Montr´
eal, and (d) Quebec.
C) Histogram (e) is for Gatineau, (f) Montr´
eal, and (d) Quebec.
D) Histogram(f) is for Gatineau, (d) Montr´
eal, and (e) Quebec.
E) Histogram(e) is for Gatineau, (d) Montr´
eal, and (f) Quebec.
Solution
(Section 7.1): We need to match each histogram with the corresponding boxplot. Note that
histogram (f) is approximately symmetrical, histogram (d) is positively asymmetrical and histogram (e) is
negatively asymmetrical. To match (f), we need to look at a box whose two halves are of roughly equal
length, so (b) matches (f). We should be able to identify a mosaic far from the center for an asymmetrical
distribution. Note that the center of the distribution of (a) is the highest of the three distributions, and
that the left portion of the box and the left whiskers are longer than on the right side, so (a) corresponds to
(e). The center of the distribution of (c) is the lowest of the three distributions, it is a positive asymmetric
3
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distribution, so (c) corresponds to (d). You can also use whiskers to make the association. The answer is
E.
5. The grizzly bear is a large animal generally found in the highlands of western North America. Its weight
depends on its location. It is estimated that an adult male grizzly bear in the Alaska Peninsula region has
an average weight of 375 kg, with a standard deviation of 21 kg. The weight of an adult male grizzly bear
is assumed to follow a normal distribution.
Use the output
R
below to find the weight
x
0
(in kg) such that 95% of adult male grizzly bears in the
Alaska Peninsula region weigh more than
x
0
. (In this output,
*
denotes multiplication.)
A) 375 + 21
*
qnorm
(0.05, 0, 1)
B) 375 - 21
*
qnorm
(0.05, 0, 1)
C)
pnorm
(0.95, 375, 21)
D)
qnorm
(0.95, 375, 21)
E)
pnorm
(0.05, 375, 21)
Solution
(Section 5.2) Let
X
be the weight
x
0
(in kg) of a randomly selected adult male grizzly bear
from the Alaska Peninsula region. Then
X
follows a normal distribution with mean
μ
= 375
and standard
deviation
σ
= 21
.
We need to find a value
x
0
such that
P
(
X > x
0
) = 0
.
95
.
This means that
P
(
X < x
0
) = 0
.
05
. The value is given by the command
R qnorm
(0.05, 375, 21) but this command is not
included among the listed answers. By standardization,
0
.
05 =
P
(
X < x
0
) =
P
X
-
375
21
<
x
0
-
375
21
=
P
(
Z < z
0
)
where
z
0
=
x
0
-
375
21
=
qnorm
(0
.
05
,
0
,
1)
. Solving for
x
0
we obtain:
x
0
= 375 + 21
z
0
. The answer is A.
6. Researchers were interested in analyzing the width of sepals of 50 samples of three species of Iris (150
observations in total). Below is the normal QQ graph and histogram of this data set.
Which of the following statements is correct? (Only one statement is correct.)
A) The distribution of the width of sepals is highly skewed to the
left
. It is inappropriate to assume that
the birth weight of babies follows a normal distribution.
B) The width of sepals follows a Student distribution with 149 degrees of freedom.
C) The width of sepals follows a normal distribution.
D) The distribution of the width of sepals is highly skewed to the
right
. It is inappropriate to assume
that the width of sepals follows a normal distribution.
E) The distribution of the width of sepals is approximately symmetrical. It is not appropriate to assume
that the width of sepals follow a normal distribution.
Solution (Sections 7.1 et 7.3) In the quantile-quantile plot, the points lie on a straight line, suggesting a
linear trend. It is therefore reasonable to assume that the width of sepals follows a normal distribution.
The normal quantile-quantile plot should not be used for the T distribution. The answer is C.
4
Part 2: Long Answer Questions
Record your answer to the long answer questions in the space provided below, specifying clearly your notation
and including a proper justification. Show the details of your calculations.
1. The seed weight of the princess bean
Phaseotus vulgaris
follows a normal distribution with mean
μ
= 450
mg and variance
σ
2
= 14161
mg. We select a random sample of size
n
from Phaseotus vulgaris seeds.
Let
X
be the average seed weight in this sample. Find the sample size
n
such that
P
(
X >
500) = 0
.
2
.
Solution
(Section 7.2) Let
X
be the weight of a seed chosen at random and
¯
X
the mean weight of the
seeds in a sample of size
n
. Since
X
follows a normal distribution,
Z
=
¯
X
-
450
119
/
√
n
∼
N
(0
,
1)
.
We want to find
n
such that
P
(
¯
X >
500) = 0
.
2
or, equivalently,
P
(
¯
X <
500) = 0
.
8
. Standardizing, it
follows that:
0
.
8 =
P
(
¯
X <
500) =
P
Z <
500
-
450
119
/
√
n
=
P
Z <
50
√
n
119
.
In Table 18.3, we look for a value
z
such that
P
(
Z < z
) = 0
.
8
. We find
z
= 0
.
845
. So
50
√
n
119
= 0
.
845
,
and
√
n
=
(119)(0
.
845)
50
= 2
.
0111
.
Therefore,
n
= (2
.
0111)
2
= 4
.
045
≈
4
.
5
2. The water in a certain lake has a salinity of around 70 mg/L. The salinity is measured for 5 water samples
taken from this lake. Here are the data.
x
1
= 60
.
15
x
2
= 73
.
24
x
3
= 69
.
03
x
4
= 105
.
58
x
5
= 80
.
04
a) (5 points) What is the geometric mean of this data set?
b) (5 points) We transform the data using the linear transformation
X
0
= 2
X
+ 3
. What is the median
of the transformed measurements
x
0
1
, x
0
2
, x
0
3
, x
0
4
, x
0
5
?
Solution (Section 7.1)
a)
Method 1.
We apply a logarithmic transformation to this data:
Y
= ln(
X
)
. We obtain the following
new data:
y
1
= ln(60
.
15)
≈
4
.
096
,
y
2
= ln(73
.
24)
≈
4
.
294
,
y
3
= ln(69
.
03)
≈
4
.
234
,
y
4
= ln(105
.
58)
≈
4
.
659
,
y
5
= ln(80
.
04)
≈
4
.
382
.
The mean of the log-transformed data is :
¯
y
=
4
.
096 + 4
.
294 + 4
.
234 + 4
.
659 + 4
.
382
5
≈
4
.
333
.
The geometric mean of the original data is :
g
=
e
¯
y
=
e
4
.
333
≈
76
.
2
.
Method 2.
The geometric mean is calculated as follows:
g
= (
x
1
×
x
2
×
x
3
×
x
4
×
x
5
)
1
/
5
= (60
.
15
×
73
.
24
×
69
.
03
×
105
.
58
×
80
.
04)
1
/
5
≈
76
.
2
.
b)
Method 1
The transformed measurements are :
x
0
1
= 2
×
x
1
+ 3 = 2
×
60
.
15 + 3 = 123
.
30
,
x
0
2
= 2
×
x
2
+ 3 = 2
×
73
.
24 + 3 = 149
.
48
,
x
0
3
= 2
×
x
3
+ 3 = 2
×
69
.
03 + 3 = 141
.
06
,
x
0
4
= 2
×
x
4
+ 3 = 2
×
105
.
58 + 3 = 214
.
16
,
x
0
5
= 2
×
x
5
+ 3 = 2
×
80
.
04 + 3 = 163
.
08
.
We arrange the transformed data in ascending order. We obtain :
y
0
1
= 123
.
30
,
y
0
2
= 141
.
06
,
y
0
3
= 149
.
48
,
y
0
4
= 163
.
08
,
y
0
5
= 214
.
16
.
The median of the transformed data is
y
0
3
= 149
.
48
.
Method 2
We first find the median of the original data set
x
1
, x
2
, x
3
, x
4
, x
5
. To do this, we arrange the
original data in ascending order and obtain :
60
.
15
,
69
.
03
,
73
.
24
,
80
.
04
,
105
.
58
.
The median of the original data is
73
.
24
. The median of the transformed data is
2
×
73
.
24 + 3 = 149
.
48
.
6
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3. Assume that the blood glucose concentration of a population of diabetic patients follows a normal distri-
bution with a mean of 150 mg/dl and a standard deviation of 20 mg/dl.
a) (5 points) What is the probability that a randomly selected diabetic patient will have a blood glucose
concentration between 125 mg/dl and 165 mg/dl?
b) (5 points) A group of researchers is testing a sample of 9 diabetic patients. What is the probability
that the average blood glucose concentration in this sample will be between 125 mg/dl and 165 mg/dl?
Solution
a) (Section 5.2) Let
X
be the concentration of glucose in the blood of a randomly selected
diabetic patient. The probability of occurrence is :
P
(125
< X <
165)
=
P
125
-
150
20
<
X
-
150
20
<
165
-
150
20
=
P
(
-
1
.
25
< Z <
0
.
75) =
P
(
Z <
0
.
75)
-
P
(
Z <
-
1
.
25)
=
0
.
7734
-
0
.
1056 = 0
.
6678
.
We use Tables 18.2 and 18.3 to obtain the probabilities.
b) (Section 7.2) Let
X
be the mean blood glucose concentration of the sample. Then
X
follows a normal
distribution with a mean of 150 mg/dl (the same as the population mean) and a standard deviation of
20
/
√
9 = 20
/
3
mg/dl. The probability we’re looking for is as follows:
P
(125
<
X <
165)
=
P
125
-
150
20
/
3
<
X
-
150
20
/
3
<
165
-
150
20
/
3
=
P
(
-
3
.
75
< Z <
2
.
25) =
P
(
Z <
2
.
25)
-
P
(
Z <
-
3
.
75)
=
0
.
9878
-
0
.
0001 = 0
.
9877
where for the last line we have again used Tables 18.2 and 18.3.
7
4. The following data give the systolic blood pressure (in mmHg) of 13 people at rest:
120
125
122
118
127
121
126
119
128
123
124
120
90
a) (5 points) Find the median (
˜
x
) and the two quartiles (
q
1
,
q
3
).
b) (5 points) Give the values of outliers (if any).
Solution
(Section 7.1) a) the data sorted in ascending order are as follows:
y
1
= 90
y
2
= 118
y
3
= 119
y
4
= 120
y
5
= 120
y
6
= 121
y
7
= 122
y
8
= 123
y
9
= 124
y
10
= 125
y
11
= 126
y
12
= 127
y
13
= 128
Since 13 is an odd number, the median is
y
n
+1
2
=
y
7
= 122
.
Note that
n
+ 1
4
=
14
4
= 3
.
5
and
3(
n
+ 1)
4
= 10
.
5
.
To calculate
q
1
:
q
1
= (0
.
5)
y
3
+ (0
.
5)
y
4
= (0
.
5)(119) + (0
.
5)(120) = 119
.
5
The third quartile is
q
3
= (0
.
5)
y
10
+ (0
.
5)
y
11
= (0
.
5)(125) + (0
.
5)(126) = 125
.
5
b) For IQR calculation:
IQR
=
q
3
-
q
1
= 125
.
5
-
119
.
5 = 6
now let’s calculate the bounds:
The fence 1:
fence 1
=
q
1
-
1
.
5(
IQR
) = 119
.
5
-
1
.
5(6) = 119
.
5
-
9 = 110
.
5
The fence 2:
fence 2
=
q
3
+ 1
.
5(
IQR
) = 125
.
5 + 1
.
5(6) = 125
.
5 + 9 = 134
.
5
Outliers are values outside the bounds (i.e. smaller than Fence 1 or larger than Fence 2).
The only outlier in this data set is 90, as it is below Fence 1 of 110.5.
8
Table 18.2 Cumulative distribution function for
N
(0
,
1)
:
Φ(
z
) =
P
(
Z
≤
z
)
0
.
09
0
.
08
0
.
07
0
.
06
0
.
05
0
.
04
0
.
03
0
.
02
0
.
01
0
.
00
z
.
0001
.0001
.0001
.0001
.0001
.0001
.0001
.0001
.0001
.0001
-
3
.
8
.0001
.0001
.0001
.0001
.0001
.0001
.0001
.0001
.0001
.0001
-
3
.
7
.0001
.0001
.0001
.0001
.0001
.0001
.0001
.0001
.0002
.0002
-
3
.
6
.0002
.0002
.0002
.0002
.0002
.0002
.0002
.0002
.0002
.0002
-
3
.
5
.0002
.0003
.0003
.0003
.0003
.0003
.0003
.0003
.0003
.0003
-
3
.
4
.0003
.0004
.0004
.0004
.0004
.0004
.0004
.0005
.0005
.0005
-
3
.
3
.0005
.0005
.0005
.0006
.0006
.0006
.0006
.0006
.0007
.0007
-
3
.
2
.0007
.0007
.0008
.0008
.0008
.0008
.0009
.0009
.0009
.0010
-
3
.
1
.0010
.0010
.0011
.0011
.0011
.0012
.0012
.0013
.0013
.0013
-
3
.
0
.0014
.0014
.0015
.0015
.0016
.0016
.0017
.0018
.0018
.0019
-
2
.
9
.0019
.0020
.0021
.0021
.0022
.0023
.0023
.0024
.0025
.0026
-
2
.
8
.0026
.0027
.0028
.0029
.0030
.0031
.0032
.0033
.0034
.0035
-
2
.
7
.0036
.0037
.0038
.0039
.0040
.0041
.0043
.0044
.0045
.0047
-
2
.
6
.0048
.0049
.0051
.0052
.0054
.0055
.0057
.0059
.0060
.0062
-
2
.
5
.0064
.0066
.0068
.0069
.0071
.0073
.0075
.0078
.0080
.0082
-
2
.
4
.0084
.0087
.0089
.0091
.0094
.0096
.0099
.0102
.0104
.0107
-
2
.
3
.0110
.0113
.0116
.0119
.0122
.0125
.0129
.0132
.0136
.0139
-
2
.
2
.0143
.0146
.0150
.0154
.0158
.0162
.0166
.0170
.0174
.0179
-
2
.
1
.0183
.0188
.0192
.0197
.0202
.0207
.0212
.0217
.0222
.0228
-
2
.
0
.0233
.0239
.0244
.0250
.0256
.0262
.0268
.0274
.0281
.0287
-
1
.
9
.0294
.0301
.0307
.0314
.0322
.0329
.0336
.0344
.0351
.0359
-
1
.
8
.0367
.0375
.0384
.0392
.0401
.0409
.0418
.0427
.0436
.0446
-
1
.
7
.0455
.0465
.0475
.0485
.0495
.0505
.0516
.0526
.0537
.0548
-
1
.
6
.0559
.0571
.0582
.0594
.0606
.0618
.0630
.0643
.0655
.0668
-
1
.
5
.0681
.0694
.0708
.0721
.0735
.0749
.0764
.0778
.0793
.0808
-
1
.
4
.0823
.0838
.0853
.0869
.0885
.0901
.0918
.0934
.0951
.0968
-
1
.
3
.0985
.1003
.1020
.1038
.1056
.1075
.1093
.1112
.1131
.1151
-
1
.
2
.1170
.1190
.1210
.1230
.1251
.1271
.1292
.1314
.1335
.1357
-
1
.
1
.1379
.1401
.1423
.1446
.1469
.1492
.1515
.1539
.1562
.1587
-
1
.
0
.1611
.1635
.1660
.1685
.1711
.1736
.1762
.1788
.1814
.1841
-
0
.
9
.1867
.1894
.1922
.1949
.1977
.2005
.2033
.2061
.2090
.2119
-
0
.
8
.2148
.2177
.2206
.2236
.2266
.2296
.2327
.2358
.2389
.242
-
0
.
7
.2451
.2483
.2514
.2546
.2578
.2611
.2643
.2676
.2709
.2743
-
0
.
6
.2776
.2810
.2843
.2877
.2912
.2946
.2981
.3015
.3050
.3085
-
0
.
5
.3121
.3156
.3192
.3228
.3264
.3300
.3336
.3372
.3409
.3446
-
0
.
4
.3483
.3520
.3557
.3594
.3632
.3669
.3707
.3745
.3783
.3821
-
0
.
3
.3859
.3897
.3936
.3974
.4013
.4052
.4090
.4129
.4168
.4207
-
0
.
2
.4247
.4286
.4325
.4364
.4404
.4443
.4483
.4522
.4562
.4602
-
0
.
1
.4641
.4681
.4721
.4761
.4801
.4840
.4880
.4920
.4960
.5000
-
0
.
0
9
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Table 18.3 Cumulative distribution function for
N
(0
,
1)
:
Φ(
z
) =
P
(
Z
≤
z
)
z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.0
.5000
.5040
.5080
.5120
.5160
.5199
.5239
.5279
.5319
.5359
0.1
.5398
.5438
.5478
.5517
.5557
.5596
.5636
.5675
.5714
.5753
0.2
.5793
.5832
.5871
.5910
.5948
.5987
.6026
.6064
.6103
.6141
0.3
.6179
.6217
.6255
.6293
.6331
.6368
.6406
.6443
.6480
.6517
0.4
.6554
.6591
.6628
.6664
.6700
.6736
.6772
.6808
.6844
.6879
0.5
.6915
.6950
.6985
.7019
.7054
.7088
.7123
.7157
.7190
.7224
0.6
.7257
.7291
.7324
.7357
.7389
.7422
.7454
.7486
.7517
.7549
0.7
.7580
.7611
.7642
.7673
.7704
.7734
.7764
.7794
.7823
.7852
0.8
.7881
.7910
.7939
.7967
.7995
.8023
.8051
.8078
.8106
.8133
0.9
.8159
.8186
.8212
.8238
.8264
.8289
.8315
.8340
.8365
.8389
1.0
.8413
.8438
.8461
.8485
.8508
.8531
.8554
.8577
.8599
.8621
1.1
.8643
.8665
.8686
.8708
.8729
.8749
.8770
.8790
.8810
.8830
1.2
.8849
.8869
.8888
.8907
.8925
.8944
.8962
.8980
.8997
.9015
1.3
.9032
.9049
.9066
.9082
.9099
.9115
.9131
.9147
.9162
.9177
1.4
.9192
.9207
.9222
.9236
.9251
.9265
.9279
.9292
.9306
.9319
1.5
.9332
.9345
.9357
.9370
.9382
.9394
.9406
.9418
.9429
.9441
1.6
.9452
.9463
.9474
.9484
.9495
.9505
.9515
.9525
.9535
.9545
1.7
.9554
.9564
.9573
.9582
.9591
.9599
.9608
.9616
.9625
.9633
1.8
.9641
.9649
.9656
.9664
.9671
.9678
.9686
.9693
.9699
.9706
1.9
.9713
.9719
.9726
.9732
.9738
.9744
.9750
.9756
.9761
.9767
2.0
.9772
.9778
.9783
.9788
.9793
.9798
.9803
.9808
.9812
.9817
2.1
.9821
.9826
.9830
.9834
.9838
.9842
.9846
.9850
.9854
.9857
2.2
.9861
.9864
.9868
.9871
.9875
.9878
.9881
.9884
.9887
.9890
2.3
.9893
.9896
.9898
.9901
.9904
.9906
.9909
.9911
.9913
.9916
2.4
.9918
.9920
.9922
.9925
.9927
.9929
.9931
.9932
.9934
.9936
2.5
.9938
.9940
.9941
.9943
.9945
.9946
.9948
.9949
.9951
.9952
2.6
.9953
.9955
.9956
.9957
.9959
.9960
.9961
.9962
.9963
.9964
2.7
.9965
.9966
.9967
.9968
.9969
.9970
.9971
.9972
.9973
.9974
2.8
.9974
.9975
.9976
.9977
.9977
.9978
.9979
.9979
.9980
.9981
2.9
.9981
.9982
.9982
.9983
.9984
.9984
.9985
.9985
.9986
.9986
3.0
.9987
.9987
.9987
.9988
.9988
.9989
.9989
.9989
.9990
.9990
3.1
.9990
.9991
.9991
.9991
.9992
.9992
.9992
.9992
.9993
.9993
3.2
.9993
.9993
.9994
.9994
.9994
.9994
.9994
.9995
.9995
.9995
3.3
.9995
.9995
.9995
.9996
.9996
.9996
.9996
.9996
.9996
.9997
3.4
.9997
.9997
.9997
.9997
.9997
.9997
.9997
.9997
.9997
.9998
3.5
.9998
.9998
.9998
.9998
.9998
.9998
.9998
.9998
.9998
.9998
3.6
.9998
.9998
.9999
.9999
.9999
.9999
.9999
.9999
.9999
.9999
3.7
.9999
.9999
.9999
.9999
.9999
.9999
.9999
.9999
.9999
.9999
3.8
.9999
.9999
.9999
.9999
.9999
.9999
.9999
.9999
.9999
.9999
10
Table 18.4
T
distribution with
ν
degrees of freedom
F
T
(
t
) =
P
(
T
≤
t
)
.6
.75
.9
.95
.975
.99
.995
ν
t
.
40
,ν
t
.
25
,ν
t
.
10
,ν
t
.
05
,ν
t
.
025
,ν
t
.
01
,ν
t
.
005
,ν
1
0.325
1.000
3.078
6.314
12.706
31.821
63.657
2
0.289
0.816
1.886
2.920
4.303
6.965
9.925
3
0.277
0.765
1.638
2.353
3.182
4.541
5.841
4
0.271
0.741
1.533
2.132
2.776
3.747
4.604
5
0.267
0.727
1.476
2.015
2.571
3.365
4.032
6
0.265
0.718
1.440
1.943
2.447
3.143
3.707
7
0.263
0.711
1.415
1.895
2.365
2.998
3.499
8
0.262
0.706
1.397
1.860
2.306
2.896
3.355
9
0.261
0.703
1.383
1.833
2.262
2.821
3.250
10
0.260
0.700
1.372
1.812
2.228
2.764
3.169
11
0.260
0.697
1.363
1.796
2.201
2.718
3.106
12
0.259
0.695
1.356
1.782
2.179
2.681
3.055
13
0.259
0.694
1.350
1.771
2.160
2.650
3.012
14
0.258
0.692
1.345
1.761
2.145
2.624
2.997
15
0.258
0.691
1.341
1.753
2.131
2.602
2.947
16
0.258
0.690
1.337
1.746
2.120
2.583
2.921
17
0.257
0.689
1.333
1.740
2.110
2.567
2.898
18
0.257
0.688
1.330
1.734
2.101
2.552
2.878
19
0.257
0.688
1.328
1.729
2.093
2.539
2.861
20
0.257
0.687
1.325
1.725
2.086
2.528
2.845
21
0.257
0.686
1.323
1.721
2.080
2.518
2.831
22
0.256
0.686
1.321
1.717
2.074
2.508
2.819
23
0.256
0.685
1.319
1.714
2.069
2.500
2.807
24
0.256
0.685
1.318
1.711
2.064
2.492
2.797
25
0.256
0.684
1.316
1.708
2.060
2.485
2.787
26
0.256
0.684
1.315
1.706
2.056
2.479
2.779
27
0.256
0.684
1.314
1.703
2.052
2.473
2.771
28
0.256
0.683
1.313
1.701
2.048
2.467
2.763
29
0.256
0.683
1.311
1.699
2.045
2.464
2.756
30
0.256
0.683
1.310
1.697
2.042
2.457
2.750
∞
0.253
0.674
1.282
1.645
1.96
2.326
2.576
Note:
z
α
=
t
α,
∞
11