University of Washington Bothell Emergency Manager Core Competencies Discussion

The Module for Week 1 contains a spreadsheet showing all the Washington State DOT contract prices for 3/8-inch hot mix asphalt mixtures from 25 September 2017 to 25 September 2018. Use this data to answer this question.

Plot a histogram of this data (45 points) using the guidance in the course notes (both lectures are attached). Based on your plot, how do you characterize the data?

– uniformly distributed?

– normally distributed?

– bimodal?

Here is the raw data download from the Washington State DOT showing the low bid prices for 3/8-inch HMA over a year’s pe
This data are organized a bit better on the next tab for your analysis.
Use this data to answer quiz questions in week 1.
Bid Item Unit Price Tabulation Standard Items
Standard Item Number
Region: State Wide
Contracts awarded from 9/25/2017 thru 9/25/2018
Unit Of Measure Item Description
Northwest
5766
TON
HMA CL. 3/8 IN. PG 64-22
5766
TON
HMA CL. 3/8 IN. PG 64-22
5766
TON
HMA CL. 3/8 IN. PG 64-22
5766
TON
HMA CL. 3/8 IN. PG 58H-22
5766
TON
HMA CL. 3/8 IN. PG 58H-22
5766
TON
HMA CL. 3/8 IN. PG 58H-22
5766
TON
HMA CL. 3/8 IN. PG
5766
TON
HMA CL. 3/8 IN. PG 58H-22
5766
TON
HMA CL. 3/8 IN. PG 58H-22
Northwest Region Average Low Bid for Item No. 5766 is $141.52 Total Quantity Planned: 37,570.00 Total Amount: $5
State Average Low Bid for Item No. 5766 is $96.06 Total Quantity Planned: 283,714.78 Total Amount: $27,253,838.05
North Central
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
North Central Region Average Low Bid for Item No. 5766 is $82.03 Total Quantity Planned: 46,313.20 Total Amount: $
State Average Low Bid for Item No. 5766 is $96.06 Total Quantity Planned: 283,714.78 Total Amount: $27,253,838.05
Southwest
5766
TON
HMA CL. 3/8 IN. PG 64-28
5766
TON
HMA CL. 3/8 IN. PG 64-28 (BRIDGE DECK)
5766
TON
HMA CL. 3/8 IN. PG 64-28 (TRANSITION)
5766
TON
HMA CL. 3/8 IN. PG 64-22
5766
TON
HMA CL. 3/8 IN. PG 58V-22
5766
TON
HMA CL. 3/8 IN. PG 58V-22 (BRIDGES)
5766
TON
HMA CL. 3/8 IN. PG 58H-22
5766
TON
HMA CL. 3/8 IN. PG 58H-22 (BRIDGE)
5766
TON
HMA CL. 3/8 IN. PG 58H-22
5766
TON
HMA CL. 3/8 IN. PG 58H-22
5766
TON
HMA CL. 3/8 IN. PG 58H-22
5766
TON
HMA CL. 3/8 IN. PG 58H-22 (BRIDGE)
5766
TON
HMA CL. 3/8 IN. PG 58H-22
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28 (BRIDGE)
5766
TON
HMA CL. 3/8 IN. PG 58H-22
Southwest Region Average Low Bid for Item No. 5766 is $94.25 Total Quantity Planned: 123,072.78 Total Amount: $1
State Average Low Bid for Item No. 5766 is $96.06 Total Quantity Planned: 283,714.78 Total Amount: $27,253,838.05
South Central
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
South Central Region Average Low Bid for Item No. 5766 is $89.74 Total Quantity Planned: 19,417.00 Total Amount: $
State Average Low Bid for Item No. 5766 is $96.06 Total Quantity Planned: 283,714.78 Total Amount: $27,253,838.05
Eastern
5766
TON
HMA CL. 3/8 IN. PG 76-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
5766
TON
HMA CL. 3/8 IN. PG 64H-28
Eastern Region Average Low Bid for Item No. 5766 is $83.64 Total Quantity Planned: 57,341.80 Total Amount: $4,796
State Average Low Bid for Item No. 5766 is $96.06 Total Quantity Planned: 283,714.78 Total Amount: $27,253,838.05
ces for 3/8-inch HMA over a year’s period.
e Tabulation Standard Items
m 9/25/2017 thru 9/25/2018
Total records found: 45
Job Number Contract Number AD Date
Planned Quantity
Low Bid
17A011
009161
17A015
009168
17A018
009183
17A040
009212
17A038
009220
17A043
009221
17A037
009227
17A021
009231
17A044
009256
nned: 37,570.00 Total Amount: $5,316,830.08
78 Total Amount: $27,253,838.05
09/25/2017
10/16/2017
11/20/2017
01/22/2018
02/05/2018
02/12/2018
02/20/2018
02/26/2018
03/26/2018
4.00
1,940.00
9,749.00
400.00
8,430.00
7,919.00
590.00
8,280.00
258.00
$404.52
$400.00
$145.00
$70.00
$97.50
$98.00
$140.00
$154.00
$550.00
18B002
009181
18B003
009209
18B005
009236
18B007
009260
18B009
009261
18B010
009270
18B014
009274
18B012
009275
18B011
009279
lanned: 46,313.20 Total Amount: $3,798,923.00
78 Total Amount: $27,253,838.05
11/20/2017
01/16/2018
03/05/2018
04/02/2018
04/02/2018
04/16/2018
04/23/2018
04/23/2018
04/30/2018
23,537.00
755.00
523.00
20,123.00
240.00
430.00
375.00
260.20
70.00
$85.00
$100.00
$125.00
$70.00
$200.00
$145.00
$150.00
$215.00
$375.00
11/27/2017
11/27/2017
11/27/2017
12/26/2017
02/12/2018
02/12/2018
02/26/2018
02/26/2018
02/26/2018
03/05/2018
04/09/2018
04/09/2018
04/02/2018
04/16/2018
400.00
160.00
70.00
1,670.00
65,636.00
1,226.00
29,827.50
35.49
4,563.79
670.00
4,555.00
235.00
1,830.00
10,650.00
$450.00
$423.00
$423.00
$105.00
$79.25
$100.00
$98.00
$110.00
$185.00
$160.00
$130.00
$160.00
$157.00
$80.00
17X307
17X316
17X316
16X311
18X304
18X304
18X317
18X317
18X322
18X337
18X320
18X320
18X336
18X308
009184
009185
009185
009197
009222
009222
009229
009229
009232
009238
009262
009262
009265
009268
18X308
009268
18X318
009291
ned: 123,072.78 Total Amount: $11,599,333.05
78 Total Amount: $27,253,838.05
04/16/2018
05/29/2018
144.00
1,400.00
$145.00
$110.00
18Y002
009250
18Y001
009257
18Y008
009286
lanned: 19,417.00 Total Amount: $1,742,555.00
78 Total Amount: $27,253,838.05
03/12/2018
03/26/2018
05/14/2018
7,610.00
691.00
11,116.00
$88.00
$137.00
$88.00
17Z010
009187
18Z010
009210
18Z009
009223
17Z004
009252
18Z011
009258
18Z005
009287
18Z014
009289
18Z025
009298
d: 57,341.80 Total Amount: $4,796,196.92
78 Total Amount: $27,253,838.05
12/11/2017
01/22/2018
02/20/2018
03/19/2018
03/26/2018
05/14/2018
05/21/2018
07/09/2018
20,400.00
415.00
9,730.00
11,762.80
8,040.00
6,350.00
81.00
563.00
$75.55
$135.00
$70.00
$84.33
$94.40
$86.80
$474.00
$315.00
Bid Item Unit Price Tabulation Standard Items
Region: State Wide
Contracts awarded from 9/25/2017 thru 9/25/2018
Standard Item Number
Unit Of Measure Item Description
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
5766
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
TON
HMA CL. 3/8 IN. PG 64-22
HMA CL. 3/8 IN. PG 64-22
HMA CL. 3/8 IN. PG 64-22
HMA CL. 3/8 IN. PG 58H-22
HMA CL. 3/8 IN. PG 58H-22
HMA CL. 3/8 IN. PG 58H-22
HMA CL. 3/8 IN. PG
HMA CL. 3/8 IN. PG 58H-22
HMA CL. 3/8 IN. PG 58H-22
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64-28
HMA CL. 3/8 IN. PG 64-28 (BRIDGE DECK)
HMA CL. 3/8 IN. PG 64-28 (TRANSITION)
HMA CL. 3/8 IN. PG 64-22
HMA CL. 3/8 IN. PG 58V-22
HMA CL. 3/8 IN. PG 58V-22 (BRIDGES)
HMA CL. 3/8 IN. PG 58H-22
HMA CL. 3/8 IN. PG 58H-22 (BRIDGE)
HMA CL. 3/8 IN. PG 58H-22
HMA CL. 3/8 IN. PG 58H-22
HMA CL. 3/8 IN. PG 58H-22
HMA CL. 3/8 IN. PG 58H-22 (BRIDGE)
HMA CL. 3/8 IN. PG 58H-22
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64H-28 (BRIDGE)
HMA CL. 3/8 IN. PG 58H-22
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 76-28
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64H-28
Total records found: 4
5766
5766
TON
TON
HMA CL. 3/8 IN. PG 64H-28
HMA CL. 3/8 IN. PG 64H-28
d Items
/2018
Total records found: 45
Planned Quantity Low Bid
4.00
1,940.00
9,749.00
400.00
8,430.00
7,919.00
590.00
8,280.00
258.00
23,537.00
755.00
523.00
20,123.00
240.00
430.00
375.00
260.20
70.00
400.00
160.00
70.00
1,670.00
65,636.00
1,226.00
29,827.50
35.49
4,563.79
670.00
4,555.00
235.00
1,830.00
10,650.00
144.00
1,400.00
7,610.00
691.00
11,116.00
20,400.00
415.00
9,730.00
11,762.80
8,040.00
6,350.00
$404.52
$400.00
$145.00
$70.00
$97.50
$98.00
$140.00
$154.00
$550.00
$85.00
$100.00
$125.00
$70.00
$200.00
$145.00
$150.00
$215.00
$375.00
$450.00
$423.00
$423.00
$105.00
$79.25
$100.00
$98.00
$110.00
$185.00
$160.00
$130.00
$160.00
$157.00
$80.00
$145.00
$110.00
$88.00
$137.00
$88.00
$75.55
$135.00
$70.00
$84.33
$94.40
$86.80
81.00
563.00
$474.00
$315.00
USE
THIS
COLUMN
Basic Statistical Measures
1 INTRODUCTION
What area of study involves taxes, war, Caesar Augustus, insurance, gambling, the Domesday
Book, Guinness, geology and pavements (to name a few of the big ones)? Well, of course, it is
statistics.
For the most part, statistical methods as we use them today are largely a product of the 20 th
century. However, that said, the origins of this topic area (the ultimate in applied mathematics)
goes back to at least the Roman Empire and further back to the Babylonians and Egyptians. As
Steel and Torrie (1960) note, the term “statistics” likely started as state arithmetic needed to
levy taxes or wage war since counts of money and subjects are needed to do both. Have we
not been reminded from time-to-time (at least one a year) that Caesar Augustus decreed that
all the world should be taxed (an interesting concept). In 1086, William the Conqueror ordered
a statistical survey of England. This was an attempt to measure the land related wealth of the
country. The recorded volumes became known as the Domesday Book (a corruption of
Doomsday since the assessment of levies / taxes were irrevocable).
By the 14th century, Flemish shipping was using an early form of probability for ship insurance
(some might say it was really more a form of gambling). This appears to have evolved into the
current insurance industry (one wonders if Lloyd’s needs a refresher course in probability).
More specifically, Pascal (1623 – 1662) the French philosopher, mathematician, and physicist
and Fermat (1601-1665) also a French mathematician together developed the theory of
probability in 1654.
The normal distribution (or curve) was first published by de Moivre in 1733 but remained
unknown until his work was discovered in a library 200 years later (one must wonder if any of
our reports will enjoy the same fate). However, more well known was the “discovery” of the
normal distribution by Laplace (1749 – 1827), a French astronomer and mathematician, and
Gauss (1777 – 1855), a German mathematician. In fact, Gauss also developed the method of
least squares which we hold so dear in regression analysis. The normal curve is often referred
to as the “Gaussian Curve” (and rightfully so).
In the period 1830 – 1833, Charles Lyell (a Scottish geologist (1797 – 1875)) published three
volumes entitled Principles of Geology. He helped to identify layers of rocks according to fossils
(and associated fossil counts). He counted the strata names based on the proportions of fossils
as Pleistocene (most recent), Pliocene (majority recent), Miocene (minority recent) and Eocene
(dawn of the recent). “Recent” referred to live mollusks he observed in the oceans. The older,
deeper strata contained fewer fossils of living species. Apparently, Charles Darwin was a fan of
Lyell and much of Darwin’s eventual work was statistically based.
1
Other well-known statisticians include Pearson (1857 – 1936, British mathematician) and W.S.
Gosset (1876 –1937). Gosset was a student of Karl Pearson and was employed by Guinness
brewers. Apparently, Gosset had a pseudonym of “student” while working at Guinness. He
developed the “t-test” hence it often referred to as “Student’s t”.
There are many others such as R.A. Fisher who introduced the concept of analysis of variation
which is basically “an arithmetic process for partitioning a total sum of squares into
components associated with recognized sources of variations” (after Steel and Torrie 1960).
2 DEFINITIONS
POPULATION (Willenbrock 1976). All measurements or counts that are obtainable from all of
the objects that process some common characteristic. Example: a “population” of data would
be the pavement condition measured on all Interstate highways in a specific state.
SAMPLE (Willenbrock 1976). A set of measurements or counts that constitute a part (or all) of
the population.
RANDON SAMPLING (Willenbrock 1976). A sampling procedure whereby any one measurement
in the population is as likely to be included as any other.
BIASED SAMPLING (after Willenbrock 1976). A sampling procedure whereby certain individual
measurements have a greater chance of being included than others. Example: biased sampling
would be taking density measurements only at places on a base course that appeared to be
well compacted.
MEAN. Average of a group of measurements. The population mean is designated  (Greek mu)
x mean by σ (Greek sigma).
and a sample
MEDIAN. The number, in a set of numbers arranged in ascending order, that divides the set so
that half of the numbers are higher and half are lower.
RANGE. The largest measurement minus the smallest measurement in a group of data.
STANDARD DEVIATION. A measure of variation or dispersion of a group of data. Specifically,
the average of the squares of the numerical differences of each measurement (or observation)
from the mean. The population standard deviation is designated by “σ” and a sample standard
deviation by “s.”
HISTOGRAM. A graphical form of data presentation. A bar chart that shows in terms of area the
relative number of measurements of different classes. The width of the bar represents the
class interval, the height represents the number of measurements.
VARIABLE (ASTM 1985). A quantity to which any of the values in a given set may be assigned,
i.e., something on which measurements are made.
2
3 HISTOGRAM CONSTRUCTION
3.1 NUMBER OF CELLS (K) AFTER BLANK (1980))
The text has its own recommendations but they are rather complex mathematically. These are
simpler:
 Use 10 to 20 cells for >50 data points
 Use 6 to 10 cells for 1474 kg/ m 3 )
= 1.0000 – 0.0047
= 0.9953 or 99.53 percent
This makes sense: 1471 kg/m3 is way to the left of the proctor density plot (Figure 6) so you
would expect there to be a high probability that the density will be greater than 1471 kg/m 3.
2.2 2.2.2 EXAMPLE 2: PORTLAND CEMENT CONCRETE STRENGTH
If a distribution of PCC strength data is μ = 40 MPa and σ = 5.0 MPa, answer the following
questions:
a.
b.
c.
d.
What is the probability strength will be less than or equal to 50.0 MPa?
What is the probability strength will be less than or equal to 40.0 MPa?
What is the probability strength will be more than 30.0 MPa?
What is the probability strength will be less than or equal to 30.0 MPa?
Refer to Figure 8 for the results.
10
Figure 8. PCC strength probabilities.
From Table 1
z 30 
30  40
 2 (area less than Z = -2 equals 0.0228)
5.0
z 40 
40  40
 0 (area less than Z = -0 equals 0.5000)
5.0
z 50 
50  40
 2 (area less than Z = +2 equals 0.9772)
5.0
(a) P (strength < 50 Mpa) = 1.000 – 0.0228 = 0.9772 or 97.72% (b) P (strength < 40 Mpa) = 1.000 – 0.5000 = 0.5000 or 50.00% (c) P (strength > 30 Mpa) = 1.000 – 0.0228 = 0.9772 or 97.72%
(d) P (strength < 30 Mpa) = 0.0228 or 2.28% 11 2.3 EXAMPLE 3: READY-MIX CONCRETE The contractor claims that the batch plant can produce PCC mix with μ  35.0 MPa σ  3.0 MPa Assume that a somewhat unrealistic job specification states that an acceptable PCC must have a compressive strength no lower than 30.0 MPa and no higher than 40.0 MPa (after seven days of cure). Question: If the contractor sends 50 truckloads of this mix to the job site, how many of the trucks should be rejected if you know the real, potential compressive strength of each truckload (of course you cannot do this but what the heck)? Solution: Using z-statistic tables, you find that the total area under the standard normal distribution between 30.0 and 40.0 MPa is about 0.90 (recall that the maximum is 1.0000 under the curve). Thus, approximately 90 percent of the population will be between 30.0 and 40.0 Mpa. Thus, 1.0000 – 0.90 = 0.10 or about 10 percent of the 50 trucks (i.e. about 5 trucks) should be rejected. Try to match this solution by using Table 1. Hint: start by computing z 30 and z 40 , then use Table 1 2.4 EXAMPLE 4: AC BINDER CONTENTS Assume that the AC binder data in the introduction reading represents a population, i.e., x μ s σ What is the probability that the binder content will be greater than 5.0% and 5.3% for Mix A? Less than 6.0% and 6.3% for Mix B? Solution Mix A: x  5.04%  μ s  0.18%  σ z x  μ 5.0  5.04   0.22 σ 0.18 12 z 5.3  5.04  1.44 0.18 P(binder > 5.0% = 1.000 – 0.4129 = 0.5871 or  59%
P(binder > 5.3% = 1.000 – 0.9251 = 0.0749 or  7%
Mix B: x  6.09%  μ
s  0.18%  σ
z
6.0  6.09
 0.50
0.18
z
6.3  6.09
 1.167
0.18
P(binder < 6.0%) = 0.3085 or  31% P(binder < 6.3%) = 0.8784 or  88% or P(binder > 6.3% = 1.0000 – 0.8784 = 0.1216 or  12%
13
(use Table 1)
3 CENTRAL LIMIT THEOREM
The Central Limit Theorem (CLT) allows use of a normal distribution to make inferences about
the population from a non-normal sampling distribution.
Central Limit Theorem
“The population may have any unknown distribution with a mean  and a variance  2 . Take
samples of size n from the population. As the size n increases, the distribution of sample means
will approach a normal distribution with mean  and a variance  2 /n”

x for any μ ,σ 2
  x  N μ,σ 
as n
2
x
increases
where σ x = σ
n
=> standard deviation of the mean

Now, we can work with a distribution of sample averages x , not x.
z 
x μ
σ
n
3.1 EXAMPLE – AC BINDER CONTENTS
Mix A:
n  10
x  5.3%
assumeμ  5.0%
s  0.18%
assume  σ 
What is the chance that the average binder content  5.3% ?
z
5.3  5.0
0.18 10
 5.270
 Probbinder mean  5.3%  1.000  1.000  0%
What if s is doubled, i.e., σ  0.36%?
5.3  5.0
 2.635  Pbinder mean  5.3%   1.000  0.9958  0.0042  0.4%
0.36
10
What if s is multiplied by 4, i.e., σ  0.72% ?
z
14
5.3  5.0
 1.318  Pbinder mean  5.3%   1.000  0.9063  0.0937  9%
0.72
10
What if n  5, not10 :
z
z
z
z
5.3  5.0
 3.73  Pbinder mean  5.3  1.000  0.999  0.0001  0.01%
0.18
5
5.3  5.0
 1.86  Pbinder mean  5.3  1.000  .9686  0.0314  3.1%
0.36
5
5.3  5.0
 0.93  Pbinder mean  5.3  1.000  0.8238  0.1762  18%
0.72
5
4 THE T-DISTRIBUTION
4.1 WHY USE THE T-DISTRIBUTION?
When using inferential statistics we often have to decide which standard statistic to use.
Previously we have discussed the standard normal distribution (remember that in such a
distribution the average is z = 0, and the standard deviation is 1). That is convenient because
with a simple mathematical manipulation we can turn any normal distribution into the standard
normal distribution. It is advantageous because all the probabilities (area under the curve) for
the standard normal distribution have already been worked out and therefore we don’t have to
concern ourselves with taking integrals of the normal distribution equation (or remembering to
buy batteries for our fancy calculator or buy some lame calculator app in the app store).
Remember this equation to obtain the z-statistic for a sample mean:
𝑧𝑐𝑎𝑙𝑐 =
(𝑠𝑎𝑚𝑝𝑙𝑒 𝑚𝑒𝑎𝑛) − (ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑧𝑒𝑑 𝑚𝑒𝑎𝑛)
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟
To use this equation to make inferences about the population based on a sample we need to
know 3 things:
1. Sample mean. Pretty straightforward, we can get that if we have the sample values.
2. Hypothesized value of the population mean. This usually comes something we are testing.
𝜎
3. Standard error. This is calculated as 𝑛. This seems to be okay: n is sample size easy to
√
obtain) BUT σ is the population standard deviation. Often, we don’t know this.
One might ask, “how can we estimate population parameters if we have to know one of them,
the standard deviation, in order to make estimates?” There are 2 answers to that.
15
1. It may be that the population is well-known and that there exists (usually in the literature) a
fairly good estimate of the population standard deviation. Wishful, but possible.
2. It may not be known at all. If this is the case, we have to estimate the population standard
deviation using the best information we have: the sample standard deviation.
If #2 is what we choose then the standard error equation becomes:
𝑠
√𝑛
. This looks identical to
the previous standard error equation except that we substituted the SAMPLE standard
deviation (s) for the POPULATION standard deviation (σ). But the sample standard deviation is
only an estimate of the population standard deviation and is not likely to be a perfect estimate.
There must be some penalty in accuracy to pay for using such a gross estimate.
There is a penalty to pay…so to speak. We no longer get to use the standard normal
distribution to figure out probabilities. Instead, we use a slightly different distribution called
“Student’s t distribution” or simply the “t distribution”. The t distribution looks very much like
the standard normal distribution for large sample sizes but as sample size gets smaller the t
distribution has noticeably fatter tails than the standard normal distribution. In other words,
there is more area under the curve at the tails in the t distribution. What happens then is that if
we use the t distribution instead of the standard normal distribution our resulting confidence
intervals (covered in the next lesson) become larger meaning that we are less certain of our
conclusion. That’s our penalty.
In other words: The t-distribution is used to test sample means when the population variance
and population standard deviation are not known. The “t-statistic” is quite similar to the “zstatistic” but also includes consideration of the sample size (n).
Where
z
t
x μ
σ
(this is the standard normal z-statistic)
x μ
s
n
(this is the equivalent t-statistic)
Table 2 is used to provide a partial listing of t-values.
16
Table 2. Partial t-Distribution Table for Two-Tailed Test (after Steel and Torrie 1960)
Degrees
of
Freedom
Probability of a larger t
1
2
3
4
5
6
7
8
9
10
15
20
25
30

0.10
0.05
0.02
0.01
6.314
2.920
2.353
2.132
2.015
1.943
1.895
1.860
1.833
1.812
1.753
1.725
1.708
1.697
1.645
12.706
4.303
3.182
2.776
2.571
2.447
2.365
2.306
2.262
2.228
2.131
2.086
2.060
2.042
1.960
31.821
6.965
4.541
3.747
3.365
3.143
2.998
2.896
2.821
2.764
2.602
2.528
2.485
2.457
2.326
63.657
9.925
5.841
4.604
4.032
3.707
3.499
3.355
3.250
3.169
2.947
2.845
2.787
2.750
2.576
Probability is one-half for a one-tail test, e.g. if df = 4 and  = 5% for one-tail, then t = 2.132
(i.e., use column with P = 10% for two-tail condition).
This concept will be illustrated in more detail in the next paragraph.
(a)
How is the t-distribution formed?
 determine sample size n
 take one sample (size n) and compute x and s 2
Calculate t 
x μ
s
n
Where s
= standard deviation of sample means


n
think of t as a conversion factor to obtain probabilities
plot t on a histogram of t values
Repeat for k samples
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Example-AC Binder Contents
Mix A n = 10
x = 5.04%
S = 0.18%
18
What is the chance that the average binder content  5.2%?
t calc 

x -μ
s
n
(assume μ = 5.0%)
5.2  5.0
 3.51
0.18
10
One-tail test and degrees of freedom (v) = n-1 = 9
t table From Table 2, v = 9, shows α = 0.01
for t = 3.250, thus Prob