University of Washington Hypothesis Testing Questions

You are testing the tension strength of steel reinforcing bars you are making. You want to know if the minimum yield strength of your reinforcing bars is greater than 67,000 psi. You run 20 tests. You set a null hypothesis (H0) of strength = 67,000 psi and an alternative hypothesis (H1) that strength > 67,000 psi. You run a one-tailed t-test (since you are concerned with strengths GREATER than 67,000 psi only). Results indicate a p-value of 0.03 for the null hypothesis (H0). What does this p-value actually mean?

Want to find out from others what they say it means?

Read this article: The ASA’s Statement on p Values Context Process and Purpose.pdf (PDF is attached)

Listen to this podcast: Planet Money – Episode 677: The Experiment ExperimentLinks to an external site.

https://www.npr.org/sections/money/2016/01/15/4632…

All that (and the course material) should arm you with enough information to say what a reported p-value actually means.

Advice: you do not need any statistical calculations to answer this question. I have given you all you need. Rather, I want you to concentrate on what the p-value actually means. I ask this because so many are incorrect on its meaning. No worries – the ASA’s statement and the Planet Money episode try and drive home its true meaning.

Lecture notes:

* Logic of Hypothesis Testing (p. 369-392) – Section XI.1 through XI.11. If you want to break it up, do this for week 4.

* Course Notes: Hypothesis Testing.pdf. Another title for this could be “Using statistics to examine the difference in means.” Traditionally, the way this is set up is with two hypotheses: the null and alternative.

HYPOTHESIS TESTING
1 INTRODUCTION
Hypothesis testing is a way in which statistical methods can be used to help in the decision making
process. Such testing considers the mean and standard deviation of a group of data, the
confidence level (a probability statement) and something about the population being sampled.
Hypothesis testing is extremely helpful in performing multiple regression analysis and hence it is
important for you to understand the basics.
Note: the methods described here assume that the data being considered are normally distributed.
While there is no certainty, it is often true that construction data are normally distributed (for
example, soil density, concrete strength, pipe tolerances). Often, we make the normal assumption
without checking. It is always good to check.
2 HYPOTHESES
Webster’s Seventh New Collegiate Dictionary defines hypothesis as “…a tentative assumption
made in order to draw out and test its logical or empirical consequences…an assumption or
concession made for the sake of argument…” You can begin to see the problem in explaining
hypothesis testing.
There are always two hypotheses for any statistical test (Blank 1980). These hypotheses are
H  null hypothesis (most important)
0
H  alternative hypothesis
1
What is about to be presented is one of the fundamental problems in statistics which is the use of
“double negative” statements. Any hypothesis must be tested statistically to be rejected or not
rejected. “Not rejected” is a statistical way of accepting something. Think of it as the equivalent of
“I can’t say no” rather than simply “yes.” The first statement is slightly less committed.
The hypotheses ( H or H ) can result in two types of errors if the wrong one is selected, as shown
0
1
in Table 1. The probability of the Type I and II errors is very important, since it determines how
carefully you must distinguish between true and false hypotheses.
(This is an area in which statistical “games” can be played, so you need to be very careful.) These
probabilities are (after Blank 1980).
Probability of a Type I error = α
1
Probability of a Type II error = β
Table 1. Types of Hypothesis Errors
“The Actual Decision”
“The Truth”
Reject H
Accept H
0
0
H true
Type I Error
Correct!
0
H
0 false
Correct!
Type II Error
The general form for calculating the z-statistic for hypothesis testing is
𝑧𝑐𝑎𝑙𝑐 =
where
sample mean
(𝑠𝑎𝑚𝑝𝑙𝑒 𝑚𝑒𝑎𝑛) − (ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑧𝑒𝑑 𝑣𝑎𝑙𝑢𝑒)
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟
=
x
hypothesized
value
=
μ (sometimes assumed to = 0 in regression
hypothesis testing)
standard error
=
σ
sample size
=
 standard deviation of means of
n
random samples of size n from a “parent”
population with standard deviation σ .
Standard error is sometimes designated
σ .
x
n
The same general form applies to the t-statistic for hypothesis testing when none of
the population statistics ( μ, σ ) are known:
𝑡𝑐𝑎𝑙𝑐 =
where
sample mean
(𝑠𝑎𝑚𝑝𝑙𝑒 𝑚𝑒𝑎𝑛) − (ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑧𝑒𝑑 𝑣𝑎𝑙𝑢𝑒)
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟
=
x
hypothesized
value
=
μ (again, sometimes an assumed or stated
value)
standard error
=
s
n
s
2
x
2.1 EXAMPLE 1: PCC MIX
For this example, use the data originally shown in the Data Distributions notes (Figure 1).
Figure 1. Histogram and the normal distribution for Portland cement concrete 28-day
compressive strength data (Willenbrock 1976).
This contractor states that the batch plant has produced a mix in the past of
μ  33.26 MPa
σ  2.67 MPa
(Since these are population statistics, you can assume that these data were collected over a long
period of time.)
On the job you take six samples (cylinders) with the result that x = 31.36 MPa.
Question: Is the contractor correct?
Solution: Assume that the data are normally distributed and use hypothesis testing.
3
H : μ  33.26 MPa
0
H : μ  33.26 MPa
1
z
calc

x μ
σ
n

31.36  33.26
2.67
6
 1.74
 1.65 (for Type I error (or α)  5%)
critical
(to get this value, use the z-statistic table at end of these notes. First, this is a one-tailed ztest since all we want to know is if the sampled values indicate a mean of 33.26. Therefore,
since the table is for a one-tailed test, go into the table and find 0.0500, or closest to it for a
negative z-statistic. This ends up being 0.0495 for a z-statistic of -1.65).
z
Since 𝑧𝑐𝑎𝑙𝑐 < 𝑧𝑐𝑟𝑖𝑡 , reject H0 To visualize this, look at the graph below. This is the standard normal curve. Here, 0 represents the mean from the PCC plant (33.26 MPa). If the sample value is within the blue shaded area, then one cannot reject the hypothesis that the sample and actual means are the same or the sample mean is MORE than the actual mean. What is left, the white area, is only 4.95% (our closest estimate of 5% α from the Table) of the area under the standard normal curve. So, if our sample value of 31.36 MPa is in this white area, then we have less than a 5% chance of a Type I (α) error (to reject H0 when it is actually true) probability. Our calucated z-statistic was -1.74, which is less than -1.65 and resides in the white part of the area under the standard normal curve. Therefore, reject H0. 4 Thus, for your job, you must judge the contractor’s claim to be incorrect. Note that z= -1.65 is equivalent to x  31.46 MPa 2.2 EXAMPLE 2: PCC MIX The PCC mix contractor claims the following: PCC mix  27.58 MPa (28-day compressive strength) You take a random sample of five cylinders and cure them for 28 days (n=5). The results: x = 25.79 MPa s = 2.69 MPa n=5 Question: If you are willing to accept a 5 percent chance of a Type I error (i.e., rejecting a true ), should you believe the contractor? Solution: 𝐻0 : 𝜇 ≥ 27.58 𝑀𝑃𝑎 (𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠) 𝐻1 : 𝜇 < 27.58 𝑀𝑃𝑎 (𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑣𝑒 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠) 𝑡𝑐𝑎𝑙𝑐 = 𝑥̅ − 𝜇 25.79 − 27.58 = = −1.48 𝑠 2.69 √𝑛 √5 𝑡𝑐𝑟𝑖𝑡 (𝛼 = 5%) = −2.132 (𝑜𝑛𝑒 𝑡𝑎𝑖𝑙 𝑤𝑖𝑡ℎ 𝑣 = 𝑛 − 1, 𝑜𝑟 4 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚) Therefore, you accept H0 since: t calc  1.48  2.132  t critical (t-calc is to the right of t-crit in the plot) You have no “statistical” reason to doubt the contractor’s claim. Refer to Figure 3.2 (a) which further illustrates this example. Visualize this in the following graph. Notice that the line in the graph is set at a t-statistic that divides the area under the curve into 0.05 (5%) and 0.95 (95%) portions. In this case, the 0 of the graph represents the claimed mean of 27.58 MPa and the line at a t-statistic of -2.132 represents the spot on the graph where a sample mean of that value represents a 5% chance of a Type I (α) error (to reject H0 when it is actually true) probability. Our calucated t-statistic was -1.48, which is more than -2.132 and resides in the red part of the area under the t distribution. Therefore, accept 5 H0. Note: if we chose to reject H0 with this evidence, there would be a higher likelihood of a Type I error. About 10.65% it turns out. Note that if the Type I error (rejecting a true H ) were reduced to a 1 percent chance, then 0 t critical(@1%)  3.747 (one  tail  1% with 4 degrees of freedom, refer to Table 7). This actually moves the critical value TO THE LEFT (see below). This would not change the conclusion since -1.48 is still in the red area of the graph (to the right of t-crit). 6 Thus, you are even more unwilling to accept the alternate hypothesis (H1) that the contractor’s claim was incorrect (note that now the red area is even farther to the left in the above plot). Note that the Type I error protects against rejecting a true null hypothesis. In other words, you can select a low Type I error level so that it is difficult to reject the null hypothesis. However, as the Type I error level decreases, the Type II error level increases (not rejecting a false H0). It is not easy to illustrate the calculation of the Type II error (β), but this example is a good case since the Type I error level of about 11 percent would be needed to reject the null hypothesis. Often the Type I error is termed the “seller’s risk” and the Type II error the “buyer’s risk”. For the example, the lower the Type I error the lower the risk of the contractor. Correspondingly, the Type II error increases the risk of the DOT accepting PCC of lower than specified quality (again not rejecting a false H0). Needless to say, a balance between Type I and Type II errors is needed (but not necessarily the same number or value because one error type may be more important than another) in developing statistically based materials “acceptance plans”. What happens when you ignore the Type II (β) error. At least one illustration of ignoring the Type II error (β) is appropriate. First, calculate the value of x , which corresponds to tcrit = -2.132 (for α = 5%). This value is the one that separates the rejection and acceptance region for H0 (µ ≥ 27.58 MPa). This is as we have done in the past. Just finding tcrit and the corresponding sample average. 7 𝑥̅ − 𝜇 𝑥̅ − 27.58 𝑡𝑐𝑟𝑖𝑡 = 𝑠 = −2.13 = 2.69 √𝑛 √5 𝑥̅ = 25.01 𝑀𝑃𝑎 Next, remember that the sample average was 25.79 MPa for the five samples taken. What if this were the actual population average (and we just didn’t know it)? What chance would there be that we would accept the argument that the population mean is 27.58 MPa even though the true population mean is 25.79 MPa? Doing this would leave us thinking that the population mean was 27.58 MPa even though it really was just 25.79 MPa. Here’s how we figure that one out. The value of the Type II error (β) is the area under the curve (or distribution) with μ  25.79 MPa and σ  269 MPa but within the acceptance region of the original H distribution μ  27.58MPa  . This is illustrated in Figure 3.2 (b). 0 t 𝑐𝑎𝑙𝑐 = 𝑥̅ − 𝜇 25.01 − 25.79 = = −0.64 𝑠 2.69 √𝑛 √5 If you look up -0.64 in a one-tailed t distribution table with 4 degrees of freedom you will find that it corresponds to about 72% of the area under the curve being to the right of it (see picture on next page). Therefore, there is a 72% chance of accepting a false H μ  27.58 MPa  if the true 0 population mean μ  = 25.79 MPa. You can see that the β will change as the correct population mean changes. Clearly, this level of β is quite high. Thus, the DOT’s risk (the “buyer’s risk”) is too high. A balance in setting , β and sample size is very important in the proper use of these kinds of 8 statistical tests. This leads to another important area of statistics, generally called acceptance testing and operating characteristic curves. β = 72% 2.2.1 Dependence Between α, βand n There exists a relationship between the two probability values α and β and n for any hypothesis. The value of any one is found from the other two. Usually α and n are most critical. Three cases are illustrated: Case 1: Given α and n, find β 9 β represents the area in the acceptance region in H , but under the H curve. 0 1 Case 2: Given β and n, find α. Case 3: Given α and β, find n.  Illustration Case 1: Rutting – gravel road H : μ  50mm (typically observed after one year) 0 H : μ  50mm 1 x  60mm  s  10mm  sample results n  16  t x  μ 60  50   4.0 s 10 n 16 For α  0.05% and ν  16 - 1  15 = the one sided critical region is t table  1.753. Thus, reject H at α  0.05% 0 The corresponding rutting value at the critical boundary:  10  x - 50  1.753  x  1.753   50  54.38mm 10  16  16 10 Find area below 54.38 for H distribution 1 54.38  60 t  2.248 10 16 β  P t  -2.24  0.021 (can use Table 2.2) Summary: H 0 is rejected at α  5% , thus rutting more than “typical” (common sense would suggest this might be the case). If true average is 60mm and H 0 not rejected, then Type II error is present. There is only a 2.1% chance of accepting a false μ  50mm hypothesis. Further, β will change as correct mean shifts from 60mm. Operating Characteristic (OC) curve is a graphical representation of the values for α , β and n for a particular H 0 and any H 1 . This will be illustrated as follows. The H 0 value μ 0 has a probability of 1-α of being accepted if μ  μ 0 is correct. As correct μ moves away from μ 0 , chances of accepting false H 0 diminish. 11 2.3 EXAMPLE 3: WSDOT/INDUSTRY PCC TESTING WSDOT and industry representatives jointly tested fresh concrete delivered to the South Seattle Community College (the testing site) a few years ago. One purpose of this activity was to see how test results compared for different testing teams. The tests performed by all of the teams included both fresh and hardened mix properties: slump, air content, unit weight and compressive strength. The ready-mix was specified to conform to a standard WSDOT mix (“AX” mix, the WSDOT Standard Specifications have since been changed). This mix was delivered to the testing site as a 5.2 sack mix with fly ash and ¾ in. maximum course aggregate size; additives included an air entraining agent and a water reducer. The per cubic yard batch weights were:        Cement: 488 lb Fly ash: 152 lb Fine aggregate: 1,130 lb Course aggregate: 1,864 lb Water: 32 gal Water reducer : 24.4 oz Air entraining agent: 6.4 oz The specified 28-day design compressive strength was 4000 psi with a maximum slump of 3 inches (vibrated concrete) and an air content of 5 percent (+ 1 ½ percent). Each team picked a number at random (1 through 20). Team 1 would then obtain their PCC from the truck, Team 2 next and so forth (i.e., there should be no bias as to when the test teams received their material for testing). The team test results are shown in Table 2 with basic summary statistics in Table 3. The hypothesis test used is a means test for two independent samples with the population standard deviation unknown and for small samples. A small sample implies the number of testing teams were fewer than 30. The hypothesis test formulas used are shown in a box and the hypothesis test results in Table 4. The results shown in Table 4 also include a comparison of Tests 1-8 and 9-20. This was done since the PCC mix, as discharged from the truck, apparently had somewhat different fresh mix properties as characterized by slump. The results shown in Table 4 indicate that there were no significant test differences between WSDOT and industry test teams for measurements of slump, air content and unit weight. There were significant differences for the compressive strength results. The strength tests were organized as follows:   Cylinders prepared by all teams and tested at the WSDOT lab. Cylinders prepared by all teams and tested at a commercial lab 12 The results in Table 4 show a significant difference between the two laboratories (the WSDOT results were higher). A final set of hypothesis tests were performed to compare the test results against “fixed” values (or limits). This was done for slump, air content and compressive strength. The associated and necessary formulas are shown in a box with the results in Table 5. The hypotheses are (illustrated in Figure 2):  Slump H0 H1  μ = 3 in. (specified maximum slump for “AX” vibrated PCC) μ  3 in. (i.e., critical condition is more slump not less) Air Content H0 H1  : : : : μ = 5% (specification target air content) μ  5% Compressive strength H0 H1 : : μ = 4000 psi (specification minimum 28-day compressive strength) μ < 4000 psi The results in Table 5 indicate that the air content is within the acceptance region which is to say that the null hypothesis H 0  is accepted (i.e., there is a statistical basis for accepting the fact that the air content is, in essence, 5 percent). The compressive strength is also in the acceptance region (i.e., accept H 0 ). Naturally, this can be observed by inspection but was included to illustrate the calculation process (i.e., all means were greater than 4000 psi). Finally, the slump results are in the critical region (accept H1 ). This indicates that the slump results exceed the maximum value of 3 inches. Again, by inspection of the data, this result is rather obvious. What the above hypothesis tests do not do is indicate whether the difference between a 3 inch slump or say a 4.44 inch slump is structurally important. 13 Table 2. Test Summary: WSDOT/Industry Concrete Testing Program Air Unit Tester Slump Content Weight (a) Affiliation (in.) (%) (pcf) No. 1 Industry 4.75 4.9 148.45 2 Industry 4.75 4.5 147.20 4 Industry 5.25 5.4 147.24 5 WSDOT 4.75 4.9 148.69 6 WSDOT 5.00 5.6 147.72 7 Industry 4.75 5.5 147.44 8 WSDOT 5.50 5.0 148.15 9 WSDOT 4.00 5.3 149.39 10 Industry 4.00 5.4 146.12 11 Industry 4.50 4.6 148.39 12 WSDOT 4.25 4.8 136.90 13 Industry 4.00 5.0 147.68 14 WSDOT 4.00 5.1 163.33 15 Industry 3.75 4.8 146.70 16 WSDOT 4.00 5.0 147.83 17 WSDOT 4.00 5.0 148.05 20 Industry 4.25 4.7 146.20 Note: no testers selected numbers 3, 18 and 19 14 Results Compressive Strength (psi) Commercial Lab WSDOT Lab No. 1 No. 2 No. 1 No. 2 4280 4040 4680 5000 4360 4210 4700 4750 4180 4310 46.50 5050 4340 4390 4770 5190 4530 4400 4700 4740 4620 4400 4700 4700 4740 4730 4770 4730 4590 4590 4980 5160 4280 4320 4520 4670 4320 4280 4050 3920 4570 4410 5140 5120 4390 4420 4610 5110 4890 4670 5390 5250 4350 4380 5110 5180 4500 4590 4820 5050 4700 4730 5280 5640 4710 4780 4680 5050 Table 3. Basic Statistics for PCC Test Results Coefficient of Variation (s/ x )100 Test Data Set Sample Size (n) Mean (x) Standard Deviation (s) Slump (in.) (Basic Groupings WSDOT Industry Overall 8 9 17 4.38 4.44 4.41 0.58 0.48 0.51 13.0% 10.8% 11.5% Slump (in.) (Sequential Groupings Tests 1-8 Tests 9-20 7 10 4.96 4.08 0.30 0.21 6.1% 5.1% Air (%) (Basic Groupings) WSDOT Industry Overall 8 9 17 5.09 4.98 5.03 0.25 0.37 0.32 5.0% 7.5% 6.3% Air (%) (Sequential Groupings) Tests 1-8 Tests 9-20 7 10 5.11 4.97 0.40 0.25 7.8% 5.1% Unit Weight (pcf) WSDOT Industry Overall 8 9 17 148.8 147.3 148.0 7.1 0.8 4.8 4.8% 0.6% 3.3% Unit Weight (pcf) with “outliers” removed WSDOT Industry Overall 6 9 15 148.03 147.3 147.7 0.6 0.8 0.9 0.4% 0.6% 0.6% Compressive Strength (psi) (Basic Groupings) WSDOT Cyl/WSDOT Lab WSDOT Cly/Commercial Lab Industry Cyl/WSDOT Lab Industry Cly/Commercial Lab Overall/WSDOT Lab Overall/Commercial Lab Overall 16 16 18 18 5046 4586 4729 4368 275 153 340 181 5.5% 3.3% 7.2% 4.1% 34 34 68 4878 4471 4674 346 199 347 7.1% 4.5% 7.4% Test 1-8 14 14 20 20 4795 4395 4936 4524 163 202 425 183 3.4% 4.6% 8.6% 4.1% Compressive Strength (psi) (Sequential Groupings) Test 920 WSDOT Lab Commercial Lab WSDOT Lab Commercial Lab 15 Formulas Used for Basic Statistical Analysis (Blank 1980 and Steel & Torrie 1960) 1. Statistical tests reported are “means test for two independent samples with population standard deviation unknown and small samples” (small sample implies that the WSDOT and Industry testers were less than 30, which was the case). 2. Null hypothesis is H 0 : μ w  μ I Alternative hypothesis is H 1 : μ w  μ I Where μ w  population mean for a specific test for WSDOT testers μ I  population mean for a specific test for Industry testers 3 t-statistic t xw  xI sd ν  nw  nI  2 where: x w  xI  nw  nI  sd  = v = sample mean for a specific test for WSDOT testers sample mean for a specific test for Industry testers sample size (WSDOT) sample size (Industry) standard deviation of the difference of the sample means 1/ 2  s w 2 n w  1  s I 2 n I  1  n w  n I      n n      n  1  n  1 w I  w I   degrees of freedom = n w  1  n I  1  n w  n I  2 References: Blank, Leland, Statistical Procedures for Engineering, Management, and Science, McGraw-Hill, 1980, p. 381-383 Steel, Robert and Torrie, James, Principles and Procedures of Statistics, McGraw-Hill, 1960, p.74. 16 Table 4. Results of Hypothesis Testing for Various PCC Tests t-statistic Critical Region (a) Calculated Test Comparison Slump (Basic) Slump (Sequential) Air Content (Basic) Air Content (Sequential) Unit Weight WSDOT = Industry -0.23 -2.131 No significant difference Tests 1-8 = Tests 9-20 +7.228 +2.131 Significant difference WSDOT = Industry +0.637 +2.131 No significant difference Tests 1-8 = Tests 9-20 +0.892 +2.131 No significant difference WSDOT = Industry +0.632 +2.131 No significant difference WSDOT = Industry +2.597 +2.160 Significant difference WSDOT = Industry +2.964 +2.038 Significant difference WSDOT = Industry +3.766 +2.038 Significant difference Tests 1-8 = Tests 9-20 -1.178 -2.038 No significant difference Tests 1-8 = Tests 9-20 -1.939 -2.038 No significant difference WSDOT Lab = Commercial Lab +5.946 +2.000 Significant difference Unit Weight Without Outliers Compressive Strength (WSDOT Lab) Compressive Strength (Commercial Lab) Compressive Strength (WSDOT Lab) Compressive Strength (Commercial Lab) Compressive Strength (All Tests) (α  0.05) 17 Conclusion(b) Formulas Used for Comparing Test Results to Fixed Values (Blank 1980 and Steel and Torrie 1960) 1. Statistical tests reported are “means test for one sample with population standard deviation unknown and a small sample”. 2. Null hypothesis is H 0 : μ  fixed value Alternative hypothesis is H 1 : μ w  fixed value (or < or > fixed value)
Where:
3.
μ
=
population mean for a specific test and tester group
Fixed value =
76 mm. for maximum specified slump ( H1 >76 mm.)
=
5% for air content (assumed based on WSDOT Spec.
6-02.3(2)A for cast-in-place concrete above the finished
ground line) H1  5%
=
27.6 MPa for 28-day compressive strength
t-statistic
t
x -μ 0
s
n
ν  n 1
where:
x
μ
0
s
n
s
n
=
sample mean for a specific test or tester groups
=
stated mean population value in H
=
=
standard deviation of the sample
sample size
=
standard error (sometimes designated s )
x
0
References:
Blank, Leland, Statistical Procedures for Engineering, Management, and Science, McGraw-Hill,
1980, p. 377.
Steel, Robert and Torrie, James, Principles and Procedures of Statistics, McGraw-Hill, 1960,
p. 19.
18
Figure 2. Illustrated results for Table 5Table 4.
19
Table 5. Results of Hypothesis Testing for Fixed Population Values
t-statistic
Critical Range (a)
Test
Air Content
(Basic)
Notes
(a)
(b)
Comparison
Overall = 5%
Calculated
+0.389
WSDOT = 5%
Industry = 5%
α  0.05
+2.120
Conclusion (b)
No significant difference
+1.018
+2.365
No significant difference
-0.161
-2.306
No significant difference
Critical region defined by the t-statistic for a two tail Type I error of 5% (α  0.05) for v  n  1 degrees of
freedom.
Conclusion based on hypothesis test described in Table 3.6
t-statistic
Critical Range (c)
Test
Slump
(Basic)
Compressive
Strength (d)
Comparison
Overall = 76 mm.
Calculated
+11.604
WSDOT = 76 mm.
α  0.05
> +1.746
Conclusion
Significant difference
+7.025
> +1.895
Significant difference
Industry = 76 mm.
+9.006
> +1.860
Significant difference
Overall = 27.6 MPa
+14.796
< -1.693 No Significant difference WSDOT = 27.6 MPa +15.215 < -1.753 No Significant difference Industry = 27.6 MPa +9.097 ÿ"%ÿ-4"4"ÿ4!"ÿ%ÿ!ÿ(.!%=."ÿ"8(.!!ÿ>ÿ!ÿ”8(“:”!.ÿ
ÿ6%”(%ÿ!=ÿ4′(4″%%ÿ”%ÿ!”ÿ::ÿ!:ÿ(!”%ÿ
!%ÿ-“..ÿ!%ÿ%”%ÿ$ÿ4″.(ÿ(”5″ÿ4″%”ÿ:%”(%;ÿ4%ÿ4!(”ÿ”%ÿÿ
:”ÿ”!.ÿ!=ÿ4″ÿ.ÿ>ÿ4′(4″%%ÿ”%ÿ4!ÿ%ÿ'(!.ÿ>ÿ!ÿ’,
.”5″.ÿ”8
?@A
012345672841ÿ
ÿÿÿÿ

 !”#$%&ÿ()ÿ*+,-.+”/ÿ0+1$&+23$+-,
45667ÿ9:;?0%1@&+2%ÿ$!%ÿ/-A+@ÿ2BÿC!+@!ÿ+$ÿ@”,ÿ2%ÿ@-,@/3D%Dÿ$!”$ÿ1-.%-,%ÿ@”,ÿD+1$+,A3+1!ÿ
2%$C%%,ÿ$C-ÿ$!+,A1
E?F$”$%ÿC!%$!%&ÿ&”,D-.ÿ”11+A,.%,$ÿ%,13&%1ÿ$!”$ÿ”//ÿ3,@-,$&-//%Dÿ1-3&@%1ÿ-Gÿ
H”&+”$+-,ÿC+//ÿ2%ÿ%I3″/
J?0%K,%ÿ#&%@+1%/BÿC!”$ÿ$!%ÿ#&-2″2+/+$Bÿ+1ÿ$!”$ÿ+1ÿ@-.#3$%Dÿ$-ÿ&%”@!ÿ$!%ÿ
@-,@/31+-,ÿ$!”$ÿ”ÿD+GG%&%,@%ÿ+1ÿ,-$ÿD3%ÿ$-ÿ@!”,@%
L?0+1$+,A3+1!ÿ2%$C%%,ÿ$!%ÿ#&-2″2+/+$Bÿ-Gÿ”,ÿ%H%,$ÿ”,Dÿ$!%ÿ#&-2″2+/+$Bÿ-Gÿ”ÿ1$”$%ÿ-Gÿ
$!%ÿC-&/D
(?0%K,%ÿM,3//ÿ!B#-$!%1+1N
O?*%ÿ”2/%ÿ$-ÿD%$%&.+,%ÿ$!%ÿ,3//ÿ!B#-$!%1+1ÿG&-.ÿ”ÿD%1@&+#$+-,ÿ-Gÿ”,ÿ%P#%&+.%,$
Q?0%K,%ÿM”/$%&,”$+H%ÿ!B#-$!%1+1Nÿ
R!%ÿ1$”$+1$+@+”,ÿS?ÿT+1!%&ÿ%P#/”+,%Dÿ$!%ÿ@-,@%#$ÿ-Gÿ!B#-$!%1+1ÿ$%1$+,AÿC+$!ÿ”ÿ1$-&Bÿ
-Gÿ”ÿ/”DBÿ$”1$+,Aÿ$%”?ÿU%&%ÿC%ÿC+//ÿ#&%1%,$ÿ”,ÿ%P”.#/%ÿ2″1%Dÿ-,ÿV”.%1ÿ*-,DÿC!-ÿ
+,1+1$%Dÿ$!”$ÿ.”&$+,+1ÿ1!-3/Dÿ2%ÿ1!”W%,ÿ&”$!%&ÿ$!”,ÿ1$+&&%D?ÿX%$Y1ÿ@-,1+D%&ÿ”ÿ
!B#-$!%$+@”/ÿ%P#%&+.%,$ÿ$-ÿD%$%&.+,%ÿC!%$!%&ÿZ&?ÿ*-,Dÿ@”,ÿ$%//ÿ$!%ÿD+GG%&%,@%ÿ
2%$C%%,ÿ”ÿ1!”W%,ÿ”,Dÿ”ÿ1$+&&%Dÿ.”&$+,+?ÿF3##-1%ÿC%ÿA”H%ÿZ&?ÿ*-,Dÿ”ÿ1%&+%1ÿ-Gÿ>Oÿ
$”1$%ÿ$%1$1?ÿ[,ÿ%”@!ÿ$%1$\ÿC%ÿ]+##%Dÿ”ÿG”+&ÿ@-+,ÿ$-ÿD%$%&.+,%ÿC!%$!%&ÿ$-ÿ1$+&ÿ-&ÿ1!”W%ÿ
$!%ÿ.”&$+,+?ÿR!%,ÿC%ÿ#&%1%,$%Dÿ$!%ÿ.”&$+,+ÿ$-ÿZ&?ÿ*-,Dÿ”,Dÿ”1W%Dÿ!+.ÿ$-ÿD%@+D%ÿ
C!%$!%&ÿ+$ÿC”1ÿ1!”W%,ÿ-&ÿ1$+&&%D?ÿX%$Y1ÿ1″BÿZ&?ÿ*-,DÿC”1ÿ@-&&%@$ÿ-,ÿ>Jÿ-Gÿ$!%ÿ>Oÿ
$”1$%ÿ$%1$1?ÿ0-%1ÿ$!+1ÿ#&-H%ÿ$!”$ÿZ&?ÿ*-,Dÿ!”1ÿ”$ÿ/%”1$ÿ1-.%ÿ”2+/+$Bÿ$-ÿ$%//ÿC!%$!%&ÿ
$!%ÿ.”&$+,+ÿC”1ÿ1!”W%,ÿ-&ÿ1$+&&%D^
R!+1ÿ&%13/$ÿD-%1ÿ,-$ÿ#&-H%ÿ$!”$ÿ!%ÿD-%1_ÿ+$ÿ@-3/Dÿ2%ÿ!%ÿC”1ÿ̀31$ÿ/3@WBÿ”,Dÿ
A3%11%Dÿ&+A!$ÿ>Jÿ-3$ÿ-Gÿ>Oÿ$+.%1?ÿ*3$ÿ!-Cÿ#/”31+2/%ÿ+1ÿ$!%ÿ%P#/”,”$+-,ÿ$!”$ÿ!%ÿC”1ÿ
3̀1$ÿ/3@WB^ÿR-ÿ”11%11ÿ+$1ÿ#/”31+2+/+$B\ÿC%ÿD%$%&.+,%ÿ$!%ÿ#&-2″2+/+$Bÿ$!”$ÿ1-.%-,%ÿ
C!-ÿC”1ÿ̀31$ÿA3%11+,AÿC-3/Dÿ2%ÿ@-&&%@$ÿ>Ja>Oÿ$+.%1ÿ-&ÿ.-&%?ÿR!+1ÿ#&-2″2+/+$Bÿ@”,ÿ
2%ÿ@-.#3$%DÿG&-.ÿ$!%ÿ2+,-.+”/ÿD+1$&+23$+-,ÿ”,Dÿ$!%ÿ2+,-.+”/ÿD+1$&+23$+-,ÿ
@”/@3/”$-&ÿ1!-C1ÿ+$ÿ$-ÿ2%ÿb?b>bO?ÿR!+1ÿ+1ÿ”ÿ#&%$$Bÿ/-Cÿ#&-2″2+/+$B\ÿ”,Dÿ$!%&%G-&%ÿ
cde
0123143ÿ61789ÿ 3ÿ1ÿ3ÿ3ÿ87ÿ1ÿ3ÿ13ÿÿ1ÿ213ÿ230ÿ17ÿ1ÿÿÿ
3ÿ633ÿ70ÿ73004ÿ1ÿ33ÿÿ149ÿ6 0ÿ3ÿ87ÿ1ÿ3ÿ4ÿ388ÿ6 3 3ÿ
3ÿ94ÿ6 0ÿ0 34ÿ1ÿ039ÿ 3ÿ1 300ÿ ÿ3ÿ6 0ÿ73004ÿ0ÿ41ÿ134ÿ
803ÿ7ÿ1409383ÿ917ÿ0ÿ0ÿ14ÿÿ 3313ÿ33ÿ0ÿ014ÿ39343ÿ ÿ
ÿ149ÿ4ÿ388ÿ6 3 3ÿÿ94ÿ6 0ÿ0 34ÿ1ÿ039
!3″0ÿ14093ÿ41 3ÿ3#2 83ÿ 3ÿ03ÿ079ÿ$040″ÿ%3140ÿ017 ÿ
1ÿ933243ÿ6 3 3ÿ 040ÿ0349ÿ8300ÿ23ÿ6ÿ1303ÿ 340ÿ$040ÿ
633ÿ02 839ÿ49128ÿ49ÿ3ÿ6 0ÿ0164ÿÿ ÿ1ÿÿ 34ÿ12 844ÿ1ÿÿ
243ÿ393ÿ 3ÿ633ÿ34ÿ039ÿ1ÿ302 3ÿ16ÿ814ÿ3ÿ61789ÿ0349ÿ
6ÿ3ÿ 34ÿ 3ÿ 0ÿ633ÿ9348ÿ3#3 ÿ ÿ1ÿ 8ÿ3ÿ 0ÿ3ÿ 34ÿ
6 0ÿ1303ÿ49ÿ1ÿ3ÿ1 3ÿ 8ÿ3ÿ 34ÿ6 0ÿ1ÿ 33ÿ63 ÿ 3ÿ ÿÿ
78ÿ 04ÿ3639ÿ6 0ÿ9332439ÿ49128ÿ & 33ÿ 040ÿ
3639ÿ 0ÿ1ÿ 33&63 ÿ 340ÿ49ÿ’ÿ 040ÿ3639ÿ 0ÿ1ÿ1303ÿ
340ÿ
 3ÿ234ÿ23ÿ 040ÿ3139ÿ ÿ3ÿ61789ÿ0349ÿ6ÿ1303ÿ
340ÿ6 0ÿ()*ÿ24730ÿ0ÿ12 39ÿ1ÿÿ234ÿ1ÿ)ÿ24730ÿ1ÿ412 8&
63 ÿ 340ÿ+16ÿ2 ÿ0ÿ93343ÿ36334ÿ2340ÿ 3ÿ1739,ÿ-43ÿ
1008ÿ0ÿ ÿ 040ÿ633ÿ4.73439ÿÿ3ÿ63 ÿ1ÿ3ÿ 340ÿ-4ÿ3ÿ
1 3ÿ 49ÿ3 0ÿÿ 43ÿ3ÿ 040ÿ6 1ÿ3639ÿ 0ÿ1ÿ3ÿ1303ÿ
340ÿ349ÿ1ÿ033ÿ 340ÿ1ÿ8300ÿ23ÿ 4ÿ3ÿ1 3ÿ 040ÿ% 4912ÿ
004234ÿ1ÿ 0ÿ9130ÿ41ÿ34073ÿ ÿ3ÿ170ÿ688ÿ3ÿ3/78ÿ4ÿ88ÿ3030ÿ
1 3ÿ 4ÿ3ÿ ÿ3ÿ3639ÿ04ÿÿÿ0ÿ34ÿ3ÿ170ÿ9339ÿ4ÿ2 4ÿ
6 0ÿÿ 43ÿ 3ÿ61ÿ170ÿ1789ÿ41ÿ 3ÿ3#8ÿ3ÿ023ÿ234ÿ3ÿ1ÿ
230739ÿ3038ÿ3417ÿ07ÿ0ÿ4ÿ902ÿ$3 0ÿÿ 04″0ÿ3ÿ30ÿ16ÿ
814ÿ 040ÿ033ÿ 340ÿ 33ÿ3ÿ4472383ÿ933430ÿ36334ÿ3ÿ170ÿ
ÿ1789ÿ3ÿ16ÿ814ÿ3ÿ36ÿ 340ÿ3ÿ0ÿ4ÿ249ÿ0ÿÿ87083ÿ ÿ
303ÿ 43ÿ933430ÿ3ÿ3014083ÿ1ÿ3ÿ93343ÿ4ÿ230,
1ÿ00300ÿ3ÿ8708ÿ1ÿ3ÿ1 300ÿ ÿ3ÿ93343ÿ4ÿ234ÿ230ÿ
0ÿ973ÿ1ÿ 43ÿ63ÿ12 73ÿ3ÿ18ÿ1ÿ34ÿÿ93343ÿ0ÿ83ÿ1ÿ
83ÿ 4ÿ3ÿ10339ÿ93343ÿ1)ÿ&ÿ()*ÿ4ÿ*ÿ247302ÿÿ3ÿ93343ÿ
633ÿ4ÿÿ973ÿ01838ÿ1ÿ 43ÿ504ÿ23 190ÿ303439ÿ4ÿ6 3ÿ(ÿ0ÿ
18ÿ4ÿ3ÿ12 739ÿ1ÿ3ÿ7778*ÿ43ÿ0ÿ0ÿ07ÿÿ816ÿ18ÿ63ÿ
3ÿ1499343ÿ ÿ3ÿ93343ÿ4ÿ230ÿ0ÿ973ÿ1ÿ3ÿ 34″0ÿ63 ÿ49ÿ0ÿ41ÿ
973ÿ1ÿ 43
:;< ÿ 123ÿ45678799 ÿ83ÿ ÿÿÿÿÿÿÿ!"ÿ"ÿ##ÿÿ$ÿÿ "ÿ%ÿ&ÿ'(ÿ"ÿÿ##ÿ)ÿ*$*+*,ÿÿ"ÿ##ÿ"ÿ !ÿ#ÿÿÿ+-ÿÿÿÿÿ.ÿ)ÿ+,/ÿ)ÿ"ÿ!ÿ0ÿ11$ 23ÿ56ÿ7869ÿ3:ÿ;5638?:@8@5A539ÿ:BÿCDCECFÿ86ÿ3=7ÿ>?:@8@5A539ÿ
=7ÿG8HH:3ÿ37AAÿ3=7ÿI5BB7?7HG7DÿJ=56ÿ56ÿH:3ÿ83ÿ8AAÿK=83ÿ53ÿ;78H6Dÿÿ
L”ÿ##ÿ)ÿ*$*+*,ÿÿ”ÿ##ÿ)ÿÿÿÿ.+-ÿÿÿÿ)ÿ
+,/ÿ1ÿÿÿÿ)ÿ”ÿ!ÿ.%ÿ&ÿ!ÿÿ11/$ÿÿÿÿ
”ÿ##ÿ”ÿÿÿ)ÿ”ÿ!ÿÿ$ÿM”1″ÿ”ÿ1″ÿÿNÿÿ
ÿ!”ÿÿ))(ÿÿ”ÿ)!1ÿ’$ÿMÿÿÿ
ÿ”ÿÿÿ#ÿÿÿ!””ÿÿÿ#ÿÿÿ
#ÿ#ÿO$ÿÿÿ’ÿ1ÿ”ÿ(ÿ”ÿ#ÿÿ1ÿÿÿ)ÿ+,ÿ
ÿ$ÿPÿ”ÿ(ÿÿ#ÿÿÿÿÿÿÿ”ÿ#ÿNÿÿ
ÿ)ÿ!ÿNÿÿÿÿ”$ÿL”ÿ#ÿÿ”ÿÿ”ÿÿ!ÿ”ÿ
”ÿ##ÿ)ÿÿ#ÿ#1ÿÿ#ÿ#ÿOÿÿ*$Q*$ÿL”ÿ#ÿÿ
ÿÿRS+,ÿ”$ÿT1ÿ”ÿ#ÿ#(ÿ!ÿÿÿ”ÿ”ÿ
##ÿ)ÿ#1ÿÿÿÿÿÿÿ)ÿ+,ÿ)ÿÿÿÿ11ÿÿ
*$U*$ÿVÿÿ#ÿ!”ÿÿÿ11ÿ!ÿÿ”ÿ!ÿU*Wÿ)ÿ”ÿ(ÿ”ÿ
ÿÿÿÿ1ÿÿ”ÿ”ÿ#ÿÿÿ”ÿ))ÿ
#!ÿ”ÿ!ÿÿ)ÿ#$ÿMÿÿ(ÿÿ!ÿ#ÿÿNÿ”ÿ
”ÿ#ÿ”ÿ”ÿ#$ÿXÿÿÿ”ÿ”ÿÿÿ*$U*ÿ##ÿ”ÿ”ÿ
#ÿÿÿ”ÿ))YÿZÿ[ÿ\ÿ!ÿ”Nÿ”ÿ##ÿÿ”ÿ
!ÿ”ÿ*$***+$
Lÿ(ÿ”ÿ##ÿÿÿ”ÿ##ÿ)ÿÿÿ.RS+,ÿÿ
#/ÿÿÿ”ÿ##ÿ)ÿÿÿÿ)ÿ”ÿ!ÿ.”ÿ#ÿ!ÿÿ
11/$ÿÿ(ÿÿÿÿÿ)ÿÿ#ÿÿ)ÿ”ÿ!ÿÿ
“”ÿÿ”ÿÿ””]ÿÿ)ÿ”ÿ!$ÿT1ÿ”ÿ1(ÿ
”ÿ##ÿÿÿ”ÿ##ÿ)ÿÿÿ1ÿ”ÿ””$ÿÿÿÿ
”ÿ##ÿ)ÿ”ÿ””ÿ1ÿ”ÿ$
L”ÿÿÿÿÿ”ÿ!ÿ1ÿ”ÿ##ÿ)ÿ”ÿ””$ÿ)ÿ”ÿ
##ÿ)ÿ”ÿÿ1ÿ”ÿ””ÿÿ)^ÿ!(ÿ!ÿ”ÿ
ÿ”ÿ”ÿ””ÿÿ)$ÿ_!(ÿ!ÿÿÿÿ”ÿ##ÿ
”ÿ”ÿ””ÿÿ)$ÿÿ”ÿ%ÿ&ÿ’(ÿ”ÿ””ÿÿ”ÿ”ÿ
`ab
012234ÿ4677ÿ486ÿ9 66206ÿ64662ÿ8162ÿ129ÿ4 69ÿ142ÿ86ÿ31 74ÿ
176ÿÿ73ÿÿ48ÿ392ÿ696206ÿ4814ÿ86ÿ012ÿ4677ÿ486ÿ9 66206ÿ
366ÿ6ÿ816ÿ234ÿ03469ÿ486ÿ31 74ÿ4814ÿ86ÿ012ÿ4677ÿ486ÿ9 66206ÿÿ
1208ÿ3ÿ41440ÿ017769ÿ!1612ÿ41440ÿ396ÿ64839ÿ3ÿ0342ÿ486ÿ
31 746ÿ3ÿ834866ÿ866ÿ0341432ÿ6″ 6ÿ4814ÿ326ÿ60ÿ486ÿ
31 74ÿ3ÿ486ÿ83486ÿ636ÿ486ÿ9141ÿ16ÿ0329669ÿ129ÿ486636ÿ16ÿ
9 #074ÿ43ÿ17ÿ2ÿ36ÿ03246$4ÿ
%&’ÿ)*++ÿ,-./0&’121ÿÿ
86ÿ83486ÿ4814ÿ12ÿ11624ÿ6 604ÿÿ96ÿ43ÿ081206ÿÿ017769ÿ486ÿ277ÿ
83486ÿ32ÿ486ÿ480125ÿ6610432ÿ6$176ÿ486ÿ277ÿ83486ÿÿ4814ÿ2ÿ486ÿ
371432ÿ3ÿ8012ÿ486ÿ612ÿ46ÿ6$60469ÿ43ÿ6ÿ624ÿ48ÿ366ÿ14624ÿÿ
6″17ÿ43ÿ486ÿ612ÿ46ÿ6$60469ÿ43ÿ6ÿ624ÿ48ÿ16167684ÿ14624ÿ8ÿ277ÿ
83486ÿ012ÿ6ÿ 4462ÿ18
9366ÿ;ÿ91616ÿ
3ÿ1ÿ
9366ÿÿ
86ÿ277ÿ83486ÿ2ÿ1ÿ03 67143217ÿ49ÿ3ÿ486ÿ6714328ÿ64662ÿ88ÿ08337ÿ
196ÿ129ÿ037766ÿ196ÿ379ÿ40177ÿ6ÿ4814ÿ486ÿ371432ÿ03 671432ÿÿÿ
8ÿ012ÿ6ÿ 4462ÿ1
?ÿ;ÿ=ÿ
866ÿ@ÿÿ486ÿ371432ÿ03 671432ÿ234ÿ43ÿ6ÿ03269ÿ48ÿÿ486ÿ03 671432ÿ2ÿ
486ÿ176ÿ
74838ÿ486ÿ277ÿ83486ÿÿ177ÿ4814ÿ486ÿ176ÿ3ÿ1ÿ11646ÿÿÿ
4866ÿ16ÿ300132ÿ2ÿ808ÿ486ÿ277ÿ83486ÿÿ1ÿ176ÿ3486ÿ4812ÿÿA3ÿ
6$176ÿÿ326ÿ66ÿ4642ÿ86486ÿ1ÿB604ÿ9 669ÿ3ÿ081206ÿ2ÿ486ÿ1 74ÿ
43ÿ964626ÿ86486ÿ1ÿC69ÿ032ÿ379ÿ036ÿÿ8619ÿ3ÿ417ÿ486ÿ277ÿ
83486ÿ379ÿ6ÿ4814ÿDÿEÿFÿ
G66ÿ2ÿ29ÿ4814ÿ486ÿ277ÿ83486ÿÿ40177ÿ486ÿ3346ÿ3ÿ486ÿ
6610865ÿ83486ÿ32ÿ486ÿ480125ÿ6610432ÿ49ÿ486ÿ661086ÿ
HIH
012340567859ÿ404ÿ201677 6ÿ39ÿ5254ÿ43ÿ6259ÿ566ÿ475ÿ740ÿ3565ÿ247546ÿ
05ÿÿ0123405676ÿ404ÿ405ÿ43ÿ41256ÿ3ÿ247546ÿ5ÿ45459ÿ79547 1ÿ76ÿ24ÿ
3 9ÿ740ÿ405ÿ0325ÿ404ÿ74ÿ ÿ5ÿ976597459ÿ 9ÿ405535ÿ55459ÿÿ405ÿÿ
0123405676ÿ55ÿ45ÿÿ9755 5ÿ6ÿ5ÿ3ÿ5ÿ40 ÿ405ÿ625ÿ9755 5ÿ3ÿ
ÿ7456ÿ39ÿ5ÿ51ÿ751ÿ43ÿ3 ÿ05535ÿ405ÿ565056ÿ55459ÿ405ÿ
ÿ0123405676ÿ3ÿ3ÿ9755 5ÿ 9ÿ3 959ÿ404ÿ7ÿ405ÿ232473ÿ201677 6ÿ
7459ÿ43ÿ6259ÿ566ÿ475ÿ740ÿ3565ÿ247546ÿ
ÿ405ÿÿ0123405676ÿ76ÿ55459ÿ405ÿ405ÿ45 475ÿ43ÿ405ÿÿ0123405676ÿ
59ÿ405ÿ45 475ÿ0123405676!ÿ76ÿ 52459ÿ05ÿ45 475ÿ0123405676ÿ76ÿ6721ÿ
405ÿ5565ÿ3ÿ405ÿÿ0123405676ÿÿ405ÿÿ0123405676ÿ
ÿ #$%&’&ÿ(ÿ#)*&+),&ÿ
76ÿ55459ÿ405ÿ4055ÿ5ÿ43ÿ45 4756ÿ #$%&’&ÿ.ÿ#)*&+),&ÿ
#$%&’&ÿ/ÿ#)*&+),&0ÿ
1 41ÿ405ÿ975473ÿ3ÿ405ÿ625ÿ5 6ÿ9545756ÿ070ÿ45 475ÿ76ÿ
932459ÿ235ÿ454336ÿ05ÿ7 3541ÿ59ÿ404ÿ5547ÿ405ÿÿ0123405676ÿ
404ÿ43ÿ2324736ÿ5 6ÿ5ÿ53ÿ9356ÿ34ÿ6471ÿÿ3 673ÿ34ÿ070ÿ
232473ÿ5 ÿ76ÿ5ÿ4 765ÿ567!ÿ60359ÿ03ÿ74ÿ76ÿ647859ÿ43ÿ9ÿÿ
3 673ÿ34ÿ405ÿ975473ÿ3ÿ405ÿ9755 5
9:;
01231456357ÿ97 132ÿ
ÿÿÿÿ

!”#$%&'(ÿ*+ÿ,-./0-$1ÿ2-3&(-45&-/.
!”#$%&'(ÿ66+ÿ7.&(/859&-/.ÿ&/ÿ:;%/&#’3-3ÿ,4ÿ;,:+:ÿ
EFG#GHÿJKLM%#N$
/OP7:65::ÿC(,+(,-ÿ-,Q,6+732ÿ3Rÿ+(,ÿ25>>ÿ(9*3+(,:7:ÿ:(35>4ÿS,ÿ)2ÿ)>>D3-D232,ÿ
*-3*3:7+732
TO=+)+,ÿ+(,ÿ5:,R5>2,::ÿ3Rÿ)ÿ:79ÿ>7W,>9ÿ+()+ÿ+(,ÿ
25>>ÿ(9*3+(,:7:ÿ3Rÿ23ÿ47RR,-,26,ÿ7:ÿR)>:,ÿ,X,2ÿS,R3-,ÿ43727+9ÿX)>5,ÿ7:ÿS,>3Cÿ+(,ÿZÿ>,X,>@ÿ+(,ÿ,RR,6+ÿ7:ÿ$%F%#$%#MF[[\ÿ$#HG#]MFG%ÿ
)24ÿ+(,ÿ25>>ÿ(9*3+(,:7:ÿ7:ÿ-,Q,6+,4Oÿ83C,X,-@ÿ23+ÿ)>>ÿ:+)+7:+76)>>9ÿ:74ÿS,ÿ+-,)+,4ÿ+(,ÿ:)V,ÿC)9Oÿ^3-ÿ,U)V*>,@ÿ935ÿ:(35>4ÿ()X,ÿ>,::ÿ632?4,26,ÿ+()+ÿ
+(,ÿ25>>ÿ(9*3+(,:7:ÿ7:ÿR)>:,ÿ7Rÿ*ÿ_ÿ̀Òabÿ+()2ÿ*ÿ_ÿ̀Ò`cOÿ;(5:@ÿ-,Q,6+72>ÿ
(9*3+(,:7:ÿ7:ÿ23+ÿ)2ÿ)>>D3-D232,ÿ*-3*3:7+732O
1Rÿ+(,ÿ25>>ÿ(9*3+(,:7:ÿ7:ÿ-,Q,6+,4@ÿ+(,2ÿ+(,ÿ)>+,-2)+7X,ÿ+3ÿ+(,ÿ25>>ÿ(9*3+(,:7:ÿ
d6)>>,4ÿ+(,ÿ)>+,-2)+7X,ÿ(9*3+(,:7:eÿ7:ÿ)66,*+,4Oÿ’32:74,-ÿ+(,ÿ32,D+)7>,4ÿ+,:+ÿ72ÿ+(,ÿ
f)V,:ÿg324ÿ6):,ÿ:+5490ÿh-Oÿg324ÿC):ÿ:ÿ32ÿC(76(ÿ(,ÿQ54>ÿ(9*3+(,:7:ÿR3-ÿ+(7:ÿ32,D+)7>,4ÿ+,:+ÿ7:ÿ+()+ÿkÿlÿ̀OmÿC(,-,ÿ
kÿ7:ÿ+(,ÿ*-3S)S7>7+9ÿ3RÿS,72>ÿ(9*3+(,:7:ÿ7:ÿ
-,Q,6+,4@ÿ+(,2ÿ+(,ÿ)>+,-2)+7X,ÿ(9*3+(,:7:ÿ+()+ÿkÿnÿ̀Omÿ7:ÿ)66,*+,4Oÿ1Rÿkÿ7:ÿ,4ÿ+,:+ÿ5:,4ÿ72ÿ+(,ÿp(9:767)2:qÿr,)6+732:ÿ6):,ÿ
:+549Oÿ;(,ÿ25>>ÿ(9*3+(,:7:ÿ7:0
ÿ t3S,:,ÿuÿt)X,-)>ÿ(9*3+(,:7:ÿ7:ÿ-,Q,6+,4@ÿ+(,2ÿ+(,-,ÿ)-,ÿ+C3ÿ)>+,-2)+7X,:0
ÿ twxyzyÿ{ÿt|}y~|yÿ
twxyzyÿ€ÿt|}y~|yvÿ
‚
0123415567ÿ29ÿ4 2ÿÿ29ÿ15ÿ 1ÿ 24ÿ9 9ÿ152412ÿÿ
12 ÿÿ29ÿ15ÿ 1ÿ4ÿ29ÿÿ122ÿÿ1256ÿ5 4ÿ291ÿ29ÿ
15ÿ 1ÿ4ÿ29ÿ141 92ÿ1227ÿ29ÿÿ935ÿ53 ÿ2912ÿ29ÿ
3512ÿ 1ÿ4ÿ29ÿÿ122ÿÿ5 4ÿ291ÿ29ÿ15ÿ 1ÿ4ÿ29ÿ
141 92ÿ122
94ÿ14ÿ16ÿ2312ÿÿ9 9ÿ2ÿÿ46ÿ3556ÿ2ÿ 2ÿ55ÿ
91ÿ1256ÿ29ÿ1 ÿ3512ÿ 1ÿ4ÿ157ÿ2ÿÿ4121556ÿ5ÿ
2912ÿ14ÿ1ÿ1 219ÿ4 ÿ1256ÿ29ÿ1 ÿ 4 ÿÿ1ÿ45ÿ
9447ÿÿ4ÿ1ÿ4 2ÿ14ÿ29 4ÿ 2ÿÿ32 7ÿ
29ÿ41494ÿÿ2912ÿ29ÿ355ÿ9629ÿÿ1256ÿÿ4 ÿÿ15ÿ
 47ÿ29ÿ41494ÿÿ2ÿÿ9 9ÿ43ÿ4ÿ4ÿ45ÿÿ1ÿ22ÿÿ
29ÿ4 ÿÿ127ÿ29ÿ29ÿ4 2ÿÿ29ÿ4 ÿÿ2159 ÿ9ÿ
2ÿÿ15ÿ1 ÿÿ29ÿ 2ÿÿ29ÿ45129ÿ2 ÿ  ÿ2415ÿ
1ÿ1 ÿ22
!”#$%&'(
) ÿ22ÿ91ÿ44 256ÿ212 ÿ2912ÿ4* 2ÿ29ÿ355ÿ9629ÿ2912ÿ2ÿ
3512ÿ 1ÿ14ÿ+315ÿÿ2ÿ*326ÿ1ÿ53ÿ132ÿ9 9ÿ3512ÿ
 1ÿÿ5144ÿ217ÿ296ÿ16ÿ2912ÿ155ÿÿ1ÿ53 ÿÿ2912ÿ29ÿ3512ÿ
 1ÿ4ÿ9ÿ15 26ÿÿ53 ÿ29ÿ4 2ÿÿ29ÿ 2ÿÿ514ÿÿ63ÿ
2ÿ2912ÿ1ÿ2215 ÿ22ÿ12ÿ29ÿ,,-ÿ55ÿÿ+3152ÿ2ÿ2ÿ1412ÿ215 ÿ
22ÿ19ÿ12ÿ29ÿ,,.-ÿ55ÿ9ÿ2ÿ355ÿ9629ÿ14ÿ29
ÿ 012343ÿ5ÿ06738693ÿ
012343ÿ:ÿ06738693;ÿ
ÿ29ÿ4 4ÿÿ29ÿÿ4* 2 7ÿ29ÿ29ÿ53ÿÿ2912ÿ29ÿ3512ÿ 1ÿ
4ÿÿ122ÿÿ5 4ÿ291ÿ2912ÿ4ÿ141 92ÿ122ÿÿ29ÿ51224ÿÿ
4* 2 7ÿ29ÿ29ÿ53ÿÿ2912ÿ29ÿ3512ÿ 1ÿ4ÿÿ122ÿÿ
994ÿ291ÿ2912ÿ4ÿ141 92ÿ122ÿ) ÿ?@,A
BCD
012345432617ÿ9 1 671612ÿ32ÿ
ÿÿÿÿ
!”#”#$%&’&(#’
)*+,-./0ÿ223ÿ45.06789.:65ÿ.6ÿ;/=.:5?
)*+,-./0ÿ223ÿ@:?5:A9,59/ÿ>/=.:5?
)*+,-./0ÿ223ÿ>5.ÿQ/4Pÿ+ÿ84567.58.ÿ>5-./:+;ÿR*.-*./ÿ+ÿ-.68+5SNTU,;+>5ÿR*@ÿ+ÿ84567.58.ÿ>5-./:+;ÿP+V.68+5-;@ÿ7>QQ./.5-ÿQ/4Pÿ2ÿ+-ÿ-*.ÿ2N2Yÿ;.:.;ÿ-*.5ÿ-*.ÿ
ZY[ÿ84567.58.ÿ>5-./:+;ÿR>;;ÿ54-ÿ845-+>5ÿ2Nÿ\;;ÿ:+;=.5ÿ-*.ÿ84567.58.ÿ>5-./:+;ÿ+/.ÿ
,;+=W;.ÿ:+;=.5-./:+;ÿR>;;ÿW.ÿ
45ÿ-*.ÿ-*./ÿ+;;ÿ,4->:.ÿ4/ÿ+;;ÿ5.B+->:.fNÿA*./.Q4/.Xÿ+ÿB5>68+5-ÿ
657>5Bÿ+;;4R45ÿ4Qÿ-*.ÿ.QQ.8-NÿA*./.ÿ+/.ÿP+5@ÿ
-=+->455ÿR*>8*ÿ>-ÿ>V.;@ÿ-R4ÿ8457>->45;;ÿ*+:.ÿ.U+8-;@ÿ-*.ÿ45ÿP.+5-ÿ>8+;;@ÿ>P,45ÿ+57ÿ
+8.-+P>54,*.5ÿ,/4:>7.ÿ.U+8-;@ÿ-*.ÿ5ÿ/.;>.QNÿA*./.Q4/.Xÿ.:.5ÿ
W.Q4/.ÿ+5ÿ.U,./>P.5-ÿ84P,+/>5Bÿ-*.>/ÿ.QQ.8->:.5.:.N
hij
01ÿ345ÿ678ÿ9 595ÿ35ÿ9 3ÿ5ÿ 5ÿ595ÿ345ÿ535ÿ
5ÿ595ÿÿ345ÿÿ4345ÿ345ÿ345ÿ511593ÿÿ 3ÿ5ÿ93ÿ3ÿ345ÿ
7ÿ55ÿ! “ÿ3ÿ #93ÿ511593ÿÿ35ÿ1ÿ9 595ÿ35ÿ
”5ÿ95ÿ4ÿ345ÿÿ4345ÿ4ÿ 3ÿ5ÿ99535ÿ45ÿ3ÿÿ 3ÿ5$5935%ÿ
&5ÿ5ÿÿ345ÿ9 595ÿ35ÿÿÿ5ÿ5ÿ1ÿ345ÿ535ÿ’95ÿ
5ÿÿÿ345ÿ35ÿ3ÿ9 3ÿ5ÿ5$5935 ÿ( 55ÿ3455ÿÿÿ 35ÿ5ÿ1ÿÿ
345ÿ5ÿÿ345ÿ35ÿÿ9 3ÿ5553ÿ ÿ 5ÿ1ÿ345ÿ
9ÿ5ÿ5$5935ÿ5345
)*+
01234536781452ÿ
ÿÿÿÿ

 !”#$%&ÿ(()ÿ*+$&,-./$0,+ÿ$,ÿ12#,$!%303ÿ4%3$0+5
 !”#$%&ÿ(()ÿ6$”$03$0/”7ÿ605+08/”+/%
 !”#$%&ÿ(()ÿ42#%ÿ*ÿ”+-ÿ**ÿ9&&,&3
:;