Statistics in Construction

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.

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
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