1 Question 3.7:

The tensile strength of Portland cement is being studied. Four different mixing techniques can be used economically. A completely randomized experiment was conducted and the following data were collected:

Observations of Tensile Strength
Mixing Technique 1 2 3 4
1 3129 3000 2865 2890
2 3200 3300 2975 3150
3 2800 2900 2985 3050
4 2600 2700 2600 2765
  1. Test the hypothesis that mixing techniques affect the strength of the cement. Use ( \(\alpha\)= 0.05)

1.1 Handwritten Solution:

3.7 Page 1

3.7 Page 2

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2 Question 3.10:

A product developer is investigating the tensile strength of a new synthetic fiber that will be used to make cloth for men’s shirts. Strength is usually affected by the percentage of cotton used in the blend of materials for the fiber. The engineer conducts a completely randomized experiment with five levels of cotton content and replicates the experiment five times. The data are shown in the following table.

Observations of Tensile Strength of cloth fibre

(a) Is there evidence to support the claim that cotton content affects the mean tensile strength? Use \(\alpha\)=0.05

Cotton Weight

Percent		 1 2 3 4 5
15 7 7 15 11 9
20 12 17 12 18 18
25 14 19 19 18 18
30 19 25 22 19 23
35 7 10 11 15 11

2.1 Handwritten Solution:

3.10 Page 1

3.10 Page 2

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3.10 Page 4

3 Question 3.20:

An article in the ACI Materials Journal (Vol. 84, 1987, pp. 213–216) describes several experiments investigating the rodding of concrete to remove entrapped air. A 3-inch & 6-inch cylinder was used, and the number of times this rod was used is the design variable. The resulting compressive strength of the concrete specimen is the response. The data are shown in the following table:

Observations for Tensile Strength

(a) Is there any difference in compressive strength due to the rodding level? Use \(\alpha\)= 0.05. (b) Find the P-value for the F statistic in part (a).
Rodding Level 1 2 3
10 1530 1530 1440
15 1610 1650 1500
20 1560 1730 1530
25 1500 1490 1510

3.1 Solution with R-Studio (As mentioned):

Reading the Data:

PART A):

\[ H_o: \mu_{1}=\mu_{2}=\mu_{3}=\mu_{4} \]

\[ H_a: Atleast \space one \space \mu_i \space Differs \space (i=1,2,3,4) \]

rod10<-c(1530,1530,1440)
rod15<-c(1610,1650,1500)
rod20<-c(1560,1730,1530)
rod25<-c(1500,1490,1510)
rodadd<-rbind(rod10,rod15,rod20,rod25)
print(rodadd)
##       [,1] [,2] [,3]
## rod10 1530 1530 1440
## rod15 1610 1650 1500
## rod20 1560 1730 1530
## rod25 1500 1490 1510
r10<-mean(rod10)
r15<-mean(rod15)
r20<-mean(rod20)
r25<-mean(rod25)

Finding SSE, SStreatment, MSE, MStreatment:

SSE10 <- (1530-r10)^2 + (1530-r10)^2 + (1440-r10)^2
SSE15 <- (1610-r15)^2 + (1650-r15)^2 + (1500-r15)^2
SSE20 <- (1560-r20)^2 + (1730-r20)^2 + (1530-r20)^2
SSE25 <- (1500-r25)^2 + (1490-r25)^2 + (1510-r25)^2

SSE & MSE:

SSE<- SSE10 + SSE15 + SSE20 + SSE25
print(SSE)
## [1] 40933.33
MSE<- SSE/(12-4)
print(MSE)
## [1] 5116.667

SStreatment & MStreatment:

mean <- c(mean(rodadd))
print(mean)
## [1] 1548.333
SStreatment <- 3*((r10 - mean)^2 + (r15 - mean)^2 + (r20 - mean)^2 + (r25 - mean)^2)
print(SStreatment)
## [1] 28633.33
MStreatment<-SStreatment/(4-1)
print(MStreatment)
## [1] 9544.444

Sum Squared Total (SST):

SST<-SSE+SStreatment
print(SST)
## [1] 69566.67

Finding F-Statistic:

Fo<-MStreatment/MSE
print(Fo)
## [1] 1.865364

Finding Critical Value of F: using F-Distribution Command (qf)

?qf
Fcritical<-qf(0.95,3,8)
print(Fcritical)
## [1] 4.066181

PART B): Finding P-Value

Using F-Distribution Command:

?pf
Pvalue<-pf(1.86536, 3, 8, lower.tail = FALSE)
print(Pvalue)
## [1] 0.2137821

Conclusion:

----> Since our F-Statistic Value (1.865) is less than critical value (4.066), so we fail to reject Ho and thus we conclude that there is no difference in compressive strength due to the rodding level.

----> P-value = 0.2138

3.2 Handwritten Solution:

3.20 Part A Page 1

3.20 Part A Page 2

4 Source Code for Problem 3.20:

getwd()

##Question 3.20::

#PART A)

#Reading the Data:
rod10<-c(1530,1530,1440)
rod15<-c(1610,1650,1500)
rod20<-c(1560,1730,1530)
rod25<-c(1500,1490,1510)
rodadd<-rbind(rod10,rod15,rod20,rod25)
print(rodadd)

r10<-mean(rod10)
r15<-mean(rod15)
r20<-mean(rod20)
r25<-mean(rod25)

#Finding SSE,SStreatment,MSE,MStreatment::
SSE10 <- (1530-r10)^2 + (1530-r10)^2 + (1440-r10)^2
SSE15 <- (1610-r15)^2 + (1650-r15)^2 + (1500-r15)^2
SSE20 <- (1560-r20)^2 + (1730-r20)^2 + (1530-r20)^2
SSE25 <- (1500-r25)^2 + (1490-r25)^2 + (1510-r25)^2

SSE<- SSE10 + SSE15 + SSE20 + SSE25
print(SSE)

#MSE:
MSE<- SSE/(12-4)
print(MSE)

#SStreatment::
mean <- c(mean(rodadd))
print(mean)

SStreatment <- 3*((r10 - mean)^2 + (r15 - mean)^2 + (r20 - mean)^2 + (r25 - mean)^2)
print(SStreatment)

#MStreatment:
MStreatment<-SStreatment/(4-1)
print(MStreatment)

#SST::
SST<-SSE+SStreatment
print(SST)

#F-Statistic:
Fo<-MStreatment/MSE
print(Fo)

#Critical Value:
?qf
Fcritical<-qf(0.95,3,8)
print(Fcritical)

#PART B)
?pf
Pvalue<-pf(1.86536, 3, 8, lower.tail = FALSE)
print(Pvalue)