0.1 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:

a. Test the hypothesis that mixing techniques affect the strength of the cement. Use ( α�= 0.05 )

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

Results - F0 (12.73) > F0.05, 3.12 (3.49)

Mixing techniques affects the strength of cement. Therefore we reject null hypothesis i.e Ho

0.2 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.

Cotton Weight % 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 
knitr::include_graphics("C:/Users/Public/Solutions Q1 Q2.pdf")

Your PDF Caption

Solutions for both the questions are in single pdf attached

0.3 Question 3.20

0.4 Solution

#Data Reading
roddinglevel10 <- c(1530,1530,1440)
roddinglevel15 <- c(1610,1650,1500)
roddinglevel20 <- c(1560,1730,1530)
roddinglevel25 <- c(1500,1490,1510)
roddinglevel <- rbind(roddinglevel10,roddinglevel15,roddinglevel20,roddinglevel25)
print(roddinglevel)
##                [,1] [,2] [,3]
## roddinglevel10 1530 1530 1440
## roddinglevel15 1610 1650 1500
## roddinglevel20 1560 1730 1530
## roddinglevel25 1500 1490 1510

Finding the mean of the rodding level

rodding10 <- mean(roddinglevel10)
rodding15 <- mean(roddinglevel15)
rodding20 <- mean(roddinglevel20)
rodding25 <- mean(roddinglevel25)

Step 1 - A) SSE B) MSE

SSE10 <- (1530-roddinglevel10)^2 + (1530-roddinglevel10)^2 + (1440-roddinglevel10)^2
SSE15 <- (1610-roddinglevel15)^2 + (1650-roddinglevel15)^2 + (1500-roddinglevel15)^2
SSE20 <- (1560-roddinglevel20)^2 + (1730-roddinglevel20)^2 + (1530-roddinglevel20)^2
SSE25 <- (1500-roddinglevel25)^2 + (1490-roddinglevel25)^2 + (1510-roddinglevel25)^2
SSE <- SSE10 + SSE15 +SSE20 +SSE25
print(SSE)
## [1]  51800 101600  92200
MSE <- SSE/(8)
print(MSE)
## [1]  6475 12700 11525

Step 2- A) SStreatment B) MStreatment

mean <- c(mean(roddinglevel))
print(mean)
## [1] 1548.333
 #A)SStreatment
SStreatment <- 3*((roddinglevel10 - mean)^2 + (roddinglevel15 - mean)^2 + (roddinglevel20 - mean)^2 + (roddinglevel25 - mean)^2)
  print(SStreatment)
## [1]  19833.33 141233.33  47633.33
  #B)MStreatment
  MStreatment <- SStreatment/3
  print(MStreatment)
## [1]  6611.111 47077.778 15877.778

Step 3 - SST

SST <- SSE + SStreatment
print(SST)
## [1]  71633.33 242833.33 139833.33

Step 4 - F0

F0 <- MStreatment/MSE
print(F0)
## [1] 1.021021 3.706912 1.377681

Step 5 - P value

?qf
## starting httpd help server ... done
Fcritical<-qf(0.95,3,8)
print(Fcritical)
## [1] 4.066181
?pf
Pvalue <- pf(1.86536, 3, 8, lower.tail = FALSE)
print(Pvalue)
## [1] 0.2137821

0.5 Complete R code

#R code
roddinglevel10 <- c(1530,1530,1440)
roddinglevel15 <- c(1610,1650,1500)
roddinglevel20 <- c(1560,1730,1530)
roddinglevel25 <- c(1500,1490,1510)
roddinglevel <- rbind(roddinglevel10,roddinglevel15,roddinglevel20,roddinglevel25)
rodding10 <- mean(roddinglevel10)
rodding15 <- mean(roddinglevel15)
rodding20 <- mean(roddinglevel20)
rodding25 <- mean(roddinglevel25)
SSE10 <- (1530-roddinglevel10)^2 + (1530-roddinglevel10)^2 + (1440-roddinglevel10)^2
SSE15 <- (1610-roddinglevel15)^2 + (1650-roddinglevel15)^2 + (1500-roddinglevel15)^2
SSE20 <- (1560-roddinglevel20)^2 + (1730-roddinglevel20)^2 + (1530-roddinglevel20)^2
SSE25 <- (1500-roddinglevel25)^2 + (1490-roddinglevel25)^2 + (1510-roddinglevel25)^2
SSE <- SSE10 + SSE15 +SSE20 +SSE25
MSE <- SSE/(8)
mean <- c(mean(roddinglevel))
 #A)SStreatment
SStreatment <- 3*((roddinglevel10 - mean)^2 + (roddinglevel15 - mean)^2 + (roddinglevel20 - mean)^2 + (roddinglevel25 - mean)^2)
  #B)MStreatment
  MStreatment <- SStreatment/3
  SST <- SSE + SStreatment
F0 <- MStreatment/MSE
?qf
Fcritical<-qf(0.95,3,8)
?pf
Pvalue <- pf(1.86536, 3, 8, lower.tail = FALSE)