3.7 (a):

null hypothesis: \(H_0: \mu_{1}=\mu_{2}=\cdots =\mu_{i}=\mu\)

alternative hypothesis: \(H_1:\) at least one \(\mu_{i}\) differs

Linear effects equation:

\(\overline{y_{ij}}=\mu_{i}+\epsilon_{ij}\)

\(\overline{y_{ij}}=\mu+\tau_{i}+\epsilon_{ij}\)

where, \(\mu_{i}\): mean of a certain population, \(\mu\): ground mean, \(\epsilon_{ij}\): error between populations, \(\tau_{i}\): effect of population.

Since we do not have a continuous range of type of mixing technique, it is a fixed effect model, hypothesis turns to:

null hypothesis: \(H_0: \epsilon_{ij}\) = 0, for all i

alternative hypothesis: \(H_1: \epsilon_{ij}\neq 0\), for some i

r1 <- c(3129,3000,2865,2890)
r2 <- c(3200,3300,2975,3150)
r3 <- c(2800,2900,2985,3050)
r4 <- c(2600,2700,2600,2765)
I1=4
J1=4
data1 <- cbind(r1,r2,r3,r4)
data1 <- as.data.frame(data1)
str(data1)

## 'data.frame':    4 obs. of  4 variables:
##  $ r1: num  3129 3000 2865 2890
##  $ r2: num  3200 3300 2975 3150
##  $ r3: num  2800 2900 2985 3050
##  $ r4: num  2600 2700 2600 2765

a1 <- mean(r1)
a2 <- mean(r2)
a3 <- mean(r3)
a4 <- mean(r4)
c1 <- c(a1,a2,a3,a4)
aveg1 <- mean(c1)

Ground mean = 2931.8125 \((lb/in^{2})\)

aver1 <- ave(r1)
aver2 <- ave(r2)
aver3 <- ave(r3)
aver4 <- ave(r4)

Ediff1 <- (r1-aver1)^2
Ediff2 <- (r2-aver2)^2
Ediff3 <- (r3-aver3)^2
Ediff4 <- (r4-aver4)^2

sum1 <- sum(Ediff1)
sum2 <- sum(Ediff2)
sum3 <- sum(Ediff3)
sum4 <- sum(Ediff4)

c2 <- c(sum1,sum2,sum3,sum4)
SSE1 <- sum(c2)
MSE1 <- SSE1/I1/(J1-1)
Trdiff1 <- (a1-aveg1)^2
Trdiff2 <- (a2-aveg1)^2
Trdiff3 <- (a3-aveg1)^2
Trdiff4 <- (a4-aveg1)^2

SSTr1 <- J1*sum(Trdiff1,Trdiff2,Trdiff3,Trdiff4)
MSTr1 <- SSTr1/(I1-1)
F1 <- MSTr1/MSE1

\[ SSE= \sum_{i} \sum_{j} (\overline{y_{ij}}-\overline{y_{i.}})^2 = 1.5390825\times 10^{5} \]

\[ MSE= \frac{SSE}{I(J-1)}=\frac{1.5390825\times 10^{5}}{4(4-1)} = 1.2825688\times 10^{4} \]

\[ SSTr= \sum_{i} J* (\overline{y_{i.}}-\overline{y_{i..}})^2 = 4.8974019\times 10^{5} \]

\[ MSTr= \frac{SSTr}{I-1}= \frac{4.8974019\times 10^{5}}{4-1} = 1.6324673\times 10^{5} \]

\[ F= \frac{MSTr}{MSE} = 12.7281075 \]

alpha1=0.05
qf(1-alpha1,I1-1,I1*(J1-1))

## [1] 3.490295

Since \(\alpha=0.05\), use code \(qf((1-\alpha), I-1, I*(J-1))\) to find critical value(CV).

Since CV = 3.4902948 < 12.7281075 = F, we will reject the null hypothesis. Mixing techiniques do affect the strength of cement.

————————————————————————————————

3.10 (a):

null hypothesis: \(H_0: \mu_{1}=\mu_{2}=\cdots =\mu_{i}=\mu\)

alternative hypothesis: \(H_1:\) at least one \(\mu_{i}\) differs

Linear effects equation:

\(\overline{y_{ij}}=\mu_{i}+\epsilon_{ij}\)

\(\overline{y_{ij}}=\mu+\tau_{i}+\epsilon_{ij}\)

where, \(\mu_{i}\): mean of a certain population, \(\mu\): ground mean, \(\epsilon_{ij}\): error between populations, \(\tau_{i}\): effect of population.

Since we do not have a continuous range of type of mixing technique, it is a fixed effect model, hypothesis turns to:

null hypothesis: \(H_0: \epsilon_{ij}\) = 0, for all i

alternative hypothesis: \(H_1: \epsilon_{ij}\neq 0\), for some i

r1 <- c(7,7,15,11,9)
r2 <- c(12,17,12,18,18)
r3 <- c(14,19,19,18,18)
r4 <- c(19,25,22,19,23)
r5 <- c(7,10,11,15,11)
I1=5
J1=5
data1 <- cbind(r1,r2,r3,r4,r5)
data1 <- as.data.frame(data1)
str(data1)

## 'data.frame':    5 obs. of  5 variables:
##  $ r1: num  7 7 15 11 9
##  $ r2: num  12 17 12 18 18
##  $ r3: num  14 19 19 18 18
##  $ r4: num  19 25 22 19 23
##  $ r5: num  7 10 11 15 11

a1 <- mean(r1)
a2 <- mean(r2)
a3 <- mean(r3)
a4 <- mean(r4)
a5 <- mean(r5)
c1 <- c(a1,a2,a3,a4,a5)
aveg1 <- mean(c1)

Ground mean = 15.04 \((lb/in^{2})\)

aver1 <- ave(r1)
aver2 <- ave(r2)
aver3 <- ave(r3)
aver4 <- ave(r4)
aver5 <- ave(r5)

Ediff1 <- (r1-aver1)^2
Ediff2 <- (r2-aver2)^2
Ediff3 <- (r3-aver3)^2
Ediff4 <- (r4-aver4)^2
Ediff5 <- (r5-aver5)^2

sum1 <- sum(Ediff1)
sum2 <- sum(Ediff2)
sum3 <- sum(Ediff3)
sum4 <- sum(Ediff4)
sum5 <- sum(Ediff5)

c2 <- c(sum1,sum2,sum3,sum4,sum5)
SSE1 <- sum(c2)
MSE1 <- SSE1/I1/(J1-1)
Trdiff1 <- (a1-aveg1)^2
Trdiff2 <- (a2-aveg1)^2
Trdiff3 <- (a3-aveg1)^2
Trdiff4 <- (a4-aveg1)^2
Trdiff5 <- (a5-aveg1)^2

SSTr1 <- J1*sum(Trdiff1,Trdiff2,Trdiff3,Trdiff4,Trdiff5)
MSTr1 <- SSTr1/(I1-1)
F1 <- MSTr1/MSE1

\[ SSE= \sum_{i} \sum_{j} (\overline{y_{ij}}-\overline{y_{i.}})^2 = 161.2 \]

\[ MSE= \frac{SSE}{I(J-1)}=\frac{161.2}{5(5-1)} = 8.06 \]

\[ SSTr= \sum_{i} J* (\overline{y_{i.}}-\overline{y_{..}})^2 = 475.76 \]

\[ MSTr= \frac{SSTr}{I-1}= \frac{475.76}{5-1} = 118.94 \]

\[ F= \frac{MSTr}{MSE} = 14.7568238 \]

alpha1=0.05
qf(1-alpha1,I1-1,I1*(J1-1))

## [1] 2.866081

Since \(\alpha=0.05\), use code \(qf((1-\alpha), I-1, I*(J-1))\) to find critical value(CV).

Since CV = 2.8660814 < 14.7568238 = F, we will reject the null hypothesis. Cotton content does affect the mean tensile strength.

————————————————————————————————

3.20 (a):

null hypothesis: \(H_0: \mu_{1}=\mu_{2}=\cdots =\mu_{i}=\mu\)

alternative hypothesis: \(H_1:\) at least one \(\mu_{i}\) differs

Linear effects equation:

\(\overline{y_{ij}}=\mu_{i}+\epsilon_{ij}\)

\(\overline{y_{ij}}=\mu+\tau_{i}+\epsilon_{ij}\)

where, \(\mu_{i}\): mean of a certain population, \(\mu\): ground mean, \(\epsilon_{ij}\): error between populations, \(\tau_{i}\): effect of population.

Since we do not have a continuous range of type of mixing technique, it is a fixed effect model, hypothesis turns to:

null hypothesis: \(H_0: \epsilon_{ij}\) = 0, for all i

alternative hypothesis: \(H_1: \epsilon_{ij}\neq 0\), for some i

r1 <- c(1530,1530,1440)
r2 <- c(1610,1650,1500)
r3 <- c(1560,1730,1530)
r4 <- c(1500,1490,1510)

I1=4
J1=3
data1 <- cbind(r1,r2,r3,r4)
data1 <- as.data.frame(data1)
str(data1)

## 'data.frame':    3 obs. of  4 variables:
##  $ r1: num  1530 1530 1440
##  $ r2: num  1610 1650 1500
##  $ r3: num  1560 1730 1530
##  $ r4: num  1500 1490 1510

a1 <- mean(r1)
a2 <- mean(r2)
a3 <- mean(r3)
a4 <- mean(r4)

c1 <- c(a1,a2,a3,a4)
aveg1 <- mean(c1)

Ground mean = 1548.3333333 \((lb/in^{2})\)

aver1 <- ave(r1)
aver2 <- ave(r2)
aver3 <- ave(r3)
aver4 <- ave(r4)

Ediff1 <- (r1-aver1)^2
Ediff2 <- (r2-aver2)^2
Ediff3 <- (r3-aver3)^2
Ediff4 <- (r4-aver4)^2

sum1 <- sum(Ediff1)
sum2 <- sum(Ediff2)
sum3 <- sum(Ediff3)
sum4 <- sum(Ediff4)

c2 <- c(sum1,sum2,sum3,sum4)
SSE1 <- sum(c2)
MSE1 <- SSE1/I1/(J1-1)
Trdiff1 <- (a1-aveg1)^2
Trdiff2 <- (a2-aveg1)^2
Trdiff3 <- (a3-aveg1)^2
Trdiff4 <- (a4-aveg1)^2

SSTr1 <- J1*sum(Trdiff1,Trdiff2,Trdiff3,Trdiff4)
MSTr1 <- SSTr1/(I1-1)
F1 <- MSTr1/MSE1

\[ SSE= \sum_{i} \sum_{j} (\overline{y_{ij}}-\overline{y_{i.}})^2 = 4.0933333\times 10^{4} \]

\[ MSE= \frac{SSE}{I(J-1)}=\frac{4.0933333\times 10^{4}}{4(3-1)} = 5116.6666667 \]

\[ SSTr= \sum_{i} J* (\overline{y_{i.}}-\overline{y_{..}})^2 = 2.8633333\times 10^{4} \]

\[ MSTr= \frac{SSTr}{I-1}= \frac{2.8633333\times 10^{4}}{4-1} = 9544.4444444 \]

\[ F= \frac{MSTr}{MSE} = 1.8653637 \]

alpha1=0.05
qf(1-alpha1,I1-1,I1*(J1-1))

## [1] 4.066181

Since \(\alpha=0.05\), use code \(qf((1-\alpha), I-1, I*(J-1))\) to find critical value(CV).

Since CV = 4.0661806 > 1.8653637 = F, we will not reject the null hypothesis. There is no difference in compressive strength due to the rodding level.

(b)

pf(F1,I1-1,I1*(J1-1),lower.tail = FALSE)

## [1] 0.2137815

Homework4

Eddie J. Liu

9/25/2021

3.7 (a):

null hypothesis: \(H_0: \mu_{1}=\mu_{2}=\cdots =\mu_{i}=\mu\)

alternative hypothesis: \(H_1:\) at least one \(\mu_{i}\) differs

Linear effects equation:

\(\overline{y_{ij}}=\mu_{i}+\epsilon_{ij}\)

\(\overline{y_{ij}}=\mu+\tau_{i}+\epsilon_{ij}\)

where, \(\mu_{i}\): mean of a certain population, \(\mu\): ground mean, \(\epsilon_{ij}\): error between populations, \(\tau_{i}\): effect of population.

Since we do not have a continuous range of type of mixing technique, it is a fixed effect model, hypothesis turns to:

null hypothesis: \(H_0: \epsilon_{ij}\) = 0, for all i

alternative hypothesis: \(H_1: \epsilon_{ij}\neq 0\), for some i

Ground mean = 2931.8125 \((lb/in^{2})\)

\[ SSE= \sum_{i} \sum_{j} (\overline{y_{ij}}-\overline{y_{i.}})^2 = 1.5390825\times 10^{5} \]

\[ MSE= \frac{SSE}{I(J-1)}=\frac{1.5390825\times 10^{5}}{4(4-1)} = 1.2825688\times 10^{4} \]

\[ SSTr= \sum_{i} J* (\overline{y_{i.}}-\overline{y_{i..}})^2 = 4.8974019\times 10^{5} \]

\[ MSTr= \frac{SSTr}{I-1}= \frac{4.8974019\times 10^{5}}{4-1} = 1.6324673\times 10^{5} \]

\[ F= \frac{MSTr}{MSE} = 12.7281075 \]

Since \(\alpha=0.05\), use code \(qf((1-\alpha), I-1, I*(J-1))\) to find critical value(CV).

Since CV = 3.4902948 < 12.7281075 = F, we will reject the null hypothesis. Mixing techiniques do affect the strength of cement.

————————————————————————————————

3.10 (a):

null hypothesis: \(H_0: \mu_{1}=\mu_{2}=\cdots =\mu_{i}=\mu\)

alternative hypothesis: \(H_1:\) at least one \(\mu_{i}\) differs

Linear effects equation:

\(\overline{y_{ij}}=\mu_{i}+\epsilon_{ij}\)

\(\overline{y_{ij}}=\mu+\tau_{i}+\epsilon_{ij}\)

where, \(\mu_{i}\): mean of a certain population, \(\mu\): ground mean, \(\epsilon_{ij}\): error between populations, \(\tau_{i}\): effect of population.

Since we do not have a continuous range of type of mixing technique, it is a fixed effect model, hypothesis turns to:

null hypothesis: \(H_0: \epsilon_{ij}\) = 0, for all i

alternative hypothesis: \(H_1: \epsilon_{ij}\neq 0\), for some i

Ground mean = 15.04 \((lb/in^{2})\)

\[ SSE= \sum_{i} \sum_{j} (\overline{y_{ij}}-\overline{y_{i.}})^2 = 161.2 \]

\[ MSE= \frac{SSE}{I(J-1)}=\frac{161.2}{5(5-1)} = 8.06 \]

\[ SSTr= \sum_{i} J* (\overline{y_{i.}}-\overline{y_{..}})^2 = 475.76 \]

\[ MSTr= \frac{SSTr}{I-1}= \frac{475.76}{5-1} = 118.94 \]

\[ F= \frac{MSTr}{MSE} = 14.7568238 \]

Since \(\alpha=0.05\), use code \(qf((1-\alpha), I-1, I*(J-1))\) to find critical value(CV).

Since CV = 2.8660814 < 14.7568238 = F, we will reject the null hypothesis. Cotton content does affect the mean tensile strength.

————————————————————————————————

3.20 (a):

null hypothesis: \(H_0: \mu_{1}=\mu_{2}=\cdots =\mu_{i}=\mu\)

alternative hypothesis: \(H_1:\) at least one \(\mu_{i}\) differs

Linear effects equation:

\(\overline{y_{ij}}=\mu_{i}+\epsilon_{ij}\)

\(\overline{y_{ij}}=\mu+\tau_{i}+\epsilon_{ij}\)

where, \(\mu_{i}\): mean of a certain population, \(\mu\): ground mean, \(\epsilon_{ij}\): error between populations, \(\tau_{i}\): effect of population.

Since we do not have a continuous range of type of mixing technique, it is a fixed effect model, hypothesis turns to:

null hypothesis: \(H_0: \epsilon_{ij}\) = 0, for all i

alternative hypothesis: \(H_1: \epsilon_{ij}\neq 0\), for some i

Ground mean = 1548.3333333 \((lb/in^{2})\)

\[ SSE= \sum_{i} \sum_{j} (\overline{y_{ij}}-\overline{y_{i.}})^2 = 4.0933333\times 10^{4} \]

\[ MSE= \frac{SSE}{I(J-1)}=\frac{4.0933333\times 10^{4}}{4(3-1)} = 5116.6666667 \]

\[ SSTr= \sum_{i} J* (\overline{y_{i.}}-\overline{y_{..}})^2 = 2.8633333\times 10^{4} \]

\[ MSTr= \frac{SSTr}{I-1}= \frac{2.8633333\times 10^{4}}{4-1} = 9544.4444444 \]

\[ F= \frac{MSTr}{MSE} = 1.8653637 \]

Since \(\alpha=0.05\), use code \(qf((1-\alpha), I-1, I*(J-1))\) to find critical value(CV).

Since CV = 4.0661806 > 1.8653637 = F, we will not reject the null hypothesis. There is no difference in compressive strength due to the rodding level.

(b)

P-value using F statistic = 0.2137815.