meana10 <- mean(c(1530,1530,1440))
meanb15 <- mean(c(1610,1650,1500))
meanc20 <- mean(c(1560,1730,1530))
meand25 <- mean(c(1500,1490,1510))#meana10=1500,meanb15=1586.6,meanc20=1606,meand25=1500
grandmean <- mean(c(meana10,meanb15,meanc20,meand25))#grandmean=1548.3
sse_10 <- sum((a10-meana10)^2)
sse_15 <- sum((b15-meanb15)^2)
sse_20 <- sum((c20-meanc20)^2)
sse_25 <- sum((d25-meand25)^2)
sse <- sum(sse_10,sse_15,sse_20,sse_25)
MSE <- sse/8#sse_10=5400,sse_15=12066.6,sse_20=23266.6,sse_25=200,sse=40933.3,mse=5116.6
sstr_10 <- sum((meana10-grandmean)^2*5)
sstr_15 <- sum((meanb15-grandmean)^2*5)
sstr_20 <- sum((meanc20-grandmean)^2*5)
sstr_25 <- sum((meand25-grandmean)^2*5)
sstr <- sum(sstr_10,sstr_15,sstr_20,sstr_25)
mstr <- sstr/4#sstr_10=7008.3,sstr_15=4408.3,sstr_20=10208.3,sstr_25=7008.3,sstr=28633.3,mstr=9544.4
F <- mstr/MSE
qf(0.95,3,8)## [1] 4.066181#f=1.86 #4.06 is the critical value less than the f statistic we reject the null hypothesis #there is evidance toto support cotton content affects the meantensile strength.
pf(1.86,3,8, lower.tail = FALSE)## [1] 0.2146797##b)the p value is 0.21.
a10 <- c(1530,1530,1440)
b15 <- c(1610,1650,1500)
c20 <- c(1560,1730,1530)
d25 <- c(1500,1490,1510)
meana10 <- mean(c(1530,1530,1440))
meanb15 <- mean(c(1610,1650,1500))
meanc20 <- mean(c(1560,1730,1530))
meand25 <- mean(c(1500,1490,1510))
grandmean <- mean(c(meana10,meanb15,meanc20,meand25))
sse_10 <- sum((a10-meana10)^2)
sse_15 <- sum((b15-meanb15)^2)
sse_20 <- sum((c20-meanc20)^2)
sse_25 <- sum((d25-meand25)^2)
sse <- sum(sse_10,sse_15,sse_20,sse_25)
MSE <- sse/8
sstr_15 <- sum((meana15-grandmean)^2*5)
sstr_20 <- sum((meanb20-grandmean)^2*5)
sstr_25 <- sum((meanc25-grandmean)^2*5)
sstr_30 <- sum((meand30-grandmean)^2*5)
sstr_35 <- sum((meane35-grandmean)^2*5)
sstr <- sum(sstr_15,sstr_20,sstr_25,sstr_30,sstr_35)
mstr <- sstr/4
F <- mstr/MSE
qf(0.95,3,8)
pf(1.86,3,8, lower.tail = FALSE)