Answer 11.9
The data
chan2 = c(3.8, 2.7, 4.9, 3.4, 3.7, 4.5, 4.2, 2.8, 3.5, 4.6)
chan4 = c(3.6, 4.0, 4.5, 5.2, 4.8, 4.3, 5.7, 3.5, 3.7, 5.8)
the.t.test <- t.test(chan2, chan4, conf.level = 0.90, var.equal = T)
the.t.test##
## Two Sample t-test
##
## data: chan2 and chan4
## t = -1.9653, df = 18, p-value = 0.06501
## alternative hypothesis: true difference in means is not equal to 0
## 90 percent confidence interval:
## -1.3176448 -0.0823552
## sample estimates:
## mean of x mean of y
## 3.81 4.51Statement:
The difference between the two means is statistically different from zero at the alpha level of 0.10, t(18) = -1.97, p = 0.06.
The 90% confidence interval
the.t.test$conf.int## [1] -1.3176448 -0.0823552
## attr(,"conf.level")
## [1] 0.9The 90% CI is (-1.32, -0.08)
Answer 11.10
new.bat <- c(3.3, 6.4, 3.9, 5.4, 5.1, 4.6, 4.9, 7.2)
old.bat <- c(4.2, 2.9, 4.5, 4.9, 5.0, 5.1, 3.2, 4.0)
the.t.test2 <- t.test(new.bat, old.bat, alternative = "greater", var.equal = T)
the.t.test2##
## Two Sample t-test
##
## data: new.bat and old.bat
## t = 1.6413, df = 14, p-value = 0.0615
## alternative hypothesis: true difference in means is greater than 0
## 95 percent confidence interval:
## -0.06395579 Inf
## sample estimates:
## mean of x mean of y
## 5.100 4.225Statement:
At the alpha level of 0.05, the mean of “new bat” is not statistically greater than the mean of the “old bat”