baseline <- c(140,138,150,148,135)
Week2 <- c(132,135,151,146,130)
t.test(Week2,baseline,alternative="two.sided",paired=TRUE)
##
## Paired t-test
##
## data: Week2 and baseline
## t = -2.2616, df = 4, p-value = 0.08652
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -7.5739122 0.7739122
## sample estimates:
## mean of the differences
## -3.4
CI <- 1100 + c(-1,1)*qt(0.975,8)*30/sqrt(9)
CI
## [1] 1076.94 1123.06
chisq.test(c(3,1),p=c(0.5,0.5))
## Warning in chisq.test(c(3, 1), p = c(0.5, 0.5)): Chi-squared approximation may
## be incorrect
##
## Chi-squared test for given probabilities
##
## data: c(3, 1)
## X-squared = 1, df = 1, p-value = 0.3173
lambda <- 0.01
t <- 1787
ppois(10,lambda*t)
## [1] 0.03237153
n_plac <- 9
n_treat <- 9
mudiff_treat <- -3
mudiff_plac <- 1
sddiff_plac <- 1.8
sddiff_treat <- 1.5
s <- sqrt(((n_plac-1)*sddiff_plac^2 + (n_treat-1)*sddiff_treat^2)/(n_plac + n_treat -2))
t <- (mudiff_treat-mudiff_plac)/(s*sqrt(1/n_plac + 1/n_treat))
2*pt(t,n_plac + n_treat-2)
## [1] 0.0001025174
# The value 1078 is in the 90% confidence interval. As the 95% is wider than the 90% interval, you would not reject.