chi-square test for a test cross 06-11-2014d1815

This test can help determine if observed ratios of discrete outcomes match the expected ratio of outcomes.

Table of critical chi-square values here
http://www.itl.nist.gov/div898/handbook/eda/section3/eda3674.htm

R code
{
chisq.test(c(75,29),p = c(3/4,1/4))
}


R code
{
green<-75
white<-29
sum<-green+white
ratio_of_green<-3/4
ratiogreen<-3/4
ratiowhite<-1/4
expectedgreen<-ratiogreen*sum
cat("Expected number of Green Flowers",expectedgreen,fill=T)
Expected number of Green Flowers 78
cat("Expected number of Green Flowers:",expectedgreen,fill=T)
Expected number of Green Flowers: 78
cat("Expected number of Green Flowers",expectedgreen,fill = T)
Expected number of Green Flowers 78
expectedwhite<-ratiowhite*sum
cat("Expected number of Green Flowers:",expectedwhite,fill = T)
Expected number of Green Flowers: 26
cat("Expected number of White Flowers:",expectedwhite,fill= T)
Expected number of White Flowers: 26
zeroone<-75
zerotwo<-29
e1=expectedgreen
e2=expectedwhite
chi<-((zeroone-e1)^2/e1)+((zerotwo-e2)^2/e2)
cat("Calculated value of Chi-square from data:",chi,fill = T)
Calculated value of Chi-square from data: 0.4615385
}





> Example 1.3.1
> green<-75
> white<-29
> sum<-green+white
> ratio-oi-green<-3/4
> ratiogreen<-3/4
> ratiowhite<-1/4
> expectedgreen<-ratiogreen*sum
> cat(“Expected number of Green Flowers:",expectedgreen,fiIl = T)
Expected number of Green Flowers: 78
> expectedwhite<-ratiowhite*sum
> cat("Expected number of Green o Fl wers:",expectedwhite.fi|l = T)
Expected number of Green Flowers: 26 .
> cat("Expected number of White Flowers:".9XD9Cl9dWhile.lll| = Tl
Expected number of White Flowers: 26
> O1<-75
> O2<-29
> e1 = expectedgreen
> e2 = expectedwhite
> ohi<-((01-e1)“ 2/e1)+((O2-e2) " 2/e2)
> oat(“Calculated value of Chi-square from data:",ohi,?l| = T)
Calculated value of Chi-square from data: 0.4615385


It can also be done in the following way.
> ohisq.test(o(75,29),p = o(3/4,1/4))
Chi-squared test for given probabilities
data: o(75, 29)
X—squared = 0.4615, df = 1, p-value = 0.4969