Kelly drinks 0.5 liters of coffee and 0.3 liters of yogurt drink at breakfast. How much did she drink in total in milliliters?
2 answers:
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Answer:
Consider the following calculations
Step-by-step explanation:
The complete R snippet is as follows
install.packages("Sleuth3")
library("Sleuth3")
attach(case0502)
data(case0502)
## plot
# plots
boxplot(Percent~ Judge, data=case0502,ylab="Values",
main="Boxplots of the Data",col=c(2:7,8),horizontal=TRUE)
# perform anova analysis
a<- aov(lm(Percent~ Judge,data=case0502))
#summarise the results
summary(a)
### we can use the independent sample t test here
sp<-case0502[which(case0502$Judge=="Spock's"),]
nsp<-case0502[which(case0502$Judge!="Spock's"),]
## perform the test
t.test(sp$Percent,nsp$Percent)
The results are CHECK THE IMAGE ATTACHED
b)
> summary(a)
Df Sum Sq Mean Sq F value Pr(>F)
Judge 6 1927 321.2 6.718 6.1e-05 *** as the p value is less than 0.05 , hence there is a significant difference in the percent of women included in the 6 judges’ venires who aren’t Spock’s judge
Residuals 39 1864 47.8
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
c)
t.test(sp$Percent,nsp$Percent)
Welch Two Sample t-test
data: sp$Percent and nsp$Percent
t = -7.1597, df = 17.608, p-value = 1.303e-06 ## as the p value is less than 0.05 , hence we reject the null hypothesis in favor of alternate hypothesis and conclude that there is a significant difference in the percent of women incuded in Spock’s venires versus the percent included in the other judges’ venires combined
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-19.23999 -10.49935
sample estimates:
mean of x mean of y
14.62222 29.49189
