(1) 10 points. Consider the cloud seeding case study in Chapter 3 of The Sleuth

Question

(1) 10 points. Consider the cloud seeding case study in Chapter 3 of The Sleuth this is case0301 in the Sleuth3 package in R. The "Statistical Conclusions" on p. 59 of The Sleuth rely on the rainfall measurements having been trans formed to the log scale. Perform a two-sample t-test on these data without the log transformation, allowing the variance of the two populations to be dif ferent (ie.. don't use the var.equal = TRUE argument). Do the conclusions change from those based on the log-scale data (clearly, the numbers will be different, but think about the statistical conclusions)? (2) 10 points. Exercise 3.28 in The Sleuth. The R function t.test has an additional argument called subset, and that provides one way to perform an analysis without the outlying observation (3) 20 points. Exercise 3.32 in The Sleuth

Explanation / Answer

1)

The complete R snippet is as follows

install.packages("Sleuth3")

library(Sleuth3)
data(package = "Sleuth3")

data("case0301")
case0301

## t test with out log

seeded <- case0301[case0301$Treatment=="Seeded",]
unseeded <- case0301[case0301$Treatment=="Unseeded",]

t.test(seeded$Rainfall,unseeded$Rainfall,var.equal = FALSE)

## t test with log
  

seeded <- case0301[case0301$Treatment=="Seeded",]
unseeded <- case0301[case0301$Treatment=="Unseeded",]

t.test(log(seeded$Rainfall),log(unseeded$Rainfall),var.equal = FALSE)

The results are

> t.test(seeded$Rainfall,unseeded$Rainfall,var.equal = FALSE)

   Welch Two Sample t-test

data: seeded$Rainfall and unseeded$Rainfall
t = 1.9982, df = 33.855, p-value = 0.05377 ## as the p value is not less than 0.05 , hence we can concldue that the result is not significant at alpha = 0.05
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-4.764295 559.556603
sample estimates:
mean of x mean of y
441.9846 164.5885

t.test(log(seeded$Rainfall),log(unseeded$Rainfall),var.equal = FALSE)

   Welch Two Sample t-test

data: log(seeded$Rainfall) and log(unseeded$Rainfall)
t = 2.5444, df = 49.966, p-value = 0.01408 ## as the p value is less than 0.05 , hence we conclude that the difference in the mean values are statistically signficant
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.2408498 2.0467125
sample estimates:
mean of x mean of y
5.134187 3.990406

as we see the results become statistically signficant when we log transform the rainfall data

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