you have said that is not complete what do you mean by that / Li Li CES BOUT POP
ID: 3170148 • Letter: Y
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you have said that is not complete what do you mean by that /
Li Li CES BOUT POPULATION VARIANCES 7.4 Tests for Comparing t 2 Population Variances Bio. 7.18 A wildlife biologist was interested in determining the effect of raising deer in captivity on the size of the deer. She decided to consider three populations: deer raised in the wild, deer raised on large hunting ranches, and deer raised in zoos. She randomly selected eight deer in each of the three environments and weighed the deer at age 1 year. The weights (in pounds) are given in the following table. Environment Weight (in pounds) of Deer Wild 114.7 128.9 111.5 116.4 134.5 126.7 120.6 129.59 Ranch 120.4 91.0 119.6 119.4 150.0 169.7 100.9 76.1 103.1 90.7 129.5 75.8 182.5 76.8 87.3 77.3 Zoo a. The biologist hypothesized that the weights of deer from captive environments would have a larger level of variability than the weights from deer raised in the wild. Do the data support her contention?Explanation / Answer
Please note that we cannot provide solutions using the paid softwares such as sas , however, we shall provide the answer using the open source alternative R , the concepts remain the same
We shall have to perform the anova analysis to check if the claim is true or not so we formulate the hypothesis as
H0 : There is no difference in the variance of wild vs captive
H1 : There is a significant difference in the variance of the wild vs captive
we shall derive a new variable called "captive" that tells whether the weights are from wild or from "ranch"/zoo
The r snippet is as follows
wild <-c(114.7,128.9,111.5,116.4,134.5,126.7,120.6,129.59)
ranch <-c(120.4,91,119.6,119.4,150,169.7,100.9,76.1)
zoo <- c(103.1,90.7,129.5,75.8,182.5,76.8,87.3,77.3)
data.df<- data.frame(cbind(wild,ranch,zoo))
library(reshape2)
data.df <- melt(data.df)
data.df$captive <- ifelse(data.df$variable=="wild",0,1)
library(stats)
var.test(data.df$value,data.df$captive)
The results
> var.test(data.df$value,data.df$captive)
F test to compare two variances
data: data.df$value and data.df$captive
F = 3417.1, num df = 23, denom df = 23, p-value < 2.2e-16 ## as the p value is less than 0.05 hence we reject the null hypothesis in favor of alternate hypothesis and conclude that the variance are differnt for captive and wild environments
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
1478.226 7899.171
sample estimates:
ratio of variances
3417.128
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