Sample_ID Value 1. 17.49 1. 20.73 1. 16.66 1. 26.38 1. 21.32 1. 16.72 1. 21.95 1

Question

Sample_ID Value

1. 17.49

1. 20.73

1. 16.66

1. 26.38

1. 21.32

1. 16.72

1. 21.95

1. 22.95

1. 22.3

2. 19.79

3. 21.06

4. 20.27

5. 19.57

6. 18.45

7. 20.79

8. 19.97

9. 19.99

1. (30 points) Testing scale difference. We have two independent samples randomly generated from two population distributions: {X1, X2, . . . , Xm} and {y, ½, . . . , ·Suppose E(X) E(Y) = , Var(X) = . Var(X) = . We are interested in testing H0 : 1 = 2 vs. Ha : 2. The data is in 'Hw6Q1.txt' (the first column is the sample ID). Perform the hypothesis test at 0.05 significance level through (a) The Siegel-Tukey test (b) The Kolmogorov-Smirnov test. For the original data, draw the empirical CDFs of the two samples in one plot (hand-drawing or through R) and indicate where the maximum difference occurs.

Explanation / Answer

## Using R

> library(jmuOutlier)
>
>
> x=c(17.49,20.73,16.66,26.38,21.32,16.72,21.95,22.95,22.3)
> y=c(19.79,21.06,20.27,19.57,18.45,20.79,19.97,19.99)
>

## a) Siegel-Tukey test
> siegel.test(x, y, alternative = "two.sided")
[[1]]
[1] " Siegel-Tukey test"

$alternative
[1] "two.sided"

$rank.x
[1] 5 15 1 2 10 4 7 3 6

$rank.y
[1] 12 11 17 9 8 14 13 16

$p.value
[1] 0.00555

## Comment: The estimated p-value is less than 0.05 level of significance. hence, we can reject the null hypothesis and conclude that the standard deviation of these two population is significantly different at 0.05 level of significance.

## b) Kolmogorov_Smirnov test

> ks.test(x, y, alternative = "two.sided")

Two-sample Kolmogorov-Smirnov test

data: x and y
D = 0.55556, p-value = 0.07857
alternative hypothesis: two-sided

## Comment: The estimated p-value is greater than 0.05 level of significance. Hence, we can not reject the null hypothesis and conclude that the standard deviation of these two population is insignificant different at 0.05 level of significance.

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