Question

In: Math

Suppose you conduct 10 significant tests and obtain the following p-values: Test 1 2 3 4...

Suppose you conduct 10 significant tests and obtain the following p-values:

Test 1 2 3 4 5 6 7 8 9 10

p-value 0.001 0.030 0.002 0.006 0.040 0.003 0.010 0.100 0.020 0.004

• Which tests’ null hypotheses will you reject if you wish to control the family-wise error rate (FWER) at a significance level of 0.05?

• Which tests’ null hypotheses will you reject if you wish to control the false discovery rate (FDR) at a level of 0.05? Use the Benjamini-Hochberg method to answer this question by hand.

• Verify the results by using the related function in R

Solutions

Expert Solution

Which tests’ null hypotheses will you reject if you wish to control the family-wise error rate (FWER) at a significance level of 0.05?

10 tests are conducted. We have p-value for each test. We need to apply the Family-wise error rate (FWER)

Bonferroni Procedure for each test:

at level of significance alpha = 0.05,

So the results of the series of tests are as follows:

test # p-value FWER alpha test result
1 0.001 0.005 reject NULL
2 0.030 0.005
3 0.002 0.005 reject NULL
4 0.006 0.005
5 0.040 0.005
6 0.003 0.005 reject NULL
7 0.010 0.005
8 0.100 0.005
9 0.020 0.005
10 0.004 0.005 reject NULL

Therefore, test # 1,3, 6, 10 have p-values that enable us to reject Null Hypothesis controlling for FWER at a significance level of 0.05.

Which tests’ null hypotheses will you reject if you wish to control the false discovery rate (FDR) at a level of 0.05? Use the Benjamini-Hochberg method to answer this question by hand.

Benjamini-Hochberg method is applied and results are given in table below for FDR level = 0.05

original test # ordered p-value [k] k/m*alpha p < k/m*alpha Reject NULL
1 0.001 1 0.005 TRUE Reject NULL
3 0.002 2 0.01 TRUE Reject NULL
6 0.003 3 0.015 TRUE Reject NULL
10 0.004 4 0.02 TRUE Reject NULL
4 0.006 5 0.025 TRUE Reject NULL
7 0.010 6 0.03 TRUE Reject NULL
9 0.020 7 0.035 TRUE Reject NULL
2 0.030 8 0.04 TRUE Reject NULL
5 0.040 9 0.045 TRUE Reject NULL
8 0.100 10 0.05 FALSE

Therefore for all tests expect test # 8, we reject the null hypothesis.

Verify the results by using the related function in R

> pvec <- c(0.001,0.03,0.002,0.006,0.04,0.003,0.01,0.1,0.02,0.004)
> length(pvec)
> alpha_FDR <- 0.05
> alpha_seq <- seq(1:10)/10*alpha_FDR
> alpha_seq
[1] 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0.045 0.050

### ordered pvalues

> pvec[order(pvec)]
[1] 0.001 0.002 0.003 0.004 0.006 0.010 0.020 0.030 0.040 0.100

> compare_pvec <- ifelse(pvec[order(pvec)] <= alpha_seq, 1,0)
> compare_pvec
[1] 1 1 1 1 1 1 1 1 1 0

### max where 1

> max(which(compare_pvec == 1))
[1] 9

### ALL Tests that are ordered from 1 to above cut-off = 9

> order(pvec)[1:9]
[1] 1 3 6 10 4 7 9 2 5

### reject null hypothesis for all these tests

### verification complete. Reject null hypothesis for all except test number 8


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