Statistics for laboratory scientists II

Solutions for the homework problems for lecture 2

1. Here is the R code for creating the table and calculating the expected counts under the null hypothesis of p_i=1/10 for all i.

       mydata <- c(98, 99, 100, 89, 107, 114, 100, 112, 85, 96)
       n <- sum(mydata)
       ex <- n * rep(1/10, 10)

   1. Here is the code for calculating the chi-square test statistic and corresponding P-value.

          chi <- sum( (mydata-ex)^2 / ex)               # value = 7.56
          1 - pchisq(chi, 9)                            # P-value = 0.58

   2. Here is the code for calculating the LRT statistic and corresponding P-value.

          lrt <- 2 * sum( mydata * log(mydata/ex) )     # value = 7.58
          1 - pchisq(lrt, 9)                            # P-value = 0.58

   3. Here is some code for using computer simulations to estimate P-values.

          n.sim <- 1000
          results <- matrix(ncol=2, nrow=n.sim)
          for(i in 1:n.sim) {	
             # simulate data
             simdat <- rmultinom(1, n, rep(0.1,10))
      
             # calculate statistics
             chi.sim <- sum( (simdat - ex)^2 / ex)
             lrt.sim <- 2 * sum( simdat * log(simdat/ex) )
      
             results[i,] <- c(chi.sim, lrt.sim)
           }
      
           # p-value for chi-square test
           mean(results[,1] >= chi)  # p-value = 0.57 (for my simulations)
       	
           # p-value for likelihood ratio test (LRT)
           mean(results[,2] >= lrt)  # p-value = 0.58 (for my simulations)

   4. We fail to reject the null hypothesis of equal frequencies. The observed differences in the counts could reasonably be ascribed to chance variation.

2. With the second version of the table, in applying either the chi-square test or likelihood ratio test, we would get identical statistics and P-values. However, these tests look for general differences from equal frequencies, and not the sort of trend in this observed table.

   With such data, we would want to use a different statistical test, such as of whether the mean (or median) is equal to 4.5 or not. Of course, one would ideally choose to perform such a test before obtaining the data rather than after having observed such a trend.

[ Main page | 4th term syllabus | R for Windows ]

Last modified: Tue Mar 28 17:52:44 EST 2006