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Workshop 4
Workshop 4
Workshop 4
Workshop 4
Workshop 4
Workshop 4
Workshop 4
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Workshop 4

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  • 1. MAS230 : Biostatistical Methods Tutorial 4SPSS Guidelines • If you have not completed Tutorial 3, please do so before beginning Tutorial 4. • Kappa Test and McNemar’s Test Revisited: – Last week, we examined how to carry out a kappa test or McNemar’s test if we have totals for a 2 × 2 table. This required that we use 0s and 1s to represent rows and columns and then weight each unique combination of 0s and 1s by the corresponding quantity in the table. In essence, these weights created replicates for each of the unique 0-1 combinations. ∗ e.g. Consider the table we examined last week. The top left cell corre- 30 20 20 40 sponds to a row value of ‘0’ and a column value of ‘0’. When we weighted by cases, this cell was given a weight of 30. In essence, SPSS created 30 separate entries with a row value of ‘0’ and a column value of ‘0’, even though this does not appear anywhere. – Suppose that, instead of a table like the one above, we are simply given 0-1 variables values for each person. These 0-1 variables correspond to a row and column of a table of total counts. Instead of having the totals counts falling in each cell, however, we have the individual data used to produce those total counts. In this case, we can replicate the analysis we did last week, but we no longer need to weight by cases since the data we have are essentially the expanded version of the table. – Select: Analyze −→ Descriptive Statistics −→ Crosstabs · · · and input the variable corresponding to the rows and variable corresponding to columns – Click the Statistics button and tick the boxes for Kappa or McNemar – These instructions apply to other common analyses for tables, including chi- square tests. • You should be able to carry out all other analyses using the instructions provided in the course reader and previous tutorials. 1
  • 2. R Guidelines • If you have not completed Tutorial 3, please do so before beginning Tutorial 4. • Recall that you will need to determine the location of data files when you save them to your computer (e.g., “C:Documents and Settings · · · dataset3.sav.” Re- member that you will need to report this file location to R as “C:Documents and Settings · · · dataset3.sav” for R to read the file). To determine this, you may need to right-click on the file and select Properties. For Mac users, you will need to command-click on the file and select Get Info. • Remember that, to open SPSS data files, you will need to load the foreign package by running the code: > library(foreign) The following code will read in the data to the variable “dataset3” after you replace my file location with the correct file location on your computer: > dataset3 <- read.spss(‘‘/Users/ryan/Documents/MAS230/dataset3.sav’’) To access the variables, run the code: > attach(dataset3) • Recall from previous tutorials that Wilcoxon signed-rank tests and Mann-Whitney U tests can be carried out using the function wilcox.test(), sign tests can be car- ried out using the binom.test() function, and t-tests (one-sample, paired, and two- sample) can be carried out using the function t.test(). Type in ?wilcox.test, ?binom.test(), or ?t.test() to see R’s help file on these particular functions. • Instructions for Q-Q plots, bar charts, kappa tests, and McNemar’s test are provided in the previous tutorial. • Tests for Independent Proportions: Tests for two independent proportions can be carried out using the prop.test() function. This requires that you specify the number of successes for the two samples as well as the sample sizes. Suppose we had two samples of size 25 and 30, and we observed 10 successes in the first sample, and 18 successes in the second sample. Further suppose that we wanted to test the hypotheses H0 : π1 = π2 H1 : π1 < π2 We could carry out the test by running the following code: 2
  • 3. > prop.test(x = c(10, 18), n = c(25, 30), alternative = "less", correct = FALSE) 2-sample test for equality of proportions without continuity correction data: c(10, 18) out of c(25, 30) X-squared = 2.1825, df = 1, p-value = 0.06979 alternative hypothesis: less 95 percent confidence interval: -1.00000000 0.01821449 sample estimates: prop 1 prop 2 0.4 0.6 Thus, the p-value is 0.06979. Note that this uses a normal approximation, but it reports a χ2 test statistic. To get the z-statistic, we can simply take the square root √ of the χ2 test statistic, so the test statistic is given by z = 2.1825 = 1.477329.• Chi-Square Tests: To carry out a chi-square test, use the function chisq.test(). Suppose we have the table of counts for assignment of males and females to treatment and control groups, and we want to determine whether assignment to treatment or control is associated with sex (i.e. probability of being assigned to treatment or control changes based on your sex). To carry out a chi-square test of independence, we first must construct Treatment Control Male 30 20 Female 20 40 a matrix (think of this as being like a table) for the counts in the table. Running the code > x <- matrix(c(30, 20, 20, 40), nrow = 2, ncol = 2, byrow = TRUE) creates a matrix with two rows and two columns, and it fills in this matrix with the numbers 30, 20, 20, and 40 by going across the rows (instead of down the columns). Running the code > chisq.test(x, correct = FALSE) Pearson’s Chi-squared test data: x X-squared = 7.8222, df = 1, p-value = 0.005161 3
  • 4. carries out a chi-square test and reports a test statistic of χ2 = 7.8222 with a p-value of 0.005161.• Fisher’s Exact Test: To perform Fisher’s exact test, use the function fisher.test(). This requires a matrix specification in the same format as for the chi-square test. Running the code > fisher.test(x) Fisher’s Exact Test for Count Data data: x p-value = 0.007045 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 1.283785 7.051709 sample estimates: odds ratio 2.968567 carries out Fisher’s exact test and produces a p-value of 0.007045.• Levene’s Test: To carry out Levene’s test, you will first need to install the car package if you are working from your home computer. (If you do not recall how to do this, refer back to Tutorial 1, where directions were given for installing the foreign package.) Once you have installed the car package, you can load it by running > library(car) Levene’s test can then be carried out by using the leveneTest() function (specified as levene.test() in older versions of R). Before using this function, however, you need to first carry out a linear regression using the lm() function. You must take all of your columns and save them to one variable (I’ll call this outcome), and then you should create a second variable that records the group to which each outcome corresponds (I’ll call this group). Then running the following code > lm(outcome ∼ factor(group)) will carry out a linear regression of outcome on the different group categories. The dependent variable is specified on the left of the ‘∼,’ and the independent variable (or variables) is specified to the right of the ‘∼.’ We will need to save the regression in a variable that is then passed as an argument to the leveneTest function, > model <- lm(outcome ∼ factor(group)) > leveneTest(model) 4
  • 5. – Note that the output from Levene’s test is slightly different for SPSS and R. This is because SPSS uses absolute deviations from the mean, whereas R uses absolute deviations from the median.• After completing the tutorial, you will want to compare your R code to mine to better understand any differences in output. You will also want to check your solutions. 5
  • 6. Questions 1. Consider Data Set 75. These data come from Judge, M.D. et al (1984). “Thermal shrinkage temperature of intramuscular collagen of bulls and steers,” Journal of Animal Science 59: 706–9, and are reproduced in Samuels and Witmer (1999), Statistics for Life Sciences, 2nd Edition, Prentice Hall, p. 357. The study is designed to assess the effect of electrical stimulation of a beef carcass in terms of improving the tenderness of the meat. In this test, beef carcasses were split in half. One side was subjected to a brief electrical current while the other was an untreated control. From each side a specimen of connective tissue (collagen) was taken, and the temperature at which shrinkage occurred was determined. Increased tenderness is related to a low shrinkage temperature. Carry out analyses to assess the impact of electrical stimulation on the meat tender- ness. Use both parametric and non-parametric methods and compare the results. Suppose acceptable tenderness corresponds to a shrinkage temperature less than 69 degrees. How would you test to see if the proportions of acceptable tenderness values differed under the two treatments? Use SPSS to create appropriate variables to enable this to be tested and carry out the analysis. Don’t forget that the sample sizes are small here. 2. Refer to Data Set 76. These data come from Mochizuki, M. et al (1984). “Effects of smoking on fetoplacentalmaternal system during pregnancy,” American J. Ob- stet. Gyn. 149: 13–20. The study considered the effects of smoking during preg- nancy by examining the placentas from 58 women after childbirth. Each mother was classified as a non-, moderate or heavy smoker during pregnancy, and the outcome measure was presence or absence of atrophied placental villi, finger-like structures that protrude from the wall to increase absorption area. Combine the two smoking classes to create a “smoker” class and carry out an appropriate test for association of villi atrophy with smoking status. (Note to SPSS users: This means that you will have to use Transform → Compute Variable. . . to create a new variable. Since smoker status is denoted by characters [H, M, N], you will need to use quotes around these in the “Numeric Expression:” box.) Given there are three ordered classes of smoking (none < moderate < heavy) think about how you might display such data. 3. An environmental scientist studying the impact of pollution on species diversity along two nearby rivers carried out a survey in which plots (quadrats) of size 30 metres by 20 metres were randomly chosen from along the banks of the rivers. 6
  • 7. Within each quadrat the numbers of different tree species were recorded. The datawere as follows: Valley River Ridge River 9 9 15 12 13 13 10 6 7 10 13 13 8 11 9 9 18 6 9 9 10 9 14 11 7 8 6 11What would you conclude from these data in terms of differences in species diversity?Think about the nature of the data, what might be the best way to compare them,what assumptions are being made in the comparison, etc. Are there any valueswhich might need special consideration? What is their effect on the various analysesif included or excluded? 7

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