Modern Statistics For The Social And Behavioral Sciences 1st Edition Rand Wilcox - Solutions

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Extend the method used in Section 3.3.1.1 to the three-factor all-random model (3.94)in Section 3.3.2.1 to derive the distributions of SS????, SSa, SSb, SSc and SSe that lead to the ANOVA table (Table 3.7).
Similar to Problem 3.1 but now for the two-factor additive crossed REM of Section 3.2.1, verify the distributions and independence of the SS, along with the E SS values given in Table 3.3.
For the two-factor crossed REM with interaction of Section 3.2.1, verify that (i) the cross terms from squaring and summing (3.42) indeed vanish, (ii) the r.h.s. terms in (3.43) are mutually independent, and (iii) the r.h.s. terms follow the stated distributions.
Using the plasma retinol data in Chapter 1, remove the former smokers as well as the leverage points from the data set and use the R function logrsm to verify that as plasma beta-carotene increases, the likelihood that a participant is a smoker decreases. Also verify that when testing at the 0.05
For the data in Table 16.6, verify that the results returned by the R function ODDSR.CI are:$odds.ratio[1] 0.4776119$ci[1] 0.1404619 1.6240215
For the data in Exercise 7, determine which if any of the claimed cell probabilities appear to be incorrect using the R function binomci. Use the Bonferroni method with the goal that the probability of one or more Type I errors will be at most 0.05.
Someone claims that the proportion of adults getting low, medium, and high amounts of exercise is 0.5, 0.3, and 0.2, respectively. To check this claim you sample 100 adults and find that 40 get low amounts of exercise, 50 get medium amounts, and 10 get high amounts. Test the claim with α = 0.05.
Compute kappa for the data in Exercise 5. What does this estimate tell you?
You observe Income (daughter) Income (father) Total High Medium Low High 30 50 20 100 Medium 50 70 30 150 Low 10 20 40 70 Total 90 140 90 320 Estimate the proportion of agreement, and compute a 0.95 confidence interval using the R function binomci.
In Exercise 1, estimate the odds ratio and interpret the results.
In Exercise 1, would you reject the hypothesis that income and outlook are independent? Use α = 0.05. Would you use the φ coefficient to measure the association between income and outlook? Why?
Referring to the previous exercise, compute a 0.95 confidence interval for δ = p1+ − p+1. Test H0: p12 = p21.
You randomly sample 200 adults, determine whether their income is high or low, and ask them whether they are optimistic about the future.The results are Yes No Total High 35 42 77 Low 80 43 123 Total 115 85 200 What is the estimated probability that a randomly sampled adult has a high income and is
R has a built-in data set called leuk. The third column indicates survival times, in weeks, for patients diagnosed with acute myelogenous leukemia. The first column indicates the patient’s white blood cell count at the time of diagnosis. Compute Pearson’s correlation and compare the results to
An exploratory method for dealing with interactions is to assume the model Y = β0 + β1X1 + β2X2 + β3X1X2 + e is true and use a partial residual plot to check the adequacy of the third term (β3X1X2) assuming that Y has a linear association between β1X1 and β2X2. Describe some reasons why this
Repeat the previous exercise, only now use the R function ancova to compare the regression lines. At which X values do you get a significant difference with α = .05?
Compare the regression slopes in the previous exercise using the R function reg2ci. Verify that the significance level is .04 when using the default settings.
For the data in Table 15.3, create a smooth using the 4000 BC data only and compare it to the smooth for the 150 AD data.
Table 15.3 contains a portion of the data reported by Thompson and Randall-Maciver (1905) dealing with skull measurements of male Egyptians from different time periods. Here, only the data from 4000 BC and 150 AD are reported. Pool the data from both periods and create a smooth using rplot, kerreg,
For the data in Exercises 2 and 3, eliminate any outliers among the predictor values found by the MVE method and for the remaining data, test H0: β1 = β2 = 0 using the methods in Section 15.7.
For the data in Exercises 2 and 3, check for curvature using a partial residual plot.
Again using the lake data, check for any outliers among the TW and NIN values using the MVE method, eliminate any outliers that are found and plot a smooth for predicting Y using the R function rplot in Section 15.4.4. Compare the results to smooths you get when the outliers are not eliminated.
Repeat the previous exercise, only now use NIN (the average influent nitrogen concentration) to predict Y.
Table 6.4 reports data on 29 lakes in Florida. Assuming that you want to predict Y (the mean annual nitrogen concentration) given TW (water retention time), plot a smooth and comment on whether a straight regression line is advisable.
For the predictor X2 in Table 15.2, MAD/0.6745=20.756. For the running interval smoother with the span taken to be f = 1, what would be the estimate of Y given that X2 = 250, based on a 20% trimmed mean?What would be the estimate based on MOM?
For the plasma data in Chapter 1, use the function mat2grp to sort the data into two groups based on whether a participant is male or female.Compare these two groups in terms of dietary beta-carotene consumed(labeled BETADIET) and dietary retinol consumed (labeled RETDIET)using the R function
For the data used in the previous three exercises, verify that the R functions outpro and outproad identify the same points as outliers.Why was this expected?
For the data used in the previous two exercises, use the R functions robpca and Ppca.summary to argue that four components are needed to capture the variability among the data.
Using the data generated in the previous exercise, compute the principal component scores and verify that corresponding correlation matrix is approximately equal to the identity matrix.
Using R, generate data and store it in the R variable x using the commands set.seed(3)x=rmul(100,p=5,mar.fun=ghdist,h=.2,rho=.0)x[,1]=rlnorm(100)^2 This creates five independent random variables, the last four of which have symmetric, heavy-tailed distributions, and the first has a skewed
For the second example in Section 14.2.1, verify that the p-value is 0.354.
Exercise 12 in Chapter 9 described a data set dealing with scents. Download the data as described and verify the p-value reported in the first example of Section 14.2.1.
Consider a bivariate data set and imagine that the data are projected onto the x-axis and then checks for outliers are made using the MADmedian rule. The same is done when projecting the data onto the y-axis and both times no outliers are found. Why might the conclusion of no outliers be
Among the five groups in the previous exercise, only one group was found to have an outlier based on the boxplot rules in Chapter 2. What might explain why more significant differences are found when comparing groups based on MOM versus the mean?
Thompson and Randall-Maciver (1905) report four measurements of male Egyptian skulls from five different time periods. The first was maximal breadth of skull and the five time periods were 4000 BC, 3300 BC, 1850 BC, 200 BC, and 150 AD.A portion of the output from lincon, when comparing means (with
Perform all pairwise comparisons of the groups in Table 10.1 using the R function rmmcppb. Use the MOM estimate of location and use difference scores. Verify that a difference between groups 2 and 3 is found with FWE set at 0.05.
Use the R function msmedse to compute the standard errors for the median corresponding to the four groups in Table 10.1. Compare the results to the estimated standard error for the 20% trimmed means returned by trimse. Now note that when comparing groups 3 and 4, you reject with 20% trimmed means
Use the R function msmed to compare all pairs of groups based on medians using the data in Table 10.1. Verify that the output from msmed is$test:Group Group test crit se[1,] 1 2 0.1443384 2.63 0.2742167[2,] 1 3 0.4313943 2.63 0.3025886[3,] 1 4 0.5795942 2.63 0.3019699[4,] 2 3 0.9624726 2.63
Use the R function lincon to verify that for the data in Table 10.1, the hypothesis of equal 20% trimmed means is rejected for groups 3 and 4, with FWE equal to 0.05, but otherwise no differences are found.
Repeat the previous exercise using the R function hochberg in Section 13.3.4. Verify that the required sample sizes are 50, 10, 18, and 17.
For the data in Table 10.1, each group has 10 observations, so when using Tamhane’s method to compute confidence intervals for all pairwise differences, the degrees of freedom are 9 and the value of h in Table 11 of Appendix B is 4.3. Verify that if the goal is to have confidence intervals with
Repeat the previous exercise but with the Tukey-Kramer and Scheff´e methods instead.
Imagine you compare four groups with Fisher’s method and you reject the hypothesis of equal means for the first two groups. If the largest observation in the fourth group is increased, what happens to MSWG?What does this suggest about power when comparing groups one and two with Fisher’s method?
Referring to the previous exercise, which would be rejected with Rom’s procedure?
You perform five tests and get p-values 0.049, 0.048, 0.045, 0.047, and 0.042. Based on the Bonferroni inequality, which would be rejected with FWE equal to 0.05?
For the previous exercise, if you use Rom’s method, which tests would be rejected?
You perform six tests and get significance levels 0.07, 0.01, 0.40, 0.001, 0.1, 0.15. Based on the Bonferroni inequality, which would be rejected with FWE equal to 0.05?
Repeat the previous exercise, but now with n1 = n2 = n3 = n4 = n5 =10, ¯X1 = 20, ¯X2 = 12, s21= 5, s22= 6, s23= 4, s24= 10, and s25= 15.
For five independent groups, assume that you plan to do all pairwise comparisons of the means and you want FWE to be 0.05. Further assume that n1 = n2 = n3 = n4 = n5 = 20, ¯X1 = 15, ¯X2 = 10, s21= 4, and s22= 9, s23= s24= s25= 15, test H0 : μ1 = μ2 using (a) Fisher’s method(assuming the ANOVA
For the data used in Exercise 1, use the R function sppbi to test the hypothesis of no interactions. Verify that the p-value is 0.35.
Compare the results from the previous three exercises and comment on finding the optimal method for detecting a true difference among groups.
In the previous exercise, you could have used averages rather than difference scores when comparing the murderers to the controls. Using the R function sppba with the argument avg set equal to T, verify that the p-value is 0.38.
Repeat Exercise 1 using one-step M-estimator via the R function sppba.Verify that the p-value is 0.092.
The last example in Section 12.6.6 described a study comparing EEG measures for murderers to a control group. The entire data set, containing measures taken at four sites in the brain, can be downloaded from the author’s web page as described in Chapter 1 and is stored in the file eegall.dat.
Repeat the illustration in Section 11.5.1, but now use pbad2way with Mahalanobis distance by setting the argument pro.dis=F. Verify that for Factor B, now the p-value is 0.166.
For the data in Exercise 11, verify that if t2way is used to compare 20% trimmed means instead, the p-values for Factors A and B are 0.145 and 0.93. Why does the p-value for Factor A drop from 0.372 when comparing means to 0.145 with 20% trimmed means instead?
For the data in the previous exercise, verify that if t2way is used to compare 20% trimmed means, the p-value when testing the hypothesis of no interaction is reported to be 0.365.
Consider the following data A B Outcome 1 1 32 1 1 21 1 3 19 1 2 21 1 3 46 1 3 33 1 3 10 1 2 11 1 2 13 1 2 12 1 2 59 1 3 28 1 1 19 1 1 72 1 1 35 2 1 33 2 1 45 2 1 31 2 2 42 2 2 67 2 2 51 2 1 19 2 2 18 2 2 21 2 3 39 2 3 63 2 3 41 2 3 10 2 3 11 2 3 34 2 3 47 2 1 21 2 1 29 2 3 26 Read this data into
Verify that when comparing the groups in Table 10.1 based on medians and with the R function pbad2way in Section 11.4.1, the significance levels for Factors A and B are 0.39 and 0.056, respectively, and that for the test of no interaction, the significance level is 0.16.
This exercise is based on a study by Atkinson and Polivy (1976) where the general goal was to study people’s reactions to unprovoked verbal abuse. In the study, 40 participants were asked to sit alone in a cubicle and answer a brief questionnaire. After waiting far longer than it took to fill out
Referring to Exercise 7, imagine the hypothesis of no interactions is rejected. Is it reasonable to conclude that the interaction is disordinal?
Use the answer to the last exercise to make a general comment on interpreting main effects when there is a disordinal interaction.
In the previous exercise, suppose the hypothesis of no interactions is rejected and in fact there is a disordinal interaction. What does this suggest about using method 1 rather than method 2.
Imagine a study where two methods are compared for treating depression.A measure of effectiveness has been developed, the higher the measure the more effective the treatment. Further assume there is reason to believe that males might respond differently to the methods compared to females. If the
Make up an example where the population means in a 3-by-3 design have no interaction effect but main effects for both factors exists.
Consider a 2-by-2 design with population means Factor B Level 1 Level 2 Level 1 μ1 = 10 μ2 = 20 Factor A Level 2 μ3 = 40 μ4 = 10 Determine whether there is an interaction. If there is an interaction, determine whether there is an ordinal interaction for rows.
For a 2-by-4 design with population means Factor B Level 1 Level 2 level 3 Level 4 Factor Level 1 μ1 μ2 μ3 μ4 A Level 2 μ5 μ6 μ7 μ8 state the hypotheses of no main effects and no interactions.
Repeat the last exercise, only now use projection distances by setting the argument op=3. Verify that the p-value is 0.004. Why does op=3 avoid the error noted in the last exercise?
For the data in Table 10.6, use pbadepth to compare the groups using default values for the arguments. You will get the message:Error in solve.default(cov, ...) :system is computationally singular:reciprocal condition number = 2.15333e-18 The difficulty is that the depth of the null vector is being
For the data in Table 10.6, use t1way to compare 20% trimmed means and verify that the significance level is less than 0.01.TABLE 10.6: Hypothetical Data Group 1 Group 2 Group 3 Group 4 Group 5 10.1 10.7 11.6 12.0 13.6 9.9 9.5 10.4 13.1 11.9 9.0 11.2 11.9 13.2 13.6 10.7 9.9 11.7 11.0 12.3 10.0 10.2
For the data in Table 10.6, verify that the p-value, based on Welch’s test, is 0.98. Use the R Function t1way. (The data are stored on the author’s web page in the file CHtable10.6.dat.)
Consider the following data.G X G X G X 1 12 1 8 1 22 3 42 3 8 3 12 3 9 3 21 2 19 2 24 2 53 2 17 2 10 2 9 2 28 2 21 1 19 1 21 1 56 1 18 1 16 1 29 1 20 3 32 3 10 3 12 3 39 3 28 3 35 2 10 2 12 Store this data in an R variable having matrix mode with two columns corresponding to the two variables
Repeat the last exercise, but now use projection distances to measure the depth of the null vector. That is, set the argument op=3. Note that now the p-value is 0.0545.
For the data in Table 10.1, use the R function pbadepth to compare the groups using the modified one-step M-estimator, MOM. That is, set the argument est=mom. Verify that the p-value is 0.123.
For the data in Table 10.1, the ANOVA F test and Welch’s test were not significant with α = .05. Imagine that you want power to be 0.9 if the mean of one of the groups differs from the others by 0.2. (In the notation of Section 10.3, a = .2.) Verify that according to the R function bdanova1, the
Consider J = 5 groups with population means 3, 4, 5, 6, and 7, and a common variance σ2 p = 2. If n = 10 observations are sampled from each group, determine the value estimated by MSBG and comment on how this differs from the value estimated by MSWG.
Verify that for the following data, MSBG = 14.4 and MSWG = 12.59.G1 G2 G3 9 16 7 10 8 6 15 13 9 6
For the data in Table 10.1, assume normality and that the groups have equal variances. As already illustrated, the hypothesis of equal means is not rejected. If the hypothesis of equal means is true, an estimate of the assumed common variance is MSBG = .12, as already explained.Describe a reason
For the data Group 1 Group 2 Group 3 Group 4 15 9 17 13 17 12 20 12 22 15 23 17 verify that when using the ANOVA F test, the estimate of the assumed common variance is MSWG = 9.5.
For the data in Exercise 1, verify that when comparing 20% trimmed means with the R function t1way, the p-value is 0.02.
For the data in Exercise 1, verify that Welch’s test statistic is Fw = 7.7 with degrees of freedom ν1 = 2 and ν2 = 13.4. Using the R function t1way, verify that you would reject the hypothesis of equal means withα = .01.
For the data in the previous exercise, test the hypothesis of equal means using the ANOVA F. Use α = .05. Verify that MSBG = 25.04, F = 6.05, and that you reject at the 0.05 level.
For the following data, Group 1 Group 2 Group 3 3 4 6 5 4 7 2 3 8 4 8 6 8 7 7 4 4 9 3 2 10 9 5 9¯X 1 = 4.75 ¯X2 = 4.62 ¯X3 = 7.75 s21= 6.214 s22= 3.982 s23= 2.214 assume that the three groups have a common population variance, σ2 p.Estimate σ2 p.
Repeat the previous exercise, only now compare the marginal 20%trimmed means using the R function yuend with the argument tr=0.2. Verify that the p-value is 0.2209.
Carnegie Mellon University maintains a site with many interesting data sets for use in the classroom. Some of these data sets are stored at http://stat.cmu.edu/DASL/allsubjects.html. Go to this web page, look under psychology, download the file called scents and read the file into an R variable
Continuing the last exercise, plot the difference scores using the R function akerd and note that the distribution is skewed to the right, suggesting that the confidence interval for the mean might be inaccurate.Using the R function trimcibt, verify that the .95 confidence interval for the 20% of
Section 9.2.1 analyzed the Indometh data stored in R using the R function trimci. Compare times 2 and 3 using means based on difference scores and verify that the p.value is 0.014.
For two dependent groups you get Group 1: 86 71 77 68 91 72 77 91 70 71 88 87 Group 2: 88 77 76 64 96 72 65 90 65 80 81 72 Apply the Wilcoxon signed rank test with α = 0.05. Verify that W =0.7565 and that you fail to reject.
For two dependent groups you get Group 1: 10 14 15 18 20 29 30 40 Group 2: 40 8 15 20 10 8 2 3 Compare the two groups with the sign test and the Wilcoxon signed rank test with α = 0.05. Verify that according to the sign test, ˆp = 0.29 and that the 0.95 confidence interval for p is (0.04, 0.71),
Repeat the last exercise, only now use a 20% trimmed mean. Verify that the p-value is 0.049.
Repeat Exercise 1, but now use a bootstrap-t method via the R function trimcibt. Verify that the p-value is 0.091.
Is it possible that the marginal trimmed means are equal but the trimmed mean based on the difference scores is not equal to zero? What does this indicate in terms of power?
Generally, why is it possible to get a different p-value comparing the marginal trimmed means versus making inferences about the trimmed mean of the difference scores?
If in Exercise 1 you compare the marginal 20% trimmed means with the R function yuend, verify that now the p-value is 0.121.
Repeat the previous exercise, but use 20% trimmed means instead using the difference scores in conjunction with the R function trimci. Verify that the p-value is 0.049.
For the data in Table 9.1, perform the paired T test for means using the weights for the east and south sides of the trees. Verify that the p-value is 0.09.
R contains data, stored in the R variable sleep, which show the effect of two soporific drugs, namely the increase in hours of sleep compared to control. Create a boxplot for both groups using the command boxplot(sleep[[1]] sleep[[2]]) and speculate about whether comparing 20% trimmed means will
For the data in the previous exercise, verify that the p-value based on Cliff’s method is .001 and for the Brunner-Munzel method the p-value is less than 0.001.
Two independent groups are given different cold medicines and the goal is to compare reaction times. Suppose that the decrease in reaction times when taking drug A versus drug B are as follows.A: 1.96, 2.24, 1.71, 2.41, 1.62, 1.93 B: 2.11, 2.43, 2.07, 2.71, 2.50, 2.84, 2.88 Verify that the
Imagine that two groups of cancer patients are compared, the first group having a rapidly progressing form of the disease and the other having a slowly progressing form instead. At issue is whether psychological factors are related to the progression of cancer. The outcome measure is one where

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Modern Statistics For The Social And Behavioral Sciences 1st Edition Rand Wilcox - Solutions

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