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Statistics Assignment Help

The document offers solutions and R code snippets for various statistical problems, mainly focused on assignment help for statistics. It includes proofs of statistical inequalities, maximum likelihood estimations for two-way layouts, and Chi-square tests for independence applied to medical data. Additionally, it presents simulation results and visualizations of different batch distributions.

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Statistics for Applications Problems from John A. Rice, Third Edition. [Chapter.Section.Problem] 1. 12.5.1 Solution: See R script/html P roblem 12 5 1.r/html 2. 12.5.6 Prove this version of the Bonferroni inequality: n i=1 i P (∩ A ) ≥ 1 − n P (A c) i=1 i ∗ n i=1 Let A = ∩ Ai. Then Σ c ∪ A = n Ac ∗ k=1 i n c n c So P (A ) = P (∪ A ) c Σ ∗ k=1 i k=1 i ≤ P (A c) Because P (A ) = 1 − P (A ), it follows that ∗ ∗ ∗ P (A ) ≥ 1 − n P (A c) k=1 i In the context of simulataneous confidence intervals Σ Ai is the the event that the ith confidence interval contains the associated parameter n i=1 ∩ Ai is the the event that all confidence intervals contains the associated parameters simultaneously. 3. 12.5.17. Find the mle’s of the parameters αi, βj, δij, and µ of the model for the two-way layout. The model for the two-way layout is given by: Yijk = µ + αi + βj + δij + Eijk where Σ Σ I i=1 J j =1 αi = 0 βj = 0 I J ij δ = Σ Σ i=1 j =1 As noted in Rice (p. 493) the log likelihood is δij = 0 statisticsassignmenthelp.com
I J K 2 1 I l = − log(2πσ ) − 2 2σ2 Σ Σ Σ J K i=1 j =1 k=1 (Yij k − µ − αi − βj − δij )2 Solving the following equations: ∂l = 0 =⇒ µ̂ = Y··· = 0 =⇒ α̂i = Yi·· − µ̂ ˆ i ·j · = 0 =⇒ β = Y − µ̂ ∂µi ∂l ∂αi ∂l ∂βi ∂l ∂δij = 0 =⇒ δ̂ij = Yij · − µ̂ − α̂i − β̂j These terms yield the answer formulas of the problem. 4. 13.8.1. See R script html P roblem 13 8 1.html 5. 13.8.25. See R script html P roblem 13 8 25.html statisticsassignmenthelp.com
Problem_12_5_1.r # Problem_12_5_1.r # Simulate 7 batches of ten normally distributed random numbers # with mean 4 and variance .0037 nbatches=7 sampsize=10 mu=4. sigmasq=.0037 set.seed(1) # Set seed to be able to replicate simulations par(mfcol=c(3,2)) for (i in c(1:6)){ batches<-matrix(rnorm(nbatches*sampsize,mean=mu, sd=sqrt(sigmasq)), nrow=sampsize, ncol=nbatches) boxplot(batches, xlab=paste("Batch for Batch Set ", i,sep="")) } statisticsassignmenthelp.com
# In this set of simulations, there are pairs of labs that appear quite different # in mean level: Batch Set 3, batches 1 and 4, Batch set 2, batches 2 and 5 # in dispersion: Batch Set 1, batches 5 and 7, Batch Set 6, batches 3 and 4. # # Note that with the smaple sample size (10), the variation will be moderate. # If we change the sampsize to 250, there should be little differences in mean # and dispersion due to the larger sample sizes. sampsize=250 set.seed(1) # Set seed to be able to replicate simulations par(mfcol=c(3,2)) for (i in c(1:6)){ batches<- matrix(rnorm(nbatches*sampsize,mean=mu, sd=sqrt(sigmasq)), nrow=sampsize, ncol=nbatches) boxplot(batches, xlab=paste("Batch for Batch Set ", i,sep="")) } # In these plots, the mean level is quite constant as is the # dispersion as measured by the inter-quartile range. statisticsassignmenthelp.com
Problem_13_8_1.r # Problem_12_5_1.r # Adult-onset diabites table: frequency data of a particular allele in a sample of # such diabetics and a sample of nondiabetics tableA=data.frame( Diabetic=rbind(Bborbb=12., BB=39), Normal=rbind(Bborbb=4., BB=49)) # # 2. Conduct ChiSquare Test of Independence ---- # Custom function implementing chisqtest fcn.chisqtest<-function(tableA){ cat("n Two-Way Table: n") print(tableA) n.total=sum(as.vector(tableA)) cat("n Total Counts in Table: ", n.total,"n") # Compute marginal probabilities of # TattooStatus and of HepCStatus probs.TattooStatus=rowSums(tableA)/n.total probs.HepCStatus=colSums(tableA)/n.total cat("n MLEs of row level probabilitiesn") print(probs.TattooStatus) cat("n MLEs of column level probabilitiesn") print(probs.HepCStatus) # Compute table of fitted cell probabilities and # expected counts assuming independence of two factors tableA.fittedprobs=as.matrix(probs.TattooStatus)%*% t( as.matrix(probs.HepCStatus) ) cat("n Fitted cell probabilities assuming independencen") print(tableA.fittedprobs) tableA.expected=n.total* tableA.fittedprobs cat("n Expected Counts assuming independence n") print(tableA.expected) # Compute standardized residuals fitted table tableA.chisqresiduals=((tableA - tableA.expected))/sqrt(tableA.expected) cat("n Table of Chi-Square Residuals by celln") print(tableA.chisqresiduals) # Compute table of chi-square test statistic contributions tableA.chisqterms=((tableA - tableA.expected)^2)/tableA.expected cat("n Table of Chi-Square statistic terms by celln") print(tableA.chisqterms) tableA.chisqStatistic=sum(as.vector(tableA.chisqterms)) cat("n Chi-Square Statistic: ",tableA.chisqStatistic,"n") df.tableA=(nrow(tableA)-1)*(ncol(tableA)-1) cat("n degrees of freedom: ", df.tableA, "n") tableA.chisqStatistic.pvalue=1- pchisq(tableA.chisqStatistic, df=df.tableA) cat("n P-Value : ", tableA.chisqStatistic.pvalue, "nn") } fcn.chisqtest(tableA) statisticsassignmenthelp.com
## ## Two-Way Table: ## Diabetic Normal ## Bborbb 12 4 ## BB 39 49 ## Total Counts in Table: 104 ## MLEs of row level probabilities ## Bborbb BB ## 0.1538462 0.8461538 ## ## MLEs of column level probabilities ## Diabetic Normal ## 0.4903846 0.5096154 ## ## Fitted cell probabilities assuming independence ## Diabetic Normal ## Bborbb 0.07544379 0.07840237 ## BB 0.41494083 0.43121302 ## ## Expected Counts assuming independence ## Diabetic Normal ## Bborbb 7.846154 8.153846 ## BB 43.153846 44.846154 ## ## Table of Chi-Square Residuals by cell ## Diabetic Normal ## Bborbb 1.4829346 -1.454686 ## BB -0.6323254 0.620280 ## Table of Chi-Square statistic terms by cell ## Diabetic Normal ## Bborbb 2.1990950 2.1161103 ## BB 0.3998355 0.3847473 ## Chi-Square Statistic: 5.099788 ## degrees of freedom: 1 ## P-Value : 0.02392877 # The p-Value is 0.0239, which is significant at nominal alpha/size level of 0.05. # From the output, we see that the expected counts assuming indepndence are # all greater than 5, so we are not overly concerned with # the chi-square distribution approximation to the chi-square statistic # under the null hypothesis of independence (of row and column factors) statisticsassignmenthelp.com
Problem_13_8_25.r #Problem_13_8_25.r # # Custom function implementing chisqtest fcn.chisqtest<-function(tableA){ cat("n Two- Way Table: n") print(tableA) n.total=sum(as.vector(tableA)) cat("n Total Counts in Table: ", n.total,"n") # Compute marginal probabilities of # TattooStatus and of HepCStatus probs.TattooStatus=rowSums(tableA)/n.total probs.HepCStatus=colSums(tableA)/n.total cat("n MLEs of row level probabilitiesn") print(probs.TattooStatus) cat("n MLEs of column level probabilitiesn") print(probs.HepCStatus) # Compute table of fitted cell probabilities and # expected counts assuming independence of two factors tableA.fittedprobs=as.matrix(probs.TattooStatus)%*% t( as.matrix(probs.HepCStatus) ) cat("n Fitted cell probabilities assuming independencen") print(tableA.fittedprobs) tableA.expected=n.total* tableA.fittedprobs cat("n Expected Counts assuming independence n") print(tableA.expected) # Compute standardized residuals fitted table tableA.chisqresiduals=((tableA - tableA.expected))/sqrt(tableA.expected) cat("n Table of Chi-Square Residuals by celln") print(tableA.chisqresiduals) # Compute table of chi-square test statistic contributions tableA.chisqterms=((tableA - tableA.expected)^2)/tableA.expected cat("n Table of Chi-Square statistic terms by celln") print(tableA.chisqterms) tableA.chisqStatistic=sum(as.vector(tableA.chisqterms)) cat("n Chi-Square Statistic: ",tableA.chisqStatistic,"n") df.tableA=(nrow(tableA)-1)*(ncol(tableA)-1) cat("n degrees of freedom: ", df.tableA, "n") tableA.chisqStatistic.pvalue=1- pchisq(tableA.chisqStatistic, df=df.tableA) cat("n P-Value : ", tableA.chisqStatistic.pvalue, "nn") } ############### # 1. Problem 25: Physicians Health Study Tattoo/HepC Two-Way Table ---- tableD=data.frame( Asprin=rbind( MCIFatal=10, MCINonFatal=129, StrokeFatal=9, StrokeNonFatal=110), Placebo=rbind( MCIFatal=26, MCINonFatal=213, StrokeFatal=6, StrokeNonFatal=92) ) tableD.colSums=colSums(tableD) tableD.1<-rbind(tableD, None=((tableD.colSums*-1) +c(11037, 11034))) # (a). Test 1 # Chisquare test of homogenity -- rates are same # for Aspirin population and Placebo population fcn.chisqtest(tableD.1) statisticsassignmenthelp.com
## ## Two-Way Table: ## Asprin Placebo ## MCIFatal 10 26 ## MCINonFatal 129 213 ## StrokeFatal 9 6 ## StrokeNonFatal 110 92 ## None 10779 10697 ## ## Total Counts in Table: 22071 ## ## MLEs of row level probabilities ## MCIFatal MCINonFatal StrokeFatal StrokeNonFatal None ## 0.0016310996 0.0154954465 0.0006796248 0.0091522813 0.9730415477 ## ## MLEs of column level probabilities ## Asprin Placebo ## 0.500068 0.499932 ## Fitted cell probabilities assuming independence ## Asprin Placebo ## MCIFatal 0.0008156607 0.0008154390 ## MCINonFatal 0.0077487764 0.0077466701 ## StrokeFatal 0.0003398586 0.0003397662 ## StrokeNonFatal 0.0045767626 0.0045755186 ## None 0.4865869042 0.4864546435 ## ## Expected Counts assuming independence ## Asprin Placebo ## MCIFatal 18.002447 17.997553 ## MCINonFatal 171.023243 170.976757 ## StrokeFatal 7.501019 7.498981 ## StrokeNonFatal 101.013728 100.986272 ## None 10739.459562 10736.540438 ## ## Table of Chi-Square Residuals by cell ## Asprin Placebo ## MCIFatal -1.8860666 1.8863230 ## MCINonFatal -3.2133793 3.2138162 ## StrokeFatal 0.5473131 -0.5473875 ## StrokeNonFatal 0.8941067 -0.8942282 statisticsassignmenthelp.com
## None 0.3815489 -0.3816008 ## ## Table of Chi-Square statistic terms by cell ## Asprin Placebo ## MCIFatal 3.5572472 3.5582143 ## MCINonFatal 10.3258068 10.3286142 ## StrokeFatal 0.2995516 0.2996331 ## StrokeNonFatal 0.7994268 0.7996441 ## None 0.1455796 0.1456192 ## ## Chi-Square Statistic: 30.25934 ## ## degrees of freedom: 4 ## ## P-Value : 4.33405e-06 # Results significant, with MCIFatal and MCINonFatal significantly # lower for Aspirin versus Placebo (see the chisq residuals) # (a). Test 2 # Analyse total incidence of myocardial infarction tableD.2=rbind( MCITotal=colSums(tableD.1[1:2,]), Other=colSums(tableD.1[3:5,])) print(tableD.2) ## Asprin Placebo ## MCITotal 139 239 ## Other 10898 10795fcn.chisqtest(tableD.2) statisticsassignmenthelp.com
## ## Two-Way Table: ## Asprin Placebo ## MCITotal 139 239 ## Other 10898 10795 ## Total Counts in Table: 22071 ## ## MLEs of row level probabilities ## MCITotal Other ## 0.01712655 0.98287345 ## ## MLEs of column level probabilities ## Asprin Placebo ## 0.500068 0.499932 ## Fitted cell probabilities assuming independence ## Asprin Placebo ## MCITotal 0.008564437 0.008562109 ## Other 0.491503525 0.491369928 ## Expected Counts assuming independence ## Asprin Placebo ## MCITotal 189.0257 188.9743 ## Other 10847.9743 10845.0257 ## Table of Chi-Square Residuals by cell ## Asprin Placebo ## MCITotal -3.6385862 3.6390808 ## Other 0.4803068 -0.4803721 ## ## Table of Chi-Square statistic terms by cell ## Asprin Placebo ## MCITotal 13.2393097 13.2429093 ## Other 0.2306947 0.2307574 ## ## Chi-Square Statistic: 26.94367 ## ## degrees of freedom: 1 ## ## P-Value : 2.09472e-07 # Total incidence of MCI is significantly reduced by aspirin # (a). Test 3 # Analyse incidence of fatal and nonfatal tableD.3=rbind( FatalNonFatal=colSums(tableD.1[c(1,2,3,4),]), Other=colSums(tableD.1[c(5),])) fcn.chisqtest(tableD.3) statisticsassignmenthelp.com
## ## Two-Way Table: ## Asprin Placebo ## FatalNonFatal 258 337 ## Other 10779 10697 ## ## Total Counts in Table: 22071 ## ## MLEs of row level probabilities ## FatalNonFatal Other ## 0.02695845 0.97304155 ## ## MLEs of column level probabilities ## Asprin Placebo ## 0.500068 0.499932 ## ## Fitted cell probabilities assuming independence ## Asprin Placebo ## FatalNonFatal 0.01348106 0.01347739 ## Other 0.48658690 0.48645464 ## ## Expected Counts assuming independence ## Asprin Placebo ## FatalNonFatal 297.5404 297.4596 ## Other 10739.4596 10736.5404 ## ## Table of Chi-Square Residuals by cell ## Asprin Placebo ## FatalNonFatal -2.2922843 2.2925959 ## Other 0.3815489 -0.3816008 ## ## Table of Chi-Square statistic terms by cell ## Asprin Placebo ## FatalNonFatal 5.2545672 5.2559958 ## Other 0.1455796 0.1456192 ## ## Chi-Square Statistic: 10.80176 ## ## degrees of freedom: 1 ## ## P-Value : 0.001014035 # Incidence of cardiovascular events is lower under Aspirin vs Placebo # but the significance is lower (though still significant relative to nominal levels) # (a). Test 4 # Among those having MCI, evaluate whether # fatality rate depends on Aspirin versus Placebo fcn.chisqtest(tableD.1[1:2,]) statisticsassignmenthelp.com
## ## Two-Way Table: ## Asprin Placebo ## MCIFatal 10 26 ## MCINonFatal 129 213 ## ## Total Counts in Table: 378 ## ## MLEs of row level probabilities ## MCIFatal MCINonFatal ## 0.0952381 0.9047619 ## ## MLEs of column level probabilities ## Asprin Placebo ## 0.3677249 0.6322751 ## ## Fitted cell probabilities assuming independence ## Asprin Placebo ## MCIFatal 0.03502142 0.06021668 ## MCINonFatal 0.33270345 0.57205845 ## ## Expected Counts assuming independence ## Asprin Placebo ## MCIFatal 13.2381 22.7619 ## MCINonFatal 125.7619 216.2381 ## ## Table of Chi-Square Residuals by cell ## Asprin Placebo ## MCIFatal -0.8899731 0.6787117 ## MCINonFatal 0.2887454 -0.2202031 ## ## Table of Chi-Square statistic terms by cell ## Asprin Placebo ## MCIFatal 0.7920521 0.46064953 ## MCINonFatal 0.0833739 0.04848942 ## ## Chi-Square Statistic: 1.384565 ## ## degrees of freedom: 1 ## ## P-Value : 0.2393251 statisticsassignmenthelp.com
# Not significant # (a). Test 5 Incidence of strokes tableD.4=rbind( Strokes=colSums(tableD.1[c(3,4),]), Other=colSums(tableD.1[c(1,2,5),])) fcn.chisqtest(tableD.4) ## ## Two-Way Table: ## Asprin Placebo ## Strokes 119 98 ## Other 10918 10936 ## ## Total Counts in Table: 22071 ## ## MLEs of row level probabilities ## Strokes Other ## 0.009831906 0.990168094 ## ## MLEs of column level probabilities ## Asprin Placebo ## 0.500068 0.499932 ## ## Fitted cell probabilities assuming independence ## Asprin Placebo ## Strokes 0.004916621 0.004915285 ## Other 0.495151341 0.495016753 ## ## Expected Counts assuming independence ## Asprin Placebo ## Strokes 108.5147 108.4853 ## Other 10928.4853 10925.5147 ## ## Table of Chi-Square Residuals by cell ## Asprin Placebo ## Strokes 1.0065480 -1.0066848 ## Other -0.1002995 0.1003132 ## ## Table of Chi-Square statistic terms by cell ## Asprin Placebo ## Strokes 1.013139 1.01341436 ## Other 0.010060 0.01006273 ## ## Chi-Square Statistic: 2.046676 ## ## degrees of freedom: 1 ## ## P-Value : 0.1525389 statisticsassignmenthelp.com
# Not significantly different # (b). Table of cariovascular mortality tableE=data.frame( Asprin=rbind( AcuteMCI=10, OtherIHD=24, SuddenDeath=22, Stroke=10, OtherCard=15), Placebo=rbind( AcuteMCI=28, OtherIHD=25, SuddenDeath=12, Stroke=7, OtherCard=11)) tableE.colSums=colSums(tableE) tableE.1<-rbind(tableE, None=((tableE.colSums*-1) +c(11037, 11034)) ) tableE.1 ## Asprin Placebo ## AcuteMCI 10 28 ## OtherIHD 24 25 ## SuddenDeath 22 12 ## Stroke 10 7 ## OtherCard 15 11 ## None 10956 10951fcn.chisqtest(tableE.1) statisticsassignmenthelp.com
## ## Two-Way Table: ## Asprin Placebo ## AcuteMCI 10 28 ## OtherIHD 24 25 ## SuddenDeath 22 12 ## Stroke 10 7 ## OtherCard 15 11 ## None 10956 10951 ## ## Total Counts in Table: 22071 ## ## MLEs of row level probabilities ## AcuteMCI OtherIHD SuddenDeath Stroke OtherCard ## 0.0017217163 0.0022201078 0.0015404830 0.0007702415 0.0011780164 ## None ## 0.9925694350 ## ## MLEs of column level probabilities ## Asprin Placebo ## 0.500068 0.499932 ## ## Fitted cell probabilities assuming independence ## Asprin Placebo ## AcuteMCI 0.0008609752 0.0008607411 ## OtherIHD 0.0011102048 0.0011099030 ## SuddenDeath 0.0007703462 0.0007701368 ## Stroke 0.0003851731 0.0003850684 ## OtherCard 0.0005890883 0.0005889281 ## None 0.4963521750 0.4962172600 ## ## Expected Counts assuming independence ## Asprin Placebo ## AcuteMCI 19.002583 18.997417 ## OtherIHD 24.503330 24.496670 ## SuddenDeath 17.002311 16.997689 ## Stroke 8.501155 8.498845 ## OtherCard 13.001767 12.998233 statisticsassignmenthelp.com
## None 10954.988854 10952.011146 ## ## Table of Chi-Square Residuals by cell ## Asprin Placebo ## AcuteMCI -2.065193737 2.065474468 ## OtherIHD -0.101681138 0.101694960 ## SuddenDeath 1.212035322 -1.212200079 ## Stroke 0.514064534 -0.514134412 ## OtherCard 0.554172450 -0.554247781 ## None 0.009660683 -0.009661996 ## ## Table of Chi-Square statistic terms by cell ## Asprin Placebo ## AcuteMCI 4.2650251718 4.266185e+00 ## OtherIHD 0.0103390539 1.034186e-02 ## SuddenDeath 1.4690296219 1.469429e+00 ## Stroke 0.2642623446 2.643342e-01 ## OtherCard 0.3071071045 3.071906e-01 ## None 0.0000933288 9.335417e-05 ## ## Chi-Square Statistic: 12.63343 ## ## degrees of freedom: 5 ## ## P-Value : 0.0270671 statisticsassignmenthelp.com
# Mortality due to AcuteMCI is significantly reduced by Aspirin # The P-value is 0.027 # (b). Test 2 # Evaluation of total cariovascular mortality: tableE.2<-rbind(TotalCard=colSums(tableE.1[c(1:5),]), None=tableE.1[6,]) fcn.chisqtest(tableE.2) ## ## Two-Way Table: ## Asprin Placebo ## TotalCard 81 83 ## None 10956 10951 ## ## Total Counts in Table: 22071 ## ## MLEs of row level probabilities ## TotalCard None ## 0.007430565 0.992569435 ## ## MLEs of column level probabilities ## Asprin Placebo ## 0.500068 0.499932 ## ## Fitted cell probabilities assuming independence ## Asprin Placebo ## TotalCard 0.003715787 0.003714777 ## None 0.496352175 0.496217260 ## ## Expected Counts assuming independence ## Asprin Placebo ## TotalCard 82.01115 81.98885 ## None 10954.98885 10952.01115 ## ## Table of Chi-Square Residuals by cell ## Asprin Placebo ## TotalCard -0.111654791 0.111669969 ## None 0.009660683 -0.009661996 ## ## Table of Chi-Square statistic terms by cell ## Asprin Placebo ## TotalCard 0.0124667923 1.247018e-02 ## None 0.0000933288 9.335417e-05 ## ## Chi-Square Statistic: 0.02512366 ## ## degrees of freedom: 1 ## ## P-Value : 0.8740593 # Not significant statisticsassignmenthelp.com

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Statistics Assignment Help

  • 1. For any help regarding Statistics Assignment Help visit : https://www.statisticsassignmenthelp.com/, Email - support@statisticsassignmenthelp.com or call us at - +1 678 648 4277 statisticsassignmenthelp.com
  • 2. Statistics for Applications Problems from John A. Rice, Third Edition. [Chapter.Section.Problem] 1. 12.5.1 Solution: See R script/html P roblem 12 5 1.r/html 2. 12.5.6 Prove this version of the Bonferroni inequality: n i=1 i P (∩ A ) ≥ 1 − n P (A c) i=1 i ∗ n i=1 Let A = ∩ Ai. Then Σ c ∪ A = n Ac ∗ k=1 i n c n c So P (A ) = P (∪ A ) c Σ ∗ k=1 i k=1 i ≤ P (A c) Because P (A ) = 1 − P (A ), it follows that ∗ ∗ ∗ P (A ) ≥ 1 − n P (A c) k=1 i In the context of simulataneous confidence intervals Σ Ai is the the event that the ith confidence interval contains the associated parameter n i=1 ∩ Ai is the the event that all confidence intervals contains the associated parameters simultaneously. 3. 12.5.17. Find the mle’s of the parameters αi, βj, δij, and µ of the model for the two-way layout. The model for the two-way layout is given by: Yijk = µ + αi + βj + δij + Eijk where Σ Σ I i=1 J j =1 αi = 0 βj = 0 I J ij δ = Σ Σ i=1 j =1 As noted in Rice (p. 493) the log likelihood is δij = 0 statisticsassignmenthelp.com
  • 3. I J K 2 1 I l = − log(2πσ ) − 2 2σ2 Σ Σ Σ J K i=1 j =1 k=1 (Yij k − µ − αi − βj − δij )2 Solving the following equations: ∂l = 0 =⇒ µ̂ = Y··· = 0 =⇒ α̂i = Yi·· − µ̂ ˆ i ·j · = 0 =⇒ β = Y − µ̂ ∂µi ∂l ∂αi ∂l ∂βi ∂l ∂δij = 0 =⇒ δ̂ij = Yij · − µ̂ − α̂i − β̂j These terms yield the answer formulas of the problem. 4. 13.8.1. See R script html P roblem 13 8 1.html 5. 13.8.25. See R script html P roblem 13 8 25.html statisticsassignmenthelp.com
  • 4. Problem_12_5_1.r # Problem_12_5_1.r # Simulate 7 batches of ten normally distributed random numbers # with mean 4 and variance .0037 nbatches=7 sampsize=10 mu=4. sigmasq=.0037 set.seed(1) # Set seed to be able to replicate simulations par(mfcol=c(3,2)) for (i in c(1:6)){ batches<-matrix(rnorm(nbatches*sampsize,mean=mu, sd=sqrt(sigmasq)), nrow=sampsize, ncol=nbatches) boxplot(batches, xlab=paste("Batch for Batch Set ", i,sep="")) } statisticsassignmenthelp.com
  • 5. # In this set of simulations, there are pairs of labs that appear quite different # in mean level: Batch Set 3, batches 1 and 4, Batch set 2, batches 2 and 5 # in dispersion: Batch Set 1, batches 5 and 7, Batch Set 6, batches 3 and 4. # # Note that with the smaple sample size (10), the variation will be moderate. # If we change the sampsize to 250, there should be little differences in mean # and dispersion due to the larger sample sizes. sampsize=250 set.seed(1) # Set seed to be able to replicate simulations par(mfcol=c(3,2)) for (i in c(1:6)){ batches<- matrix(rnorm(nbatches*sampsize,mean=mu, sd=sqrt(sigmasq)), nrow=sampsize, ncol=nbatches) boxplot(batches, xlab=paste("Batch for Batch Set ", i,sep="")) } # In these plots, the mean level is quite constant as is the # dispersion as measured by the inter-quartile range. statisticsassignmenthelp.com
  • 6. Problem_13_8_1.r # Problem_12_5_1.r # Adult-onset diabites table: frequency data of a particular allele in a sample of # such diabetics and a sample of nondiabetics tableA=data.frame( Diabetic=rbind(Bborbb=12., BB=39), Normal=rbind(Bborbb=4., BB=49)) # # 2. Conduct ChiSquare Test of Independence ---- # Custom function implementing chisqtest fcn.chisqtest<-function(tableA){ cat("n Two-Way Table: n") print(tableA) n.total=sum(as.vector(tableA)) cat("n Total Counts in Table: ", n.total,"n") # Compute marginal probabilities of # TattooStatus and of HepCStatus probs.TattooStatus=rowSums(tableA)/n.total probs.HepCStatus=colSums(tableA)/n.total cat("n MLEs of row level probabilitiesn") print(probs.TattooStatus) cat("n MLEs of column level probabilitiesn") print(probs.HepCStatus) # Compute table of fitted cell probabilities and # expected counts assuming independence of two factors tableA.fittedprobs=as.matrix(probs.TattooStatus)%*% t( as.matrix(probs.HepCStatus) ) cat("n Fitted cell probabilities assuming independencen") print(tableA.fittedprobs) tableA.expected=n.total* tableA.fittedprobs cat("n Expected Counts assuming independence n") print(tableA.expected) # Compute standardized residuals fitted table tableA.chisqresiduals=((tableA - tableA.expected))/sqrt(tableA.expected) cat("n Table of Chi-Square Residuals by celln") print(tableA.chisqresiduals) # Compute table of chi-square test statistic contributions tableA.chisqterms=((tableA - tableA.expected)^2)/tableA.expected cat("n Table of Chi-Square statistic terms by celln") print(tableA.chisqterms) tableA.chisqStatistic=sum(as.vector(tableA.chisqterms)) cat("n Chi-Square Statistic: ",tableA.chisqStatistic,"n") df.tableA=(nrow(tableA)-1)*(ncol(tableA)-1) cat("n degrees of freedom: ", df.tableA, "n") tableA.chisqStatistic.pvalue=1- pchisq(tableA.chisqStatistic, df=df.tableA) cat("n P-Value : ", tableA.chisqStatistic.pvalue, "nn") } fcn.chisqtest(tableA) statisticsassignmenthelp.com
  • 7. ## ## Two-Way Table: ## Diabetic Normal ## Bborbb 12 4 ## BB 39 49 ## Total Counts in Table: 104 ## MLEs of row level probabilities ## Bborbb BB ## 0.1538462 0.8461538 ## ## MLEs of column level probabilities ## Diabetic Normal ## 0.4903846 0.5096154 ## ## Fitted cell probabilities assuming independence ## Diabetic Normal ## Bborbb 0.07544379 0.07840237 ## BB 0.41494083 0.43121302 ## ## Expected Counts assuming independence ## Diabetic Normal ## Bborbb 7.846154 8.153846 ## BB 43.153846 44.846154 ## ## Table of Chi-Square Residuals by cell ## Diabetic Normal ## Bborbb 1.4829346 -1.454686 ## BB -0.6323254 0.620280 ## Table of Chi-Square statistic terms by cell ## Diabetic Normal ## Bborbb 2.1990950 2.1161103 ## BB 0.3998355 0.3847473 ## Chi-Square Statistic: 5.099788 ## degrees of freedom: 1 ## P-Value : 0.02392877 # The p-Value is 0.0239, which is significant at nominal alpha/size level of 0.05. # From the output, we see that the expected counts assuming indepndence are # all greater than 5, so we are not overly concerned with # the chi-square distribution approximation to the chi-square statistic # under the null hypothesis of independence (of row and column factors) statisticsassignmenthelp.com
  • 8. Problem_13_8_25.r #Problem_13_8_25.r # # Custom function implementing chisqtest fcn.chisqtest<-function(tableA){ cat("n Two- Way Table: n") print(tableA) n.total=sum(as.vector(tableA)) cat("n Total Counts in Table: ", n.total,"n") # Compute marginal probabilities of # TattooStatus and of HepCStatus probs.TattooStatus=rowSums(tableA)/n.total probs.HepCStatus=colSums(tableA)/n.total cat("n MLEs of row level probabilitiesn") print(probs.TattooStatus) cat("n MLEs of column level probabilitiesn") print(probs.HepCStatus) # Compute table of fitted cell probabilities and # expected counts assuming independence of two factors tableA.fittedprobs=as.matrix(probs.TattooStatus)%*% t( as.matrix(probs.HepCStatus) ) cat("n Fitted cell probabilities assuming independencen") print(tableA.fittedprobs) tableA.expected=n.total* tableA.fittedprobs cat("n Expected Counts assuming independence n") print(tableA.expected) # Compute standardized residuals fitted table tableA.chisqresiduals=((tableA - tableA.expected))/sqrt(tableA.expected) cat("n Table of Chi-Square Residuals by celln") print(tableA.chisqresiduals) # Compute table of chi-square test statistic contributions tableA.chisqterms=((tableA - tableA.expected)^2)/tableA.expected cat("n Table of Chi-Square statistic terms by celln") print(tableA.chisqterms) tableA.chisqStatistic=sum(as.vector(tableA.chisqterms)) cat("n Chi-Square Statistic: ",tableA.chisqStatistic,"n") df.tableA=(nrow(tableA)-1)*(ncol(tableA)-1) cat("n degrees of freedom: ", df.tableA, "n") tableA.chisqStatistic.pvalue=1- pchisq(tableA.chisqStatistic, df=df.tableA) cat("n P-Value : ", tableA.chisqStatistic.pvalue, "nn") } ############### # 1. Problem 25: Physicians Health Study Tattoo/HepC Two-Way Table ---- tableD=data.frame( Asprin=rbind( MCIFatal=10, MCINonFatal=129, StrokeFatal=9, StrokeNonFatal=110), Placebo=rbind( MCIFatal=26, MCINonFatal=213, StrokeFatal=6, StrokeNonFatal=92) ) tableD.colSums=colSums(tableD) tableD.1<-rbind(tableD, None=((tableD.colSums*-1) +c(11037, 11034))) # (a). Test 1 # Chisquare test of homogenity -- rates are same # for Aspirin population and Placebo population fcn.chisqtest(tableD.1) statisticsassignmenthelp.com
  • 9. ## ## Two-Way Table: ## Asprin Placebo ## MCIFatal 10 26 ## MCINonFatal 129 213 ## StrokeFatal 9 6 ## StrokeNonFatal 110 92 ## None 10779 10697 ## ## Total Counts in Table: 22071 ## ## MLEs of row level probabilities ## MCIFatal MCINonFatal StrokeFatal StrokeNonFatal None ## 0.0016310996 0.0154954465 0.0006796248 0.0091522813 0.9730415477 ## ## MLEs of column level probabilities ## Asprin Placebo ## 0.500068 0.499932 ## Fitted cell probabilities assuming independence ## Asprin Placebo ## MCIFatal 0.0008156607 0.0008154390 ## MCINonFatal 0.0077487764 0.0077466701 ## StrokeFatal 0.0003398586 0.0003397662 ## StrokeNonFatal 0.0045767626 0.0045755186 ## None 0.4865869042 0.4864546435 ## ## Expected Counts assuming independence ## Asprin Placebo ## MCIFatal 18.002447 17.997553 ## MCINonFatal 171.023243 170.976757 ## StrokeFatal 7.501019 7.498981 ## StrokeNonFatal 101.013728 100.986272 ## None 10739.459562 10736.540438 ## ## Table of Chi-Square Residuals by cell ## Asprin Placebo ## MCIFatal -1.8860666 1.8863230 ## MCINonFatal -3.2133793 3.2138162 ## StrokeFatal 0.5473131 -0.5473875 ## StrokeNonFatal 0.8941067 -0.8942282 statisticsassignmenthelp.com
  • 10. ## None 0.3815489 -0.3816008 ## ## Table of Chi-Square statistic terms by cell ## Asprin Placebo ## MCIFatal 3.5572472 3.5582143 ## MCINonFatal 10.3258068 10.3286142 ## StrokeFatal 0.2995516 0.2996331 ## StrokeNonFatal 0.7994268 0.7996441 ## None 0.1455796 0.1456192 ## ## Chi-Square Statistic: 30.25934 ## ## degrees of freedom: 4 ## ## P-Value : 4.33405e-06 # Results significant, with MCIFatal and MCINonFatal significantly # lower for Aspirin versus Placebo (see the chisq residuals) # (a). Test 2 # Analyse total incidence of myocardial infarction tableD.2=rbind( MCITotal=colSums(tableD.1[1:2,]), Other=colSums(tableD.1[3:5,])) print(tableD.2) ## Asprin Placebo ## MCITotal 139 239 ## Other 10898 10795fcn.chisqtest(tableD.2) statisticsassignmenthelp.com
  • 11. ## ## Two-Way Table: ## Asprin Placebo ## MCITotal 139 239 ## Other 10898 10795 ## Total Counts in Table: 22071 ## ## MLEs of row level probabilities ## MCITotal Other ## 0.01712655 0.98287345 ## ## MLEs of column level probabilities ## Asprin Placebo ## 0.500068 0.499932 ## Fitted cell probabilities assuming independence ## Asprin Placebo ## MCITotal 0.008564437 0.008562109 ## Other 0.491503525 0.491369928 ## Expected Counts assuming independence ## Asprin Placebo ## MCITotal 189.0257 188.9743 ## Other 10847.9743 10845.0257 ## Table of Chi-Square Residuals by cell ## Asprin Placebo ## MCITotal -3.6385862 3.6390808 ## Other 0.4803068 -0.4803721 ## ## Table of Chi-Square statistic terms by cell ## Asprin Placebo ## MCITotal 13.2393097 13.2429093 ## Other 0.2306947 0.2307574 ## ## Chi-Square Statistic: 26.94367 ## ## degrees of freedom: 1 ## ## P-Value : 2.09472e-07 # Total incidence of MCI is significantly reduced by aspirin # (a). Test 3 # Analyse incidence of fatal and nonfatal tableD.3=rbind( FatalNonFatal=colSums(tableD.1[c(1,2,3,4),]), Other=colSums(tableD.1[c(5),])) fcn.chisqtest(tableD.3) statisticsassignmenthelp.com
  • 12. ## ## Two-Way Table: ## Asprin Placebo ## FatalNonFatal 258 337 ## Other 10779 10697 ## ## Total Counts in Table: 22071 ## ## MLEs of row level probabilities ## FatalNonFatal Other ## 0.02695845 0.97304155 ## ## MLEs of column level probabilities ## Asprin Placebo ## 0.500068 0.499932 ## ## Fitted cell probabilities assuming independence ## Asprin Placebo ## FatalNonFatal 0.01348106 0.01347739 ## Other 0.48658690 0.48645464 ## ## Expected Counts assuming independence ## Asprin Placebo ## FatalNonFatal 297.5404 297.4596 ## Other 10739.4596 10736.5404 ## ## Table of Chi-Square Residuals by cell ## Asprin Placebo ## FatalNonFatal -2.2922843 2.2925959 ## Other 0.3815489 -0.3816008 ## ## Table of Chi-Square statistic terms by cell ## Asprin Placebo ## FatalNonFatal 5.2545672 5.2559958 ## Other 0.1455796 0.1456192 ## ## Chi-Square Statistic: 10.80176 ## ## degrees of freedom: 1 ## ## P-Value : 0.001014035 # Incidence of cardiovascular events is lower under Aspirin vs Placebo # but the significance is lower (though still significant relative to nominal levels) # (a). Test 4 # Among those having MCI, evaluate whether # fatality rate depends on Aspirin versus Placebo fcn.chisqtest(tableD.1[1:2,]) statisticsassignmenthelp.com
  • 13. ## ## Two-Way Table: ## Asprin Placebo ## MCIFatal 10 26 ## MCINonFatal 129 213 ## ## Total Counts in Table: 378 ## ## MLEs of row level probabilities ## MCIFatal MCINonFatal ## 0.0952381 0.9047619 ## ## MLEs of column level probabilities ## Asprin Placebo ## 0.3677249 0.6322751 ## ## Fitted cell probabilities assuming independence ## Asprin Placebo ## MCIFatal 0.03502142 0.06021668 ## MCINonFatal 0.33270345 0.57205845 ## ## Expected Counts assuming independence ## Asprin Placebo ## MCIFatal 13.2381 22.7619 ## MCINonFatal 125.7619 216.2381 ## ## Table of Chi-Square Residuals by cell ## Asprin Placebo ## MCIFatal -0.8899731 0.6787117 ## MCINonFatal 0.2887454 -0.2202031 ## ## Table of Chi-Square statistic terms by cell ## Asprin Placebo ## MCIFatal 0.7920521 0.46064953 ## MCINonFatal 0.0833739 0.04848942 ## ## Chi-Square Statistic: 1.384565 ## ## degrees of freedom: 1 ## ## P-Value : 0.2393251 statisticsassignmenthelp.com
  • 14. # Not significant # (a). Test 5 Incidence of strokes tableD.4=rbind( Strokes=colSums(tableD.1[c(3,4),]), Other=colSums(tableD.1[c(1,2,5),])) fcn.chisqtest(tableD.4) ## ## Two-Way Table: ## Asprin Placebo ## Strokes 119 98 ## Other 10918 10936 ## ## Total Counts in Table: 22071 ## ## MLEs of row level probabilities ## Strokes Other ## 0.009831906 0.990168094 ## ## MLEs of column level probabilities ## Asprin Placebo ## 0.500068 0.499932 ## ## Fitted cell probabilities assuming independence ## Asprin Placebo ## Strokes 0.004916621 0.004915285 ## Other 0.495151341 0.495016753 ## ## Expected Counts assuming independence ## Asprin Placebo ## Strokes 108.5147 108.4853 ## Other 10928.4853 10925.5147 ## ## Table of Chi-Square Residuals by cell ## Asprin Placebo ## Strokes 1.0065480 -1.0066848 ## Other -0.1002995 0.1003132 ## ## Table of Chi-Square statistic terms by cell ## Asprin Placebo ## Strokes 1.013139 1.01341436 ## Other 0.010060 0.01006273 ## ## Chi-Square Statistic: 2.046676 ## ## degrees of freedom: 1 ## ## P-Value : 0.1525389 statisticsassignmenthelp.com
  • 15. # Not significantly different # (b). Table of cariovascular mortality tableE=data.frame( Asprin=rbind( AcuteMCI=10, OtherIHD=24, SuddenDeath=22, Stroke=10, OtherCard=15), Placebo=rbind( AcuteMCI=28, OtherIHD=25, SuddenDeath=12, Stroke=7, OtherCard=11)) tableE.colSums=colSums(tableE) tableE.1<-rbind(tableE, None=((tableE.colSums*-1) +c(11037, 11034)) ) tableE.1 ## Asprin Placebo ## AcuteMCI 10 28 ## OtherIHD 24 25 ## SuddenDeath 22 12 ## Stroke 10 7 ## OtherCard 15 11 ## None 10956 10951fcn.chisqtest(tableE.1) statisticsassignmenthelp.com
  • 16. ## ## Two-Way Table: ## Asprin Placebo ## AcuteMCI 10 28 ## OtherIHD 24 25 ## SuddenDeath 22 12 ## Stroke 10 7 ## OtherCard 15 11 ## None 10956 10951 ## ## Total Counts in Table: 22071 ## ## MLEs of row level probabilities ## AcuteMCI OtherIHD SuddenDeath Stroke OtherCard ## 0.0017217163 0.0022201078 0.0015404830 0.0007702415 0.0011780164 ## None ## 0.9925694350 ## ## MLEs of column level probabilities ## Asprin Placebo ## 0.500068 0.499932 ## ## Fitted cell probabilities assuming independence ## Asprin Placebo ## AcuteMCI 0.0008609752 0.0008607411 ## OtherIHD 0.0011102048 0.0011099030 ## SuddenDeath 0.0007703462 0.0007701368 ## Stroke 0.0003851731 0.0003850684 ## OtherCard 0.0005890883 0.0005889281 ## None 0.4963521750 0.4962172600 ## ## Expected Counts assuming independence ## Asprin Placebo ## AcuteMCI 19.002583 18.997417 ## OtherIHD 24.503330 24.496670 ## SuddenDeath 17.002311 16.997689 ## Stroke 8.501155 8.498845 ## OtherCard 13.001767 12.998233 statisticsassignmenthelp.com
  • 17. ## None 10954.988854 10952.011146 ## ## Table of Chi-Square Residuals by cell ## Asprin Placebo ## AcuteMCI -2.065193737 2.065474468 ## OtherIHD -0.101681138 0.101694960 ## SuddenDeath 1.212035322 -1.212200079 ## Stroke 0.514064534 -0.514134412 ## OtherCard 0.554172450 -0.554247781 ## None 0.009660683 -0.009661996 ## ## Table of Chi-Square statistic terms by cell ## Asprin Placebo ## AcuteMCI 4.2650251718 4.266185e+00 ## OtherIHD 0.0103390539 1.034186e-02 ## SuddenDeath 1.4690296219 1.469429e+00 ## Stroke 0.2642623446 2.643342e-01 ## OtherCard 0.3071071045 3.071906e-01 ## None 0.0000933288 9.335417e-05 ## ## Chi-Square Statistic: 12.63343 ## ## degrees of freedom: 5 ## ## P-Value : 0.0270671 statisticsassignmenthelp.com
  • 18. # Mortality due to AcuteMCI is significantly reduced by Aspirin # The P-value is 0.027 # (b). Test 2 # Evaluation of total cariovascular mortality: tableE.2<-rbind(TotalCard=colSums(tableE.1[c(1:5),]), None=tableE.1[6,]) fcn.chisqtest(tableE.2) ## ## Two-Way Table: ## Asprin Placebo ## TotalCard 81 83 ## None 10956 10951 ## ## Total Counts in Table: 22071 ## ## MLEs of row level probabilities ## TotalCard None ## 0.007430565 0.992569435 ## ## MLEs of column level probabilities ## Asprin Placebo ## 0.500068 0.499932 ## ## Fitted cell probabilities assuming independence ## Asprin Placebo ## TotalCard 0.003715787 0.003714777 ## None 0.496352175 0.496217260 ## ## Expected Counts assuming independence ## Asprin Placebo ## TotalCard 82.01115 81.98885 ## None 10954.98885 10952.01115 ## ## Table of Chi-Square Residuals by cell ## Asprin Placebo ## TotalCard -0.111654791 0.111669969 ## None 0.009660683 -0.009661996 ## ## Table of Chi-Square statistic terms by cell ## Asprin Placebo ## TotalCard 0.0124667923 1.247018e-02 ## None 0.0000933288 9.335417e-05 ## ## Chi-Square Statistic: 0.02512366 ## ## degrees of freedom: 1 ## ## P-Value : 0.8740593 # Not significant statisticsassignmenthelp.com