## DATA GENERATION n <- 9 + sample(1:6, 4) dat <- data.frame(Evaluation = rnorm(sum(n), mean = rep(sample(seq(from = sample(25:55, 1), by = 1, length.out = sample(15:20, 1)), 4)/10, n), sd = rep(sample(70:80, 4)/100, n)), Occupation = factor(rep(1:4, n), labels = c("Student", "Employee", "Self-Employed", "Retired"))) dat$Evaluation[dat$Evaluation > 10] <- 10 dat$Evaluation[dat$Evaluation < 0] <- 0 ## QUESTION/ANSWER GENERATION questions <- character(5) solutions <- logical(5) explanations <- character(5) fm0 <- lm(Evaluation ~ 1, data = dat) fm1 <- lm(Evaluation ~ Occupation, data = dat) myanova <- anova(fm0, fm1) myanova_print <- matrix(nrow = 2, ncol = 6) dimnames(myanova_print) <- dimnames(myanova) myanova_print[, 1] <- round(myanova[, 1]) myanova_print[, 3] <- fmt(myanova[, 3], 0) rss <- round(myanova[,2], digits = max(0, min(3, 5-max(nchar(round(myanova[, 2], digits = 0)))))) myanova_print[, 2] <- format(rss, nsmall = max(0, min(3, 5-max(nchar(round(myanova[, 2], digits = 0)))))) myanova_print[2, 4] <- format(-diff(rss), nsmall = max(0, min(3, 5-max(nchar(round(myanova[, 2], digits = 0)))))) myanova_print[1, 3:6] <- "" myanova_print[2, 5] <- round(myanova[2, 5], digits = 3) myanova_print[2, 6] <- format.pval(myanova[2, 6]) r2 <- 1 - as.numeric(rss[2])/as.numeric(rss[1]) f2 <- sample(10:250, 1)/10 if(runif(1) > 0.5) { questions[1] <- paste0("The test statistic is smaller than $", f2, "$.", sep = "") solutions[1] <- myanova[2,5] < f2 explanations[1] <- paste0("The test statistic is $F = ", fmt(myanova[2,5], 3), "$ and hence ", ifelse(solutions[1], "", "_not_"), " smaller than $", f2, "$.", sep = "") } else { questions[1] <- paste0("The test statistic is larger than $", f2, "$.", sep = "") solutions[1] <- myanova[2,5] > f2 explanations[1] <- paste0("The test statistic is $F = ", fmt(myanova[2,5], 3), "$ and hence ", ifelse(solutions[1], "", "_not_"), " larger than $", f2, "$.", sep = "") } questions[2] <- "A one-sided alternative was tested for the mean values." solutions[2] <- FALSE explanations[2] <- paste("An ANOVA always tests the null hypothesis,", "that all mean values are equal against the alternative hypothesis that they are different.") r2a <- sample(10:60, 1)/100 questions[3] <- paste0("The fraction of explained variance is larger than $", 100 * r2a, "$%.", sep = "") solutions[3] <- r2 > r2a explanations[3] <- paste0("The fraction of explained variance is $", fmt(r2, 3), "$ and hence ", ifelse(solutions[3], "", "_not_"), " larger than ", r2a, ".", sep = "") questions[4] <- paste("It can be shown that the evaluation of the respondents depends on their occupation.", "(Significance level $5$%)") solutions[4] <- myanova[2,6] < 0.05 explanations[4] <- paste0("The $p$ value is $", format.pval(myanova[2,6], digits = 3), "$ and hence", ifelse(solutions[4], "", "_not_"), " significant. It can ", ifelse(solutions[4], "", "_not_"), " be shown that the evaluations differ with respect to the occupation of the respondents.") r2b <- sample(10:60, 1)/100 questions[5] <- paste0("The fraction of explained variance is smaller than $", 100 * r2b, "$%.", sep = "") solutions[5] <- r2 < r2b explanations[5] <- paste0("The fraction of explained variance is $", fmt(r2, 3), "$ and hence ", ifelse(solutions[5], "", "_not_"), " smaller than ", r2b, ".", sep = "") ## permute order of solutions/questions o <- sample(1:5) questions <- questions[o] solutions <- solutions[o] explanations <- explanations[o]
A survey with r sum(n)
persons was conducted to analyze the
design of an advertising campaign. Each respondent was asked to
evaluate the overall impression of the advertisement on an
eleven-point scale from 0 (bad) to 10 (good). The evaluations are
summarized separately with respect to type of occupation of the
respondents in the following figure.
\
par(mar = c(4, 4, 1, 1)) plot(Evaluation ~ Occupation, data = dat)
To analyze the influence of occupation on the evaluation of the advertisement an analysis of variance was performed:
options(show.signif.stars = FALSE) print(myanova_print, quote = FALSE, right = TRUE)
Which of the following statements are correct?
answerlist(questions, markup = "markdown")
In order to be able to answer the questions the fraction of
explained variance has to be determined. The residual sum of squares
when using only a single overall mean value ($\mathit{RSS}_0$) as
well as the residual sum of squares when allowing different mean
values given occupation ($\mathit{RSS}_1$) are required. Both are
given in the RSS column of the ANOVA table. The
fraction of explained variance is given by
$1 - \mathit{RSS}_1/\mathit{RSS}_0 = 1 - r rss[2]
/r rss[1]
=
r round(r2, digits = 3)
$.
The statements above can now be evaluated as right or wrong.
answerlist(ifelse(solutions, "True", "False"), explanations, markup = "markdown")
extype: mchoice
exsolution: r mchoice2string(solutions)
exname: Analysis of variance
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