Artificial Responses from Doignon and Falmagne (1999)

Description

Fictitious data set from Doignon and Falmagne (1999, chap. 7). Response patterns of 1000 respondents to five problems. Each respondent is assumed to be in one of nine possible states of the knowledge structure K.

Usage

data(DoignonFalmagne7)

Format

A list consisting of two components:

K

a state-by-problem indicator matrix representing the hypothetical knowledge structure. An element is one if the problem is contained in the state, and else zero.

N.R

a named numeric vector. The names denote response patterns, the values denote their frequencies.

Source

Doignon, J.-P., & Falmagne, J.-C. (1999). Knowledge spaces. Berlin: Springer.

Examples

data(DoignonFalmagne7)
DoignonFalmagne7$K    # knowledge structure
DoignonFalmagne7$N.R  # response patterns

Responses and Knowledge Structures from Taagepera et al. (1997)

Description

Taagepera et al. (1997) applied knowledge space theory to specific science problems. The density test was administered to 2060 students, the conservation of matter test to 1620 students. A subtest of five items each is included here. The response frequencies were reconstructed from histograms in the paper.

Usage

data(Taagepera)

Format

Two lists, each consisting of two components:

density97

a list with components K and N.R for the density test.

matter97

a list with components K and N.R for the conservation of matter test.

K

a state-by-problem indicator matrix representing the hypothetical knowledge structure. An element is one if the problem is contained in the state, and else zero.

N.R

a named numeric vector. The names denote response patterns, the values denote their frequencies.

Source

Taagepera, M., Potter, F., Miller, G.E., & Lakshminarayan, K. (1997). Mapping students' thinking patterns by the use of knowledge space theory. International Journal of Science Education, 19(3), 283–302. doi:10.1080/0950069970190303

Examples

data(Taagepera)
density97$K     # density test knowledge structure
density97$N.R   # density test response patterns
matter97$K      # conservation of matter knowledge structure
matter97$N.R    # conservation of matter response patterns

Basic Local Independence Models (BLIMs)

Description

Fits a basic local independence model (BLIM) for probabilistic knowledge structures by minimum discrepancy maximum likelihood estimation.

Usage

blim(K, N.R, method = c("MD", "ML", "MDML"), R = as.binmat(N.R),
     P.K = rep(1/nstates, nstates),
     beta = rep(0.1, nitems), eta = rep(0.1, nitems),
     betafix = rep(NA, nitems), etafix = rep(NA, nitems),
     betaequal = NULL, etaequal = NULL,
     randinit = FALSE, incradius = 0,
     tol = 1e-07, maxiter = 10000, zeropad = 16)

blimMD(K, N.R, R = as.binmat(N.R),
       betafix = rep(NA, nitems), etafix = rep(NA, nitems),
       incrule = c("minimum", "hypblc1", "hypblc2"), m = 1)

## S3 method for class 'blim'
anova(object, ..., test = c("Chisq", "none"))

Arguments

Details

See Doignon and Falmagne (1999) for details on the basic local independence model (BLIM) for probabilistic knowledge structures.

Minimum discrepancy (MD) minimizes the number of expected response errors (careless errors or lucky guesses). Maximum likelihood maximizes the likelihood, possibly at the expense of inflating the error and guessing parameters. Minimum discrepancy maximum likelihood (MDML) maximizes the likelihood subject to the constraint of minimum response errors. See Heller and Wickelmaier (2013) for details on the parameter estimation methods.

If randinit is TRUE, initial parameter values are sampled uniformly with the constraint beta + eta < 1 (Weisstein, 2013) for the error parameters, and with sum(P.K) == 1 (Rubin, 1981) for the probabilities of knowledge states. Setting randinit to TRUE overrides any values given in the P.K, beta, and eta arguments.

The degrees of freedom in the goodness-of-fit test are calculated as number of possible response patterns minus one or number of respondents, whichever is smaller, minus number of parameters.

blimMD uses minimum discrepancy estimation only. Apart from the hyperbolic inclusion rules, all of its functionality is also provided by blim. It may be removed in the future.

Value

An object of class blim having the following components:

References

Doignon, J.-P., & Falmagne, J.-C. (1999). Knowledge spaces. Berlin: Springer.

Heller, J., & Wickelmaier, F. (2013). Minimum discrepancy estimation in probabilistic knowledge structures. Electronic Notes in Discrete Mathematics, 42, 49–56. doi:10.1016/j.endm.2013.05.145

Rubin, D.B. (1981). The Bayesian bootstrap. The Annals of Statistics, 9(1), 130–134. doi:10.1214/aos/1176345338

Weisstein, E.W. (2013, August 29). Triangle point picking. In MathWorld – A Wolfram Web Resource. Retrieved from https://mathworld.wolfram.com/TrianglePointPicking.html.

See Also

simulate.blim, plot.blim, residuals.blim, logLik.blim, delineate, jacobian, endm, probability, chess.

Examples

data(DoignonFalmagne7)
K   <- DoignonFalmagne7$K    # knowledge structure
N.R <- DoignonFalmagne7$N.R  # frequencies of response patterns

## Fit basic local independence model (BLIM) by different methods
blim(K, N.R, method = "MD")    # minimum discrepancy estimation
blim(K, N.R, method = "ML")    # maximum likelihood estimation by EM
blim(K, N.R, method = "MDML")  # MDML estimation

## Parameter restrictions: beta_a = beta_b = beta_d, beta_c = beta_e
##                          eta_a =  eta_b = 0.1
m1 <- blim(K, N.R, method = "ML",
           betaequal = list(c(1, 2, 4), c(3, 5)),
              etafix = c(0.1, 0.1, NA, NA, NA))
m2 <- blim(K, N.R, method = "ML")
anova(m1, m2)

## See ?endm, ?probability, and ?chess for further examples.

Basic Local Independence Model Identification Analysis

Description

Tests the local identifiability of a basic local independence model (BLIM).

Usage

blimit(K, beta = NULL, eta = NULL, pi = NULL, file_name = NULL)

Arguments

Details

See Stefanutti et al. (2012) for details.

The blimit function has been adapted from code provided by Andrea Brancaccio, Debora de Chiusole, and Luca Stefanutti. It contains a function to compute the reduced row echelon form based on an implementation in the pracma package.

Value

A list having the following components:

References

Stefanutti, L., Heller, J., Anselmi, P., & Robusto, E. (2012). Assessing the local identifiability of probabilistic knowledge structures. Behavior Research Methods, 44(4), 1197–1211. doi:10.3758/s13428-012-0187-z

See Also

blim, jacobian.

Examples

K <- as.binmat(c("0000", "1000", "0100", "1110", "1101", "1111"))

set.seed(1234)
info <- blimit(K)

Responses to Chess Problems and Knowledge Structures

Description

Held, Schrepp and Fries (1995) derive several knowledge structures for the representation of 92 responses to 16 chess problems. See Schrepp, Held and Albert (1999) for a detailed description of these problems.

Usage

data(chess)

Format

A list consisting of five components:

dst1

a state-by-problem indicator matrix representing the knowledge structure DST1.

dst3

the knowledge structure DST3.

dst4

the knowledge structure DST4.

N.R

a named integer vector. The names denote response patterns, the values denote their frequencies.

R

a person-by-problem indicator matrix representing the responses. Column names hdbgXX and grazYY identify responses collected in Heidelberg and Graz, respectively.

Note

The graphs of the precedence relations for DST1 and DST4 in Held et. al (1995) contain mistakes that have been corrected. See examples.

Source

Held, T., Schrepp, M., & Fries, S. (1995). Methoden zur Bestimmung von Wissensstrukturen – eine Vergleichsstudie. Zeitschrift fuer Experimentelle Psychologie, 42(2), 205–236.

References

Schrepp, M., Held, T., & Albert, D. (1999). Component-based construction of surmise relations for chess problems. In D. Albert & J. Lukas (Eds.), Knowledge spaces: Theories, empirical research, and applications (pp. 41–66). Mahwah, NJ: Erlbaum.

Examples

data(chess)
chess$dst1  # knowledge structure DST1

## Precedence relation (Held et al., 1995, p. 215) and knowledge space
P <- as.binmat(c("1111011101111001",   # s
               # "0100000000000000",   # gs   mistake in Abb. 3
                 "0111010100111000",   # gs   correction
                 "0011010000011000",   # egs
                 "0011010000011000",   # eegs
                 "0000110000000000",   # cs
                 "0000010000000000",   # gcs
                 "0011011100111000",   # ts
                 "0011010100011000",   # ges
                 "1111111111111111",   # f
                 "0111010101111000",   # gf
                 "0011010000111000",   # gff
                 "0000000000010000",   # ggff
                 "0000000000001000",   # ggf
                 "0111011101111101",   # ff
                 "0111011101111011",   # tf
                 "0011010100111001"),  # tff
               as.logical = TRUE)
dimnames(P) <- list("<" = colnames(chess$R), ">" = colnames(chess$R))
K <- rbind(0L, binary_closure(t(P)))
identical(sort(as.pattern(K)),
          sort(as.pattern(chess$dst1)))

blim(chess$dst1, chess$N.R)  # Tab. 1

Conversion between Representations of Responses or States

Description

Converts between binary matrix and pattern representations of response patterns or knowledge states.

Usage

as.pattern(R, freq = FALSE, useNames = FALSE, as.set = FALSE,
           sep = "", emptyset = "{}", as.letters = NULL)

as.binmat(N.R, uniq = TRUE, col.names = NULL, as.logical = FALSE)

is.subset(R)

Arguments

Value

as.pattern returns a vector of integers named by the response patterns if freq is TRUE, else a character vector. If as.set is TRUE, the return value is of class set.

as.binmat returns an indicator matrix. If as.logical is TRUE, it returns a logical matrix.

is.subset returns a logical incidence matrix of the subset relation among states.

See Also

blim, set in package sets.

Examples

data(DoignonFalmagne7)
K <- DoignonFalmagne7$K
as.pattern(K, freq = TRUE)
as.pattern(K)
as.pattern(K, useNames = TRUE)
as.pattern(K, as.set = TRUE)

N.R <- DoignonFalmagne7$N.R
dim(as.binmat(N.R))
dim(as.binmat(N.R, uniq = FALSE))

## Knowledge structure as binary matrix
as.binmat(c("000", "100", "101", "111"))
as.binmat(set(set(), set("a"), set("a", "c"), set("a", "b", "c")))
as.binmat(c("000", "100", "101", "111"), as.logical = TRUE)

## Subset relation incidence matrix
is.subset(K)

## Plotting the knowledge structure
if(requireNamespace("relations") &&
   requireNamespace("Rgraphviz")) {
  rownames(K) <- as.pattern(K, useNames = TRUE)
  plot(relations::as.relation(is.subset(K)), main = "")
}

Delineate a Knowledge Structure by a Skill Function

Description

Computes the knowledge structure delineated by a skill function.

Usage

delineate(skillfun, itemID = 1)

Arguments

Details

The skill function (Q, S, \mu) indicates for each item in Q which subsets of skills in S are required to solve the item. Thus, \mu(q) is a set containing sets of skills. An item may have multiple entries in skillfun, each in a separate row identified by the same itemID.

See Doignon and Falmagne (1999, Chap. 4).

Value

A list of two components:

References

Doignon, J.-P., & Falmagne, J.-C. (1999). Knowledge spaces. Berlin: Springer.

See Also

blim.

Examples

# Skill function
# mu(e) = {{s, t}, {s, u}},  mu(f) = {{u}}
# mu(g) = {{s}, {t}},        mu(h) = {{t}}
sf <- read.table(header = TRUE, text = "
  item s t u
     e 1 1 0
     e 1 0 1
     f 0 0 1
     g 1 0 0
     g 0 1 0
     h 0 1 0
")
delineate(sf)

## See ?probability for further examples.

Responses and Knowledge Structures from Heller and Wickelmaier (2013)

Description

Knowledge structures and 200 artificial responses to four problems are used to illustrate parameter estimation in Heller and Wickelmaier (2013).

Usage

data(endm)

Format

A list consisting of three components:

K

a state-by-problem indicator matrix representing the true knowledge structure that underlies the model that generated the data.

K2

a slightly misspecified knowledge structure.

N.R

a named numeric vector. The names denote response patterns, the values denote their frequencies.

Source

Heller, J., & Wickelmaier, F. (2013). Minimum discrepancy estimation in probabilistic knowledge structures. Electronic Notes in Discrete Mathematics, 42, 49–56. doi:10.1016/j.endm.2013.05.145

Examples

data(endm)
endm$K    # true knowledge structure
endm$K2   # misspecified knowledge structure
endm$N.R  # response patterns

## Generate data from BLIM based on K
blim0 <- list(
     P.K = setNames(c(.1, .15, .15, .2, .2, .1, .1), as.pattern(endm$K)),
    beta = rep(.1, 4),
     eta = rep(.1, 4),
       K = endm$K,
  ntotal = 200)
class(blim0) <- "blim"
simulate(blim0)

## Fit BLIM based on K2
blim1 <- blim(endm$K2, endm$N.R, "MD")

Outer and Inner Fringes of a Knowledge Structure

Description

Returns the outer or inner fringe for each state in a knowledge structure.

Usage

getKFringe(K, nstates = nrow(K), nitems = ncol(K), outer = TRUE)

Arguments

Details

The outer fringe of a knowledge state is the set of all items that can be learned from that state, such that adding an outer-fringe item to the state results in another state in K,

K^O = \{q \notin K | K \cup \{q\} \in \mathcal{K}\}.

The inner fringe of a knowledge state is the set of all items that have been learned most recently to reach that state, such that deleting an inner-fringe item from the state results in another state in K,

K^I = \{q \in K | K - \{q\} \in \mathcal{K}\}.

Value

A state-by-problem indicator matrix representing the outer or inner fringe for each knowledge state in K.

See Also

slm, simulate.blim.

Examples

data(DoignonFalmagne7)

## Which items can be learned from each state?
getKFringe(DoignonFalmagne7$K)

## Which items in each state have been recently learned?
getKFringe(DoignonFalmagne7$K, outer = FALSE)

Forward-/Backward-Gradedness and Downgradability of a Knowledge Structure

Description

Checks if a knowledge structure is

• forward- or backward-graded in any item;

• downgradable.

Usage

is.forward.graded(K)

is.backward.graded(K)

is.downgradable(K)

Arguments

Details

A knowledge structure K is forward-graded in item q, if S \cup \{q\} is in K for every state S \in K. A knowledge structure K is backward-graded in item q, if S - \{q\} is in K for every state S \in K. See Spoto, Stefanutti, and Vidotto (2012).

A knowledge structure K is downgradable, if its inner fringe is empty only for a single state (the empty set). See Doignon and Falmagne (2015).

Value

For forward- and backward-gradedness, a named logical vector with as many elements as columns in K.

For downgradability, a single logical value.

References

Doignon, J.-P., & Falmagne, J.-C. (2015). Knowledge spaces and learning spaces. arXiv. doi:10.48550/arXiv.1511.06757

Spoto, A., Stefanutti, L., & Vidotto, G. (2012). On the unidentifiability of a certain class of skill multi map based probabilistic knowledge structures. Journal of Mathematical Psychology, 56(4), 248–255. doi:10.1016/j.jmp.2012.05.001

See Also

blim, jacobian, getKFringe.

Examples

K <- as.binmat(c("0000", "1000", "1100", "1010", "0110", "1110", "1111"))
is.forward.graded(K)                  # forward-graded in a
is.backward.graded(K)                 # not backward-graded in a
is.downgradable(K)                    # not downgradable
all(K[, "a"] | getKFringe(K)[, "a"])  # every K or outer fringe contains a

Item Tree Analysis (ITA)

Description

Item tree analysis (ITA) on a set of binary responses.

Usage

ita(R, L = NULL, makeK = FALSE, search = c("local", "global"))

Arguments

Details

ITA seeks to establish a precedence relation among a set of binary items. For each pair of items (p, q), it counts how often p is not solved if q is solved, which constitutes a violation of the relation. ITA searches for a threshold L for the maximum number of violations consistent with a (transitive) precedence relation. Its attempts to minimize the total discrepancy between R and K.

See van Leeuwe (1974) and Schrepp (1999) for details.

Value

An object of class ita having the following components:

References

Schrepp, M. (1999). On the empirical construction of implications between bi-valued test items. Mathematical Social Sciences, 38(3), 361–375. doi:10.1016/S0165-4896(99)00025-6

Van Leeuwe, J.F. (1974). Item tree analysis. Nederlands Tijdschrift voor de Psychologie en haar Grensgebieden, 29(6), 475–483.

See Also

blim.

Examples

data(chess)

ita(chess$R)  # find (locally) optimal threshold L

i <- ita(chess$R, L = 6, makeK = TRUE)
identical(sort(as.pattern(i$K)),
          sort(as.pattern(chess$dst1)))

## Plotting the precedence relation
if(requireNamespace("relations") &&
   requireNamespace("Rgraphviz")) {
  plot(relations::as.relation(i$I))
}

Jacobian Matrix for Basic Local Independence Model

Description

Computes the Jacobian matrix for a basic local independence model (BLIM).

Usage

jacobian(object, P.K = rep(1/nstates, nstates),
         beta = rep(0.1, nitems), eta = rep(0.1, nitems),
         betafix = rep(NA, nitems), etafix = rep(NA, nitems))

Arguments

Details

This is a draft version. It may change in future releases.

Value

The Jacobian matrix. The number of rows equals 2^(number of items) - 1, the number of columns equals the number of independent parameters in the model.

References

Heller, J. (2017). Identifiability in probabilistic knowledge structures. Journal of Mathematical Psychology, 77, 46–57. doi:10.1016/j.jmp.2016.07.008

Stefanutti, L., Heller, J., Anselmi, P., & Robusto, E. (2012). Assessing the local identifiability of probabilistic knowledge structures. Behavior Research Methods, 44(4), 1197–1211. doi:10.3758/s13428-012-0187-z

See Also

blim, simulate.blim, gradedness.

Examples

data(endm)
m <- blim(endm$K2, endm$N.R)

## Test of identifiability
J <- jacobian(m)
dim(J)
qr(J)$rank

Diagnostic Plot for Basic Local Independence Models

Description

Plots BLIM residuals against fitted values.

Usage

## S3 method for class 'blim'
plot(x, xlab = "Predicted response probabilities",
     ylab = "Deviance residuals", ...)

Arguments

Details

The deviance residuals are plotted against the predicted response probabilities for each response pattern.

See Also

blim, residuals.blim.

Examples

## Compare MD and MDML estimation

data(DoignonFalmagne7)
blim1 <- blim(DoignonFalmagne7$K, DoignonFalmagne7$N.R, method="MD")
blim2 <- blim(DoignonFalmagne7$K, DoignonFalmagne7$N.R, method="MDML")

par(mfrow = 1:2)      # residuals versus fitted values
plot(blim1, main = "MD estimation",   ylim = c(-4, 4))
plot(blim2, main = "MDML estimation", ylim = c(-4, 4))

Print a blim Object

Description

Prints the output of a blim model object.

Usage

## S3 method for class 'blim'
print(x, P.Kshow = FALSE, errshow = TRUE,
      digits=max(3, getOption("digits") - 2), ...)

Arguments

Value

Returns the blim object invisibly.

See Also

blim.

Examples

data(DoignonFalmagne7)
 
blim1 <- blim(DoignonFalmagne7$K, DoignonFalmagne7$N.R)
print(blim1, showP.K = TRUE)

Problems in Elementary Probability Theory

Description

This data set contains responses to problems in elementary probability theory observed before and after some instructions (the so-called learning object) were given. Data were collected both in the lab and via an online questionnaire. Of the 1127 participants eligible in the online study, 649 were excluded because they did not complete the first set of problems (p101, ..., p112) or they responded too quickly or too slowly. Based on similar criteria, further participants were excluded for the second set of problems, indicated by missing values in the variables b201, ..., b212. Problems were presented in random order.

Participants were randomized to two conditions: an enhanced learning object including instructions with examples and a basic learning object without examples. Instructions were given on four concepts: how to calculate the classic probability of an event (pb), the probability of the complement of an event (cp), of the union of two disjoint events (un), and of two independent events (id).

The questionnaire was organized as follows:

Page 1

Welcome page.

Page 2

Demographic data.

Page 3

First set of problems.

Page 4 to 8

Instructions (learning object).

Page 9

Second set of problems.

Page 10

Feedback about number of correctly solved problems.

Usage

data(probability)

Format

A data frame with 504 cases and 68 variables:

• case a factor giving the case id, a five-digits code the fist digit denoting lab or online branch of the study, the last four digits being the case number.

• lastpage Which page of the questionnaire was reached before quitting? The questionnaire consisted of ten pages.

• mode a factor; lab or online branch of study.

• started a timestamp of class POSIXlt. When did participant start working on the questionnaire?

• sex a factor coding sex of participant.

• age age of participant.

• educat education as a factor with three levels: 1 secondary school or below; 2 higher education entrance qualification; 3 university degree.

• fos field of study. Factor with eight levels: ecla economics, business, law; else miscellaneous; hipo history, politics; lang languages; mabi mathematics, physics, biology; medi medical science; phth philosophy, theology; psco psychology, computer science, cognitive science.

• semester ordered factor. What semester are you in?

• learnobj a factor with two levels: enhan learning object enhanced with examples; basic learning object without examples.

The twelve problems of the first part (before the learning object):

• p101 A box contains 30 marbles in the following colors: 8 red, 10 black, 12 yellow. What is the probability that a randomly drawn marble is yellow? (Correct: 0.40)

• p102 A bag contains 5-cent, 10-cent, and 20-cent coins. The probability of drawing a 5-cent coin is 0.35, that of drawing a 10-cent coin is 0.25, and that of drawing a 20-cent coin is 0.40. What is the probability that the coin randomly drawn is not a 5-cent coin? (0.65)

• p103 A bag contains 5-cent, 10-cent, and 20-cent coins. The probability of drawing a 5-cent coin is 0.20, that of drawing a 10-cent coin is 0.45, and that of drawing a 20-cent coin is 0.35. What is the probability that the coin randomly drawn is a 5-cent coin or a 20-cent coin? (0.55)

• p104 In a school, 40% of the pupils are boys and 80% of the pupils are right-handed. Suppose that gender and handedness are independent. What is the probability of randomly selecting a right-handed boy? (0.32)

• p105 Given a standard deck containing 32 different cards, what is the probability of not drawing a heart? (0.75)

• p106 A box contains 20 marbles in the following colors: 4 white, 14 green, 2 red. What is the probability that a randomly drawn marble is not white? (0.80)

• p107 A box contains 10 marbles in the following colors: 2 yellow, 5 blue, 3 red. What is the probability that a randomly drawn marble is yellow or blue? (0.70)

• p108 What is the probability of obtaining an even number by throwing a dice? (0.50)

• p109 Given a standard deck containing 32 different cards, what is the probability of drawing a 4 in a black suit? (Responses that round to 0.06 were considered correct.)

• p110 A box contains marbles that are red or yellow, small or large. The probability of drawing a red marble is 0.70 (lab: 0.30), the probability of drawing a small marble is 0.40. Suppose that the color of the marbles is independent of their size. What is the probability of randomly drawing a small marble that is not red? (0.12, lab: 0.28)

• p111 In a garage there are 50 cars. 20 are black and 10 are diesel powered. Suppose that the color of the cars is independent of the kind of fuel. What is the probability that a randomly selected car is not black and it is diesel powered? (0.12)

• p112 A box contains 20 marbles. 10 marbles are red, 6 are yellow and 4 are black. 12 marbles are small and 8 are large. Suppose that the color of the marbles is independent of their size. What is the probability of randomly drawing a small marble that is yellow or red? (0.48)

The twelve problems of the second part (after the learning object):

• p201 A box contains 30 marbles in the following colors: 10 red, 14 yellow, 6 green. What is the probability that a randomly drawn marble is green? (0.20)

• p202 A bag contains 5-cent, 10-cent, and 20-cent coins. The probability of drawing a 5-cent coin is 0.25, that of drawing a 10-cent coin is 0.60, and that of drawing a 20-cent coin is 0.15. What is the probability that the coin randomly drawn is not a 5-cent coin? (0.75)

• p203 A bag contains 5-cent, 10-cent, and 20-cent coins. The probability of drawing a 5-cent coin is 0.35, that of drawing a 10-cent coin is 0.20, and that of drawing a 20-cent coin is 0.45. What is the probability that the coin randomly drawn is a 5-cent coin or a 20-cent coin? (0.80)

• p204 In a school, 70% of the pupils are girls and 10% of the pupils are left-handed. Suppose that gender and handedness are independent. What is the probability of randomly selecting a left-handed girl? (0.07)

• p205 Given a standard deck containing 32 different cards, what is the probability of not drawing a club? (0.75)

• p206 A box contains 20 marbles in the following colors: 6 yellow, 10 red, 4 green. What is the probability that a randomly drawn marble is not yellow? (0.70)

• p207 A box contains 10 marbles in the following colors: 5 blue, 3 red, 2 green. What is the probability that a randomly drawn marble is blue or red? (0.80)

• p208 What is the probability of obtaining an odd number by throwing a dice? (0.50)

• p209 Given a standard deck containing 32 different cards, what is the probability of drawing a 10 in a red suit? (Responses that round to 0.06 were considered correct.)

• p210 A box contains marbles that are green or red, large or small The probability of drawing a green marble is 0.40, the probability of drawing a large marble is 0.20. Suppose that the color of the marbles is independent of their size. What is the probability of randomly drawing a large marble that is not green? (0.12)

• p211 In a garage there are 50 cars. 15 are white and 20 are diesel powered. Suppose that the color of the cars is independent of the kind of fuel. What is the probability that a randomly selected car is not white and it is diesel powered? (0.28)

• p212 A box contains 20 marbles. 8 marbles are white, 4 are green and 8 are red. 15 marbles are small and 5 are large. Suppose that the color of the marbles is independent of their size. What is the probability of randomly drawing a large marble that is white or green? (0.15)

Further variables:

• time01, ..., time10 the time (in s) spent on each page of the questionnaire. In the lab branch of the study, participants started directly on Page 2.

• b101, ..., b112 the twelve problems of the first part coded as correct (1) or error (0).

• b201, ..., b212 the twelve problems of the second part coded as correct (1) or error (0).

Source

Data were collected by Pasquale Anselmi and Florian Wickelmaier at the Department of Psychology, University of Tuebingen, in February and March 2010.

Examples

data(probability)

## "Completer" sample
pb <- probability[!is.na(probability$b201), ]

## Response frequencies for first and second part
N.R1 <- as.pattern(pb[, sprintf("b1%.2i", 1:12)], freq = TRUE)
N.R2 <- as.pattern(pb[, sprintf("b2%.2i", 1:12)], freq = TRUE)

## Conjunctive skill function, one-to-one problem function
sf1 <- read.table(header = TRUE, text = "
  item cp id pb un
     1  0  0  1  0
     2  1  0  0  0
     3  0  0  0  1
     4  0  1  0  0
     5  1  0  1  0
     6  1  0  1  0
     7  0  0  1  1
     8  0  0  1  1
     9  0  1  1  0
    10  1  1  0  0
    11  1  1  1  0
    12  0  1  1  1
")

## Extended skill function
sf2 <- rbind(sf1, read.table(header = TRUE, text = "
  item cp id pb un
     2  0  0  0  1
     3  1  0  0  0
     6  0  0  1  1
     7  1  0  1  0
    12  1  1  1  0
"))

## Delineated knowledge structures
K1 <- delineate(sf1)$K
K2 <- delineate(sf2)$K

## After instructions, fit of knowledge structures improves
sapply(list(N.R1, N.R2), function(n) blim(K1, n)$discrepancy)
sapply(list(N.R1, N.R2), function(n) blim(K2, n)$discrepancy)

Residuals for Basic Local Independence Models

Description

Computes deviance and Pearson residuals for blim objects.

Usage

## S3 method for class 'blim'
residuals(object, type = c("deviance", "pearson"), ...)

Arguments

Details

See residuals.glm for details.

Value

A named vector of residuals having as many elements as response patterns.

See Also

blim, residuals.glm, plot.blim.

Examples

data(DoignonFalmagne7)
blim1 <- blim(DoignonFalmagne7$K, DoignonFalmagne7$N.R)

sum( resid(blim1)^2 )                # likelihood ratio G2
sum( resid(blim1, "pearson")^2 )     # Pearson X2

Arithmetic Problems for Elementary and Middle School Students

Description

The 23 fraction problems were presented to 191 first-level middle school students (about 11 to 12 years old). A subset of 13 problems is included in Stefanutti and de Chiusole (2017).

The eight subtraction problems were presented to 294 elementary school students and are described in de Chiusole and Stefanutti (2013).

Usage

data(schoolarithm)

Format

fraction17

a person-by-problem indicator matrix representing the responses of 191 persons to 23 problems. The responses are classified as correct (0) or incorrect (1).

The 23 problems were:

• p01 % \big(\frac{1}{3} + \frac{1}{12}\big) : \frac{2}{9} = ?

• p02 % \big(\frac{3}{2} + \frac{3}{4}\big) \times \frac{5}{3} - 2 = ?

• p03 % \big(\frac{5}{6} + \frac{3}{14}\big) \times \big(\frac{19}{8} - \frac{3}{2}\big) = ?

• p04 % \big(\frac{1}{6} + \frac{2}{9}\big) - \frac{7}{36} = ?

• p05 % \frac{7}{10} + \frac{9}{10} = ?

• p06 % \frac{8}{13} + \frac{5}{2} = ?

• p07 % \frac{8}{12} + \frac{4}{15} = ?

• p08 % \frac{2}{9} + \frac{5}{6} = ?

• p09 % \frac{7}{5} + \frac{1}{5} = ?

• p10 % \frac{2}{7} + \frac{3}{14} = ?

• p11 % \frac{5}{9} + \frac{1}{6} = ?

• p12 % \big(\frac{1}{12} + \frac{1}{3}\big) \times \frac{24}{15} = ?

• p13 % 2 - \frac{3}{4} = ?

• p14 % \big(4 + \frac{3}{4} - \frac{1}{2}\big) \times \frac{8}{6} = ?

• p15 % \frac{4}{7} + \frac{3}{4} = \frac{?}{28}

• p16 % \frac{5}{8} - \frac{3}{16} = \frac{? - ?}{16}

• p17 % \frac{3}{8} + \frac{5}{12} = \frac{? \times 3 + ? \times 5}{24}

• p18 % \frac{2}{7} + \frac{3}{5} = \frac{5 \times ? + 7 \times ?}{35}

• p19 % \frac{2}{3} + \frac{6}{9} = \frac{?}{9} = \frac{?}{?}

• p20 Least common multiple lcm(6, 8) = ?

• p21 % \frac{7}{11} \times \frac{2}{3} = ?

• p22 % \frac{2}{5} \times \frac{15}{4} = ?

• p23 % \frac{9}{7} : \frac{2}{3} = ?

subtraction13 is a data frame consisting of the following components:

School

factor; school id.

Classroom

factor; class room id.

Gender

factor; participant gender.

Age

participant age.

R

a person-by-problem indicator matrix representing the responses of 294 persons to eight problems.

The eight problems were:

• p1 73 - 58

• p2 317 - 94

• p3 784 - 693

• p4 507 - 49

• p5 253 - 178

• p6 2245 - 418

• p7 156 - 68

• p8 3642 - 753

Source

The data were made available by Debora de Chiusole, Andrea Brancaccio, and Luca Stefanutti.

References

de Chiusole, D., & Stefanutti, L. (2013). Modeling skill dependence in probabilistic competence structures. Electronic Notes in Discrete Mathematics, 42, 41–48. doi:10.1016/j.endm.2013.05.144

Stefanutti, L., & de Chiusole, D. (2017). On the assessment of learning in competence based knowledge space theory. Journal of Mathematical Psychology, 80, 22–32. doi:10.1016/j.jmp.2017.08.003

Examples

data(schoolarithm)

## Fraction problems used in Stefanutti and de Chiusole (2017)
R <- fraction17[, c(4:8, 10:11, 15:20)]
colnames(R) <- 1:13
N.R <- as.pattern(R, freq = TRUE)

## Conjunctive skill function in Table 1
sf <- read.table(header = TRUE, text = "
  item  a  b  c  d  e  f  g  h
     1  1  1  1  0  1  1  0  0
     2  1  0  0  0  0  0  1  1
     3  1  1  0  1  1  0  0  0
     4  1  1  0  0  1  1  1  1
     5  1  1  0  0  1  1  0  0
     6  1  1  1  0  1  0  1  1
     7  1  1  0  0  1  1  0  0
     8  1  1  0  0  1  0  1  1
     9  0  1  0  0  1  0  0  0
    10  0  1  0  0  0  0  0  0
    11  0  0  0  0  1  0  0  0
    12  1  1  0  0  1  0  1  1
    13  0  0  0  0  0  1  0  0
")
K <- delineate(sf)$K  # delineated knowledge structure
blim(K, N.R)

## Subtraction problems used in de Chiusole and Stefanutti (2013)
N.R <- as.pattern(subtraction13$R, freq = TRUE)

# Skill function in Table 1
# (f) mastering tens and hundreds; (g) mastering thousands; (h1) one borrow;
# (h2) two borrows; (h3) three borrows; (i) mastering the proximity of
# borrows; (j) mastering the presence of the zero; (k) mental calculation
sf <- read.table(header = TRUE, text = "
  item  f  g h1 h2 h3  i  j  k
     1  0  0  1  0  0  0  0  0
     2  1  0  1  0  0  0  0  0
     3  1  0  1  0  0  1  0  0
     4  1  0  1  1  1  0  1  0
     4  0  0  0  0  0  0  0  1
     5  1  0  1  1  1  1  0  0
     6  1  1  1  1  0  0  0  0
     7  1  0  1  1  1  1  0  0
     8  1  1  1  1  1  0  0  0
")
K <- delineate(sf)$K
blim(K, N.R)

Simulate Responses from Basic Local Independence Models (BLIMs)

Description

Simulates responses from the distribution corresponding to a fitted blim model object.

Usage

## S3 method for class 'blim'
simulate(object, nsim = 1, seed = NULL, ...)

Arguments

Details

Responses are simulated in two steps: First, a knowledge state is drawn with probability P.K. Second, responses are generated by applying rbinom with probabilities computed from the model object's beta and eta components.

Value

A named vector of frequencies of response patterns.

See Also

blim, endm.

Examples

data(DoignonFalmagne7)
 
m1 <- blim(DoignonFalmagne7$K, DoignonFalmagne7$N.R)
simulate(m1)

## Parametric bootstrap for the BLIM
disc <- replicate(200, blim(m1$K, simulate(m1))$discrepancy)

hist(disc, col = "lightgray", border = "white", freq = FALSE, breaks = 20,
     main = "BLIM parametric bootstrap", xlim = c(.05, .3))
abline(v = m1$discrepancy, lty = 2)

## Parameter recovery for the SLM
m0 <- list( P.K = getSlmPK( g = rep(.8, 5),
                            K = DoignonFalmagne7$K,
                           Ko = getKFringe(DoignonFalmagne7$K)),
           beta = rep(.1, 5),
            eta = rep(.1, 5),
              K = DoignonFalmagne7$K,
         ntotal = 800)
class(m0) <- c("slm", "blim")

pars <- replicate(20, coef(slm(m0$K, simulate(m0), method = "ML")))
boxplot(t(pars), horizontal = TRUE, las = 1,
        main = "SLM parameter recovery")

## See ?endm for further examples.

Simple Learning Models (SLMs)

Description

Fits a simple learning model (SLM) for probabilistic knowledge structures by minimum discrepancy maximum likelihood estimation.

Usage

slm(K, N.R, method = c("MD", "ML", "MDML"), R = as.binmat(N.R),
    beta = rep(0.1, nitems), eta = rep(0.1, nitems),
    g = rep(0.1, nitems),
    betafix = rep(NA, nitems), etafix = rep(NA, nitems),
    betaequal = NULL, etaequal = NULL,
    randinit = FALSE, incradius = 0,
    tol = 1e-07, maxiter = 10000, zeropad = 16,
    checkK = TRUE)

getSlmPK(g, K, Ko)

## S3 method for class 'slm'
print(x, P.Kshow = FALSE, parshow = TRUE,
      digits=max(3, getOption("digits") - 2), ...)

Arguments

Details

See Doignon and Falmagne (1999) for details on the simple learning model (SLM) for probabilistic knowledge structures. The model requires a well-graded knowledge space K.

An slm object inherits from class blim. See blim for details on the function arguments. The helper function getSlmPK returns the distribution of knowledge states P.K.

Value

An object of class slm and blim. It contains all components of a blim object. In addition, it includes:

References

Doignon, J.-P., & Falmagne, J.-C. (1999). Knowledge spaces. Berlin: Springer.

See Also

blim, simulate.blim, getKFringe, is.downgradable

Examples

data(DoignonFalmagne7)
K   <- DoignonFalmagne7$K     # well-graded knowledge space
N.R <- DoignonFalmagne7$N.R   # frequencies of response patterns

## Fit simple learning model (SLM) by different methods
slm(K, N.R, method = "MD")    # minimum discrepancy estimation
slm(K, N.R, method = "ML")    # maximum likelihood estimation by EM
slm(K, N.R, method = "MDML")  # MDML estimation

## Compare SLM and BLIM
m1 <-  slm(K, N.R, method = "ML")
m2 <- blim(K, N.R, method = "ML")
anova(m1, m2)