Signal Detection Theory, Lecture 3
Eugenie Roudaia
Chapter 5: Classification Experiments For One-Dimensional Stimulus Sets
Classification experiments:
- Yes/No: N = 2 stimuli and M = 2 responses (e.g., old/new, present/absent)
- Category scaling/categorization: N > 2 stimuli, M = 2 responses (category A vs B)
- Identification: N >2 stimuli and M = N responses
Perceptual one-dimensionality
- Physically one-dimensional: light intensity, sound frequency
- Perceptually one-dimensional: e.g.
Under SDT, if stimuli \(S_1, S_2, S_3\) are perceptually one dimensional and the means of the distributions they give rise to can be ordered such as \(\mu_1< \mu_2< \mu_3\), then \[ d'(1,3) = d'(1,2)+d'(2,3) \] * cumulative d’: d’ for any stimulus S to the end-point.
- total d’: cumulative d’ between the two endpoints.
Thurstone (1927) developed models of one-dimensional classification based on the normal distribution, known as “Thurstonian scaling”.
False memory example and pseudo-d’
Recognition memory experiment:
- Study items thematically related (bed, night, dream)
- Test items: Old (bed, night, dream), New (flower, tree), New-lures (sleep, pillow)
- Responses: “Old” or “New”
## items pYes zPyes
## 1 Old 0.85 1.0364334
## 2 New 0.30 -0.5244005
## 3 New-Lures 0.80 0.8416212
Can calculate \(d'_{(Old, New)}\) as usual:
- \(d'_{(Old, New)}= z(p(\text{"old"}|\text{Old})) - z(p(\text{"old"}|\text{New})) = 1.03 -(-0.52) = 1.55\)
Can calculate \(d'_{(Old, New-Lures)}\) as usual:
- \(d'_{(Old, New-Lures)}= z(p(\text{"old"}|\text{Old})) - z(p(\text{"old"}|\text{New-Lures})) = 1.03 -(0.84) = 0.19\)
And can also calculate \(d'_{(New-Lures, New)}\) : \(pseudo-d' = z(p(\text{"old"}|\text{New-Lures})) - z(p(\text{"old"}|\text{New})) = 0.84 - (-0.52) = 1.36\)
Two-response Classification
- Auditory detection experiment
- 7 sound intensities presented randomly on each trial: \(S_1\) intensity = 0. \(S_2\) to \(S_7\) arranged with increasing intensity.
- “Yes” is sound is heard, “No” if not heard.
## Stimulus Yes No
## 1 1 0.02 0.98
## 2 2 0.06 0.94
## 3 3 0.15 0.85
## 4 4 0.40 0.60
## 5 5 0.80 0.20
## 6 6 0.92 0.08
## 7 7 0.96 0.04Can calculate d’ directly for each adjacent or non-adjacent stimulus pair:
- \(d'(1,2) = z(p(\text{Yes}|\text{S2})) - z(p(\text{Yes}|\text{S1})\)
round(qnorm(0.06) - qnorm(0.02),3)## [1] 0.499- \(d'(1,3) = z(p(\text{Yes}|\text{S3})) - z(p(\text{Yes}|\text{S1})\)
round(qnorm(0.15) - qnorm(0.02),3)## [1] 1.017Can calculate d’ indirectly for non-adjacent stimulus pairs:
- \(d'(1,3) = d'(1,2) + d'(2,3)\)
Here, we calculate d’ for each adjacent pair, and calculate the cumulative sum:
audExpt$zYes = round(qnorm(audExpt$Yes),3)
audExpt$dPrime = c(NA,(diff(audExpt$zYes)))
audExpt$cumulativeDp = c(NA, cumsum(audExpt$dPrime[2:7]))
audExpt## Stimulus Yes No zYes dPrime cumulativeDp
## 1 1 0.02 0.98 -2.054 NA NA
## 2 2 0.06 0.94 -1.555 0.499 0.499
## 3 3 0.15 0.85 -1.036 0.519 1.018
## 4 4 0.40 0.60 -0.253 0.783 1.801
## 5 5 0.80 0.20 0.842 1.095 2.896
## 6 6 0.92 0.08 1.405 0.563 3.459
## 7 7 0.96 0.04 1.751 0.346 3.805Cumulative d’ on line 3 is the indirectly-calculated d’(1,3) = 1.018, very similar to directly obtained d’=1.017.
Representation of the stimuli on decision space: 
Typically, the mean of an endpoint stimulus is set to 0 (here, S1). Cumulative d’ from that endpoint specifies the means of the distributions S1..S7.
The criterion location can be found from any pair using \(\lambda = -z(FA)\).
- For S1 and S2, the “FA” = p(Yes|S1) =0.02, z(0.02)= -2.05, so \(\lambda = 2.05\) above \(\mu_{S1} = 0\).
- For S3 and S2, the “FA” = p(Yes|S2) =0.06, z(0.06)= -1.56, so \(\lambda = 1.56\) above \(\mu_{S2}\), which is \(\text{cumulative } d'_{S2} = 0.50\), so the criterion location is at 0.5+1.56 = 2.06.
- For S5 and S6, the FA = p(Yes|S5) = 0.8, z(0.8) = 0.85. Since \(\mu_5 = 2.90\), criterion location is 2.9 + -0.85 = 2.05.
If we plot the p(“yes”) or cumulative d’ as a function of the stimulus, we get a psychometric function.
Threshold:
- stimulus leading to a certain level of d’ or
- stimulus leading to a certain level p(yes).
\(d'\)-based definition is response-bias free, while p(yes) definition only depends on ‘hits’, so in not free of response bias.
Classification with a standard
- A standard line of length \(L_5\) is presented, followed by a test stimulus of \(L_1\) to \(L_7\).
Participant response whether the second line was ‘longer’ or ‘shorter’ than the standard.
- PSE is defined as the stimulus yielding p(‘longer’) = 0.5.
- This is a measure of bias.
- JND is the half of the difference in stimulus levels yielding p(longer) = 0.75 and p(longer) = 0.25
- This is a measure of sensitivity: stimulus difference yielding a d’ = 0.674
round(qnorm(0.25),3)## [1] -0.674Speech identification
- Lengthening of voice-onset time (VOT) produces stimuli on a /ga/ /ka/ continuum
- Stimuli with different VOT levels are presented in random order
- Participant responds: /ka/ or /ga/
- speech researchers call this ‘identification’, but this is still a one-dimensional classification design.
Typically interested in the ‘category boundary’, which is, like the PSE, the stimulus on which p(/ga/) = 0.5.
## Stimulus ka ga zKa dPrime cumulativeDp
## 1 1 0.02 0.98 -2.054 NA NA
## 2 2 0.06 0.94 -1.555 0.499 0.499
## 3 3 0.15 0.85 -1.036 0.519 1.018
## 4 4 0.40 0.60 -0.253 0.783 1.801
## 5 5 0.80 0.20 0.842 1.095 2.896
## 6 6 0.92 0.08 1.405 0.563 3.459
## 7 7 0.96 0.04 1.751 0.346 3.805
- Can also present a ‘standard’ /ga/ sound, followed by the same question: /ka/ or /ga/?
- This typically just moves the criterion location, but does not affect sensitivity.
Experiments with More Than 2 Responses
- 4 pure-tone stimuli are presented and participants respond with 1 of 4 response buttons.
- Videos are shown with one of 4 emotional levels, participants respond with 1 of 4 responses on an ‘emotion’ scale
Experimental data yield a confusion matrix:
Stimuli = c(1,2,3,4)
Resp2 = c(11, 12, 21, 0)
Resp1 = c(39, 17, 0, 0)
Resp3 = c(0, 21, 12, 17)
Resp4 = c(0, 0, 17, 33)
intensity = data.frame(Stimuli, Resp1, Resp2, Resp3, Resp4)
intensity## Stimuli Resp1 Resp2 Resp3 Resp4
## 1 1 39 11 0 0
## 2 2 17 12 21 0
## 3 3 0 21 12 17
## 4 4 0 0 17 33cat(' proportions\n')## proportionsintensityProportions = intensity[,2:5]/apply(intensity[,2:5], 1, sum)
data.frame(Stimuli, intensityProportions)## Stimuli Resp1 Resp2 Resp3 Resp4
## 1 1 0.78 0.22 0.00 0.00
## 2 2 0.34 0.24 0.42 0.00
## 3 3 0.00 0.42 0.24 0.34
## 4 4 0.00 0.00 0.34 0.66cat('cumulative proportions\n')## cumulative proportionsintensityCumulativeProportions = t(apply(intensityProportions,1,cumsum))
data.frame(Stimuli, intensityCumulativeProportions)## Stimuli Resp1 Resp2 Resp3 Resp4
## 1 1 0.78 1.00 1.00 1
## 2 2 0.34 0.58 1.00 1
## 3 3 0.00 0.42 0.66 1
## 4 4 0.00 0.00 0.34 1cat('z-transforms\n')## z-transformsintensityZ = round(qnorm(intensityCumulativeProportions),3)
intensityZ## Resp1 Resp2 Resp3 Resp4
## [1,] 0.772 Inf Inf Inf
## [2,] -0.412 0.202 Inf Inf
## [3,] -Inf -0.202 0.412 Inf
## [4,] -Inf -Inf -0.412 Infcat('d-primes\n')## d-primesdP_1vs2 = intensityZ[1,]-intensityZ[2,]
dP_2vs3 = intensityZ[2,]-intensityZ[3,]
dP_3vs4 = intensityZ[3,]-intensityZ[4,]
dPrimes = rbind(dP_1vs2, dP_2vs3, dP_3vs4)
rownames(dPrimes) = c('d(1,2)', 'd(2,3)', 'd(3,4)')
(dPrimes)## Resp1 Resp2 Resp3 Resp4
## d(1,2) 1.184 Inf NaN NaN
## d(2,3) Inf 0.404 Inf NaN
## d(3,4) NaN Inf 0.824 NaNcat('Can estimate d(1,3) using d(1,2) + d(2,3) = -1.18 + - 0.404 \n')## Can estimate d(1,3) using d(1,2) + d(2,3) = -1.18 + - 0.4041.18 + 0.404 ## [1] 1.584cat('# Total d\': d(1,2)+d(2,3)+d(3,4) = -1.18 + - 0.404 + -0.824\n')## # Total d': d(1,2)+d(2,3)+d(3,4) = -1.18 + - 0.404 + -0.824 1.18 + 0.404 + 0.824## [1] 2.408As long as we have at least one response for which the rate is non-0 or non-1 for a pair of stimuli, we can calculate their d’. If not, then we can use the indirect method.
Psychological scaling of the 4 stimuli based on cumulative d’:
x = seq(-3,6,len = 100)
plot(x, dnorm(x = x, mean = 0, sd = 1), col = 1, type = "l", ylab = c(''))
lines(x, dnorm(x = x, mean = 1.18, sd = 1), col = 1, lty =2, type = "l")
lines(x, dnorm(x = x, mean = 1.18+0.404, sd = 1), col = 1, lty =3, type = "l")
lines(x, dnorm(x = x, mean = 1.18+0.404+0.824, sd = 1), col = 1, lty =4, type = "l")
legend("topleft", legend = c('S1', 'S2', "S3", "S4"), lty = c(1,2,3,4), bty = "n")
Non-parametric measure: ‘Mean category scaling’
Most common analysis method is to calculate the mean response rating for each stimulus.
intensityProportions## Resp1 Resp2 Resp3 Resp4
## 1 0.78 0.22 0.00 0.00
## 2 0.34 0.24 0.42 0.00
## 3 0.00 0.42 0.24 0.34
## 4 0.00 0.00 0.34 0.66plot(Stimuli,as.matrix(intensityProportions)%*%c(1,2,3,4), type = "b",
xlim = c(1,4), ylim = c(1,4), ylab = 'Average response rating')
abline(a = 0, b = 1, lty = 2)
Since this only takes into account responses for one stimulus, it is a measure of response bias.
The slope, or difference, between average response ratings, is often used as a measure of sensitivity. However, it is flawed as well.
\[\text{Mean rating for S1} = 1\times P(\text{"1"}|S1) + 2\times P(\text{"2"}|S1)\\ \text{considering S2 as 'stimulus' and S1 as noise, then P("1"|S1) = CR = (1-FA) and FA = P("2"|S1), so }\\ \text{Mean rating for S1} = 1 \times (1-F) + 2 \times (FA) = 1+F\]
\[\text{Mean rating for S2} = 1\times P(\text{"1"}|S2) + 2\times P(\text{"2"}|S2)\\ \text{still considering S2 as 'stimulus' and S1 as noise, P("1"|S2) = Miss = (1-H) and P("2"|S2) = H, so}\\ \text{Mean rating for S2} = 1 \times (1-H) + 2 \times (H) = 1+H\]
\[ \text{Mean rating S2 - Mean rating S1} = 1+H - (1+F) = H - F \text{, which is not bias-free.} \]
If the stimuli are not uni-dimensional, they cannot be represented on a single axis, and will live instead in a multidimensional space. The field of multidimensional scaling seeks to analyze such situations.
Chapter 6: Multi-dimensional stimuli
This chapter deals with detection and discrimination of compound stimuli, i.e., those composed of two or more distinct perceptual components and serves as an introduction to Multidimensionsal Detection Theory.
Examples of compound stimuli: * Is an aircraft a plane or a helicopter? * Detecting a light-beep signal or vibrating cell phone ringtone.
Typical experimental question of interest: how much is detection of a multidimensional stimulus better than detection of the individual components? Signal detection theory provides a quantitative framework for comparing these situations.
Bivariate distributions
3D and 2D representations of the joint distribution of two variables, loudness and brightness, produced when a tone and a light are presented together.
- height of 3D graph shows the likelihood of a particular value of loudness and brightness (located on the 2D surface).
- highest point lies over the means \(\mu_x\) and \(\mu_y\).
- The 2D version shows cross-sections of the 3D plot at points of equal likelihood. Diameters represent multiples of standard deviations.
- When variablility in unequal in the two dimentions, equal likelihood contours are elliptical.
- The criterion is defined by a line or curve in space termed decision boundary.
Here, the variability in the two dimensions differs, generating elliptical 2D representation.
- Volume to the right of a line boundary in the joint distribution is the same as the area to the right of the criterion in the marginal distribution.
- Margination distributions are projections of the bivariate distributions onto a single axis
- The height of the marginal distribution at some value of x is the sum of heights of joint distribution at that x-value across all values of y.
- The marginal distribution for y has the same \(\sigma\) as \(\sigma_y\) in the 2D plot.
The axes above are perpendicular and the ellipse long and short axes are oriented parallel to the main axes. This represents perceptually independent variables.
Perceptually dependent variables are shown using non-perpendicular axes and/or elliptical distributions oriented at an angle to the main axes, as shown below:
Product rule
In the maximum rule, an observer says “yes” if both loudness and brightness are above some criteria.
- The “yes” rate is shown in the hashed upper-right square
- The product rule is used to calculate this rate (intersection of two areas): The proportion of the y-marginal distribution above the horizontal criterion multiplied by the proportion of the x-marginal distribution to the right of the vertical criterion.
- The “L” shaped “no” space (“miss” rate) is the complement (1 - “yes” rate).
The min rule occurs when the observer says “yes” if either the tone or the light are above the criterion.
- The “no” rate can be obtained by the product rule (\(\Phi(z_x)\Phi(z_y)\))
- the “yes” rate is 1 - \(\Phi(z_x)\Phi(z_y)\)
Arbitrary decision axes
So far, the decision axis has always been either one or the other variable, but it does not have to be. Below, you see a decision axis as a line at 45 deg to both axes. This corresponds to an observer that adds the values of loudness and brightness and compares the sum to their decision criterion to decide if the tone-light pair was presented.
- Increasing values of x and y correspond to higher values on the decision axis, decreasing values of x and y correspond to lower values, with (0,0) projecting to 0.
- The decision boundary is always perpendicular to the decision axis.
- The probability of “yes” response is the area under the marginal distribution to the right of the boundary.
Compound detection problem
In multidimensional tasks, observers may adopt a number of possible response strategies. To compare decision strategies, we can use the proportion correct of an unbiased observer. For a yes/no task, \(p(c)_{max} = \Phi(d'/2)\).
Decisional separability
This strategy consists of ignoring one component and basing one’s decision on only one component.
Assuming equal presentation probabilities, \(p(c) = 0.5*(H + CR) = 0.5*(\Phi(d'_x-k_x) + \Phi(k_x))\). An unbiased observer would place the criterion at \(d'_x/2\), so \(p(c)_{max} = \Phi(d'_x/2)\). If \(d'_x = 1\), \(p(c)_{max} = 0.69\).
The Max and Min rules with non-parametric assumptions
Notice that the min rule has a much higher FA and Hit rate than the max rule, although both lead to the same p(c).
Calculating the compound H and FA rates from SDT using the product rule, we can obtain expressions for the hits and false alarms of compound stimuli wrt any general criterion location, k:
\[\begin{align} \text{maximum rule: } &H_{xy} = \Phi(d'-k)^2 \text{ and }\\ &F_{xy} = \Phi(-k)^2\\ \text{minimum rule: } &H_{xy} = 1-\Phi(-d'+k)^2 \text{ and }\\ &F_{xy} = 1-\Phi(k)^2 \end{align}\]Diagonal decision (optimal rule):
Using the Pythagorean Theorem, we can derive distance between the null and compound stimulus on the decision axis:
\[ d'_{compound} = \sqrt{{d'_x}^2+{d'_y}^2 } \]
If both \(d'_x = d'_y\), then \(d'_{compound} = \sqrt{2{d'}^2} = \sqrt{2} d'\).
For d’ = 1, \(p(c)_{max} = \Phi(d'_{compound}/2) = \Phi(\sqrt(2)d'/2) = \Phi(\sqrt(2)*0.5) = 0.76\). Clearly this strategy yields the highest proportion correct (for an unbiased observer).
Expressions for the ROC curve can be obtained from the decision axis..
\[ H_{xy} = \Phi(\sqrt{2}d'-k) \\ F_{xy} = \Phi(k) \]
This figure compares the ROC curves for these different decision strategies. The min and max rule have similar sensitivity, both of which are worse that the optimal rule. This improvement in accuracy for the optimal rule is because this strategy combines information before making a decision about the individual components.
Chapter 7: 2AFC or ‘Forced-choice designs’
MacMillan and Creelman classify 2AFC as a comparison designs which require a comparison between two stimulus presentations and the possible stimuli are only of 2 classes.
There is nothing new about the ‘forced choice’ part. The title of the task design is not very diagnostic. The distinction between Yes/No and 2AFC is that both stimulus classes are presented on every trial, in random (spatial or temporal) order. The observer reports on the order of the stimuli (signal in the first or second interval).
Word recognition example:
- Participants read list of words to remember (study phase).
- Test phase: two words presented on each trial, one above the other.
- Response: “top is old” or “bottom is old”.
Data for such an experiment can be represented as
The 2AFC model:
- \(X_s\) is a random variable for event from signal distribution
- \(X_n\) is a random variable for event from noise distribution
The unbiased decision rule is to pick the interval with the larger value.
- Compute \(X_{1} \text{and } X_2\) and their difference.
- If \(X_{1}- X_2>0\), say “old on top”.
- If \(X_{1}- X_2<0\), say “old on bottom.
For \(X_a \sim N(\mu_a, \sigma_a^2)\) and \(X_b \sim N(\mu_b, \sigma_b^2)\), if the observations are independent, then \(Y_{a-b} = X_a -X_b\sim N(\mu_a-\mu_b, \sigma_a^2+ \sigma_b^2)\).
\[ \text{For <Old, New> trials, } Y_{<Old, New>} = X_1 - X_2 = X_s - X_n \sim N(\mu_s, 1+\sigma_s^2)\\ \text{For <New, Old> trials, } Y_{<New, Old>} = X_1 - X_2 = X_n - X_s \sim N(-\mu_s, 1+\sigma_s^2)\]
Best performance is obtained by an unbiased observer (shown in blue) that compares \(X_1- X_2\) to 0. However, the observer could have a ‘position bias’ (e.g. shown in green), whereby
- If \(X_{1}- X_2>=\lambda_{FC}\), say “old on top”.
- If \(X_{1}- X_2<\lambda_{FC}\), say “old on bottom.
Notice that \(X_1-X_2\) and \(X_2 - X_1\) have the same variance (assuming no interactions between position and stimulus). This means that we can use the equal variance model equations for d’ here even when \(X_s\) and \(X_n\) do not have equal variance.
\[ d'_{2AFC} = \frac{\text{difference in means}}{\text{common standard deviation}} = \frac{2\mu_s}{\sqrt{1+\sigma_s^2}} \]
Under the equal variance mode, \(\mu_s\) = d’ and \(\sigma_s^2 = 1\).
\[ d'_{2AFC} = \frac{2d'}{\sqrt{1+1}} = \sqrt{2}d'\]
The 2AFC model using 2-D representation:
We can also represent the same situation using 2 axes. The internal value generated by the top word is plotted on the x- axis and bottom word on the y-axis. The plot shows the two distributions associated with \(<Old,New>\) trials (bottom right) and \(<New,Old>\) trials, top left. For each trial, the observer responds depending on which axis (x or y) is closer, i.e., uses the positive diagonal as a boundary.
The distance between the two distribution means is $ d’_{2AFC} = = d’$, as we saw before.
So, \[ d'_{yes/no} = \frac{z(H) - z(FA)}{\sqrt{2}} \]
Let’s say “signal” is \(<Old, New>\) and noise is \(<New, Old>\), then \[ H = P("\text{old on top}"|<Old, New>) = A\\ FA = P("\text{old on top}"|<New,Old>) = C\\\]
(H = 16/25)## [1] 0.64(FA = 7/25)## [1] 0.28\[ d'_{2AFC} = z(A) - z(C)\\ \lambda_{2AFC} = -0.5 * (z(A) + z(C))\\ log(\beta)_{2AFC} = -0.5 * (z^2(C) - z^2(A))\\ \]
(dprime2AFC = qnorm(H) - qnorm(FA))## [1] 0.9413003i.e., the same formulas as with Yes/No equal variance model.
\(d'_{YesNo} = d'_{2AFC}/\sqrt(2)\)
(dprimeYesNo = dprime2AFC/sqrt(2))## [1] 0.6655998We can also recast these in terms of P(c), using
\[ PC_{<Old, New>} = P(\text{old on top }|<Old, New>) = A = H \\
PC_{<New, Old>} = P(\text{old on bottom}|<New, Old>) = D = 1-C = 1-FA\].
Since \(z(x) = -z(1-x)\), \(-z(C) = +z(1-C) = +z(D)\), we can replace \(z(A)\) with \(-z(PC_{New,Old})\) and \(z(A)\) with \(-z(PC_{Old,New})\).
\[ d'_{2AFC} = z(PC_{Old,New>}) + z(PC_{<New,Old>})\\ \lambda_{2AFC} = 0.5(z(PC_{<New,Old>}) - z(PC_{Old, New}))\\ log(\beta)_{2AFC} = 0.5(z^2(PC_{<New,Old>}) - z^2(PC_{Old, New}))\\ \]
# calculate values for the example
PCnewold = 18/25
PColdnew = 16/25
# qnorm(x) = -qnorm(1-x)
(dprime2AFC_2 = qnorm(PColdnew) + qnorm(PCnewold))## [1] 0.9413003(lambda2AFC = 0.5*(qnorm(PCnewold) - qnorm(PColdnew))) ## [1] 0.1121914# lambda is slightly positive, so bias to say 'old at the bottom'2AFC and Proportion correct and 2AFC Area theorem
Overall \(P_C\) is a combination of correct-response probabilities for each order, \(PC_{Old, New}\) and \(PC_{New, Old}\), weighted by the response probabilities of each trial type.
\[ P_C = P_{Old,New} PC_{Old,New}+ P_{New, Old} PC_{New, Old}\] For an unbiased observer, \(PC_{<old,new>} = PC_{<new,old>}\) and \(\lambda\) should be \(0\) when probabilities of each order are equal. When they are unequal, \(\lambda_{FC}\) should shift to increase PC for the most frequent type.
For an unbiased observer, the 2AFC Area Theorem states that \[ p(c)_{2AFC} = \text{Area under the Yes/No isosensitivity curve} = A_z\] Under the equal variance model, the unbiased observer places their criterion on the minor diagonal and proportion correct at this point is found by
\[ p(c)_{max} = \Phi( \frac{z(H) - z(F)}{2} )\] In the Yes/No task, \[ p(c)_{max, YN} = \Phi(d'/ 2)\]
In the 2AFC task, \[ p(c)_{max, 2AFC} = \Phi(\frac{\sqrt{2}d'}{2}) = \Phi(\frac{d'}{\sqrt{2}}) \] According to the Area theorem,
\[ p(c)_{max, 2AFC} = \Phi(\frac{d'}{\sqrt{2}}) = A_z\]
We can solve the above for d’, to give
\[ d' = \sqrt(2) * z(p(c)_{max,2AFC})\]
# for the example above, p(c) = 0.68.
(dprime = sqrt(2)*qnorm(0.5*(16/25 + 18/25)))## [1] 0.6614262AFC vs Yes/No
- 2AFC discourages bias for a particular stimulus.
- If unbiased, p(c) is generally a good measure
- Caveat: observers are not always unbiased in 2AFC! If biased, p(c) is not good.
- 2AFC performance levels are higher than in Yes/No
- Studies don’t always find the \(\sqrt(2)d'\) relationship, but there are many reasons why that can happen, without the need to reject SDT.