Stop Signal Reaction Time


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KathyB
KathyB
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Hi,
I'm using the Stop Signal Task and noticed that in the code for the stop signal reaction time (SSRT) it says: "for participants who inhibited significantly more or less than 50% of the time
the subtraction method to calculate SSRT (see expressions.SSRT) cannot be used". What can I do with these participants? Is there another way I can calculate SSRT for them? Or do I need to exclude them?

Alternatively, if I just want an overall measure of inhibitory control, am I best to just use the z-score and corresponding p value and not worry about SSRT?

Any advice appreciated.
Thanks,
Kath
Dave
Dave
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KathyB - Monday, April 10, 2017
Hi,
I'm using the Stop Signal Task and noticed that in the code for the stop signal reaction time (SSRT) it says: "for participants who inhibited significantly more or less than 50% of the time
the subtraction method to calculate SSRT (see expressions.SSRT) cannot be used". What can I do with these participants? Is there another way I can calculate SSRT for them? Or do I need to exclude them?

Alternatively, if I just want an overall measure of inhibitory control, am I best to just use the z-score and corresponding p value and not worry about SSRT?

Any advice appreciated.
Thanks,
Kath

Verbruggen et al. reference Logan (1994) and Logan & Cowan (1984) for alternative calculation methods.

KathyB
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Dave - Monday, April 10, 2017
KathyB - Monday, April 10, 2017

Hi,

I'm using the Stop Signal Task and noticed that in the code for the stop signal reaction time (SSRT) it says: "for participants who inhibited significantly more or less than 50% of the time the subtraction method to calculate SSRT (see expressions.SSRT) cannot be used". What can I do with these participants? Is there another way I can calculate SSRT for them? Or do I need to exclude them?

Alternatively, if I just want an overall measure of inhibitory control, am I best to just use the z-score and corresponding p value and not worry about SSRT?

Any advice appreciated.
Thanks,
Kath

Verbruggen et al. reference Logan (1994) and Logan & Cowan (1984) for alternative calculation methods.

Hi Dave,
Thanks for the response. I would like to report the z-score. I have a question about the sign of the z-score generated by the program: does a negative z-score mean that the person inhibited more than 50% of the time or less than 50% of the time? My understanding is that the higher the z-score, the more responses made to stop signals (i.e. more errors). Is this correct?
Thank you,
Kath




Dave
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KathyB - Sunday, May 7, 2017
Dave - Monday, April 10, 2017
KathyB - Monday, April 10, 2017

Hi,

I'm using the Stop Signal Task and noticed that in the code for the stop signal reaction time (SSRT) it says: "for participants who inhibited significantly more or less than 50% of the time the subtraction method to calculate SSRT (see expressions.SSRT) cannot be used". What can I do with these participants? Is there another way I can calculate SSRT for them? Or do I need to exclude them?

Alternatively, if I just want an overall measure of inhibitory control, am I best to just use the z-score and corresponding p value and not worry about SSRT?

Any advice appreciated.
Thanks,
Kath

Verbruggen et al. reference Logan (1994) and Logan & Cowan (1984) for alternative calculation methods.

Hi Dave,
Thanks for the response. I would like to report the z-score. I have a question about the sign of the z-score generated by the program: does a negative z-score mean that the person inhibited more than 50% of the time or less than 50% of the time? My understanding is that the higher the z-score, the more responses made to stop signals (i.e. more errors). Is this correct?
Thank you,
Kath




> does a negative z-score mean that the person inhibited more than 50% of the time or less than 50% of the time

So, the z-score is calculated like this:
 
z = (X -Np)/sqrt(Npq)

where

X is the observed number of signal-respond responses,
N is the total number of stop-signal trials,
p is the "target" probability (.5), i.e. 50%,
and q is 1 - p (also .5).
sqrt(Npq) is the standard deviation of the binomial distribution.

Example:
If there were 64 stop trials and 28 of them were signal-respond responses (i.e. failed inhibitions), the z score would be

z = (28 - (64*0.5)) / sqrt(64*0.5*0.5) = (28 - 32) / sqrt(16) = -4 / 4 = -1

i.e., negative.

The parallel case, if there were 64 stop trials and 36 of them were signal-respond responses, the z score would be

z = (36 - (64*0.5)) / sqrt(64*0.5*0.5) = (36 - 32) / sqrt(16) = 4 / 4 = 1

i.e. positive.

Hope this clarifies.
KathyB
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Dave - Monday, May 8, 2017
KathyB - Sunday, May 7, 2017
Dave - Monday, April 10, 2017
KathyB - Monday, April 10, 2017

Hi,

I'm using the Stop Signal Task and noticed that in the code for the stop signal reaction time (SSRT) it says: "for participants who inhibited significantly more or less than 50% of the time the subtraction method to calculate SSRT (see expressions.SSRT) cannot be used". What can I do with these participants? Is there another way I can calculate SSRT for them? Or do I need to exclude them?

Alternatively, if I just want an overall measure of inhibitory control, am I best to just use the z-score and corresponding p value and not worry about SSRT?

Any advice appreciated.
Thanks,
Kath

Verbruggen et al. reference Logan (1994) and Logan & Cowan (1984) for alternative calculation methods.

Hi Dave,
Thanks for the response. I would like to report the z-score. I have a question about the sign of the z-score generated by the program: does a negative z-score mean that the person inhibited more than 50% of the time or less than 50% of the time? My understanding is that the higher the z-score, the more responses made to stop signals (i.e. more errors). Is this correct?
Thank you,
Kath




> does a negative z-score mean that the person inhibited more than 50% of the time or less than 50% of the time

So, the z-score is calculated like this:
 
z = (X -Np)/sqrt(Npq)

where

X is the observed number of signal-respond responses,
N is the total number of stop-signal trials,
p is the "target" probability (.5), i.e. 50%,
and q is 1 - p (also .5).
sqrt(Npq) is the standard deviation of the binomial distribution.

Example:
If there were 64 stop trials and 28 of them were signal-respond responses (i.e. failed inhibitions), the z score would be

z = (28 - (64*0.5)) / sqrt(64*0.5*0.5) = (28 - 32) / sqrt(16) = -4 / 4 = -1

i.e., negative.

The parallel case, if there were 64 stop trials and 36 of them were signal-respond responses, the z score would be

z = (36 - (64*0.5)) / sqrt(64*0.5*0.5) = (36 - 32) / sqrt(16) = 4 / 4 = 1

i.e. positive.

Hope this clarifies.

Very helpful, thank you.
Kath
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