RE: lexical decision task: items showing up multiple times

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RE: lexical decision task: items showing up multiple times

lexical decision task: items showing up multiple times

Author	Message
juliank	juliank posted 11 Years Ago Topic Details Share Topic
Group: Forum Members Posts: 9, Visits: 45	I'm using the lexical decision task from the task library, but I'm finding many items are being shown more than once, and some not at all. I'm not sure if this is how it was designed to work, but I need to it only show each item once. Unfortunately though I can't work out why it's showing items multiple times, or how to stop it from doing so. In case it's helpful here's the block I've got: <block lexicalDecisionTask> / bgstim = (reminderleft, reminderright) / trials = [1-254=noreplace(ldtmainitem, ldtcontrol, ldtungrammatical, cphrase)] </block> Thank you in advance, Julian
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Dave	Dave posted 11 Years Ago ANSWER Post Details Share Post
Group: Administrators Posts: 13K, Visits: 108K	/ trials = [1-254=noreplace(ldtmainitem, ldtcontrol, ldtungrammatical, cphrase)] means run 254 trials consisting of ldtmainitem, ldtcontrol, ldtungrammatical and cphrase in equal proportions, which, of course, doesn't work out mathematically (254/4=63.5). You are oversampling some elements, i.e., running more trials of a given type than the trial has items to present -- hence you see repetitions. You need to (a) work out the correct numbers for your trials and (b) properly check when you are sampling from what and how often.
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juliank	juliank posted 11 Years Ago ANSWER Post Details Share Post
Group: Forum Members Posts: 9, Visits: 45	OK. I see - that makes sense. Each set of element has a different set of trials... in this case, is there any way to specify a certain number from each element? I.e. "5 trials from element1", "20 trials from element2" etc... Or would I need to merge items into two equal elements? ldtmainitem, ldtcontrol, ldtungrammatical, cphrase ---> "grammatical, ungrammatical" Julian
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Dave	Dave posted 11 Years Ago ANSWER Post Details Share Post
Group: Administrators Posts: 13K, Visits: 108K	What matters at the <block>-level is the proportion of elements. E.g. / trials = [1-8 = noreplace(a,b)] -> 4 a-trials, 4 b-trials in random order / trials = [1-8 = noreplace(a,a,a,b)] -> 6 a-trials, 2 b-trials in random order You need to work out the ratio you need for your particular case and enter elements into the pool in that ratio.
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juliank	juliank posted 11 Years Ago ANSWER Post Details Share Post
Group: Forum Members Posts: 9, Visits: 45	Thank you. Julian
	Reply Like 19