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## groups in inquisit 6 (updating tutorial for simple IAT)

https://www.millisecond.com/forums/Topic29505.aspx

 By RobinW - 7/29/2020 Hello,working through the tutorials I noticed that the tutorial for the simple IAT is not working for Inquisit 6 (apparently it was written for Inquisit 3). I managed to adapt it to inquisit 6, but I can't figure out how to make groups like is done in the tutorial: / group = (1 of 2) (block2=fi_practice, block3=incompatible, block4=if_practice, block5=compatible)/ group = (2 of 2) (block2=if_practice, block3=compatible, block4=fi_practice, block5=incompatible)How can I achieve this in Inquisit 6?Thanks for your help!Robin By Dave - 7/29/2020 +x RobinW - 7/29/2020Hello,working through the tutorials I noticed that the tutorial for the simple IAT is not working for Inquisit 6 (apparently it was written for Inquisit 3). I managed to adapt it to inquisit 6, but I can't figure out how to make groups like is done in the tutorial: / group = (1 of 2) (block2=fi_practice, block3=incompatible, block4=if_practice, block5=compatible)/ group = (2 of 2) (block2=if_practice, block3=compatible, block4=fi_practice, block5=incompatible)How can I achieve this in Inquisit 6?Thanks for your help!RobinThe element does not exist anymore under Inquisit 6. The way to do this then is to define two elements with the respective block orders:/ groups = (1 of 2)/ blocks = [...; 2=fi_practice; 3=incompatible; 4=if_practice; 5=compatible; ...]/ groups = (2 of 2)/ blocks = [...; 2=if_practice; 3=compatible, 4=fi_practice, 5=incompatible; ...] By RobinW - 7/29/2020 +x Dave - 7/29/2020+x RobinW - 7/29/2020Hello,working through the tutorials I noticed that the tutorial for the simple IAT is not working for Inquisit 6 (apparently it was written for Inquisit 3). I managed to adapt it to inquisit 6, but I can't figure out how to make groups like is done in the tutorial: / group = (1 of 2) (block2=fi_practice, block3=incompatible, block4=if_practice, block5=compatible)/ group = (2 of 2) (block2=if_practice, block3=compatible, block4=fi_practice, block5=incompatible)How can I achieve this in Inquisit 6?Thanks for your help!RobinThe element does not exist anymore under Inquisit 6. The way to do this then is to define two elements with the respective block orders:/ groups = (1 of 2)/ blocks = [...; 2=fi_practice; 3=incompatible; 4=if_practice; 5=compatible; ...]/ groups = (2 of 2)/ blocks = [...; 2=if_practice; 3=compatible, 4=fi_practice, 5=incompatible; ...]Thanks four your quick response! In an experiment of mine I'm using a different approach to assigning subjects to one of two groups within a survey:/ pages = [1=questions1; 2=questions2; 3=noreplace(imaginationtastk1, imaginationtastk1); 4=questions3]/ responsefontstyle = ("Verdana", -12, false, false, false, false, 5, 0)/ itemfontstyle = ("Verdana", -13, false, false, false, false, 5, 0)/ itemspacing = 2%/ showpagenumbers = falseI just changed that to the 'group'-approach, but I just realized that I would have to enter a group for every subject in this case. Does the approach I take in the example above have any disadvantages? With this approach I wouldn't be able to control randomization but Inquisit would approximately assign 50% to one and 50% to the other condition, right? By Dave - 7/29/2020 +x RobinW - 7/29/2020+x Dave - 7/29/2020+x RobinW - 7/29/2020Hello,working through the tutorials I noticed that the tutorial for the simple IAT is not working for Inquisit 6 (apparently it was written for Inquisit 3). I managed to adapt it to inquisit 6, but I can't figure out how to make groups like is done in the tutorial: / group = (1 of 2) (block2=fi_practice, block3=incompatible, block4=if_practice, block5=compatible)/ group = (2 of 2) (block2=if_practice, block3=compatible, block4=fi_practice, block5=incompatible)How can I achieve this in Inquisit 6?Thanks for your help!RobinThe element does not exist anymore under Inquisit 6. The way to do this then is to define two elements with the respective block orders:/ groups = (1 of 2)/ blocks = [...; 2=fi_practice; 3=incompatible; 4=if_practice; 5=compatible; ...]/ groups = (2 of 2)/ blocks = [...; 2=if_practice; 3=compatible, 4=fi_practice, 5=incompatible; ...]Thanks four your quick response! In an experiment of mine I'm using a different approach to assigning subjects to one of two groups within a survey:/ pages = [1=questions1; 2=questions2; 3=noreplace(imaginationtastk1, imaginationtastk1); 4=questions3]/ responsefontstyle = ("Verdana", -12, false, false, false, false, 5, 0)/ itemfontstyle = ("Verdana", -13, false, false, false, false, 5, 0)/ itemspacing = 2%/ showpagenumbers = falseI just changed that to the 'group'-approach, but I just realized that I would have to enter a group for every subject in this case. Does the approach I take in the example above have any disadvantages? With this approach I wouldn't be able to control randomization but Inquisit would approximately assign 50% to one and 50% to the other condition, right?I don't see any between-subjects condition in that survey code you posted. I see that you are sampling one random page here3=noreplace(imaginationtastk1, imaginationtastk1);but the noreplace() pool seems to contain the same, single page (imaginationtastk1) twice.Other than that, no, this makes no guarantee that the distribution would be 50/50 across subjects, although just by virtue of random sampling theory it should be close to that given a large number of participants. Ultimately, it's like a coin toss: If you perform 100 coin tosses, there's no guarantee that it'll come up 50 times heads and 50 times tails -- each coin toss is an independent event. By RobinW - 7/29/2020 +x Dave - 7/29/2020+x RobinW - 7/29/2020+x Dave - 7/29/2020+x RobinW - 7/29/2020Hello,working through the tutorials I noticed that the tutorial for the simple IAT is not working for Inquisit 6 (apparently it was written for Inquisit 3). I managed to adapt it to inquisit 6, but I can't figure out how to make groups like is done in the tutorial: / group = (1 of 2) (block2=fi_practice, block3=incompatible, block4=if_practice, block5=compatible)/ group = (2 of 2) (block2=if_practice, block3=compatible, block4=fi_practice, block5=incompatible)How can I achieve this in Inquisit 6?Thanks for your help!RobinThe element does not exist anymore under Inquisit 6. The way to do this then is to define two elements with the respective block orders:/ groups = (1 of 2)/ blocks = [...; 2=fi_practice; 3=incompatible; 4=if_practice; 5=compatible; ...]/ groups = (2 of 2)/ blocks = [...; 2=if_practice; 3=compatible, 4=fi_practice, 5=incompatible; ...]Thanks four your quick response! In an experiment of mine I'm using a different approach to assigning subjects to one of two groups within a survey:/ pages = [1=questions1; 2=questions2; 3=noreplace(imaginationtastk1, imaginationtastk1); 4=questions3]/ responsefontstyle = ("Verdana", -12, false, false, false, false, 5, 0)/ itemfontstyle = ("Verdana", -13, false, false, false, false, 5, 0)/ itemspacing = 2%/ showpagenumbers = falseI just changed that to the 'group'-approach, but I just realized that I would have to enter a group for every subject in this case. Does the approach I take in the example above have any disadvantages? With this approach I wouldn't be able to control randomization but Inquisit would approximately assign 50% to one and 50% to the other condition, right?I don't see any between-subjects condition in that survey code you posted. I see that you are sampling one random page here3=noreplace(imaginationtastk1, imaginationtastk1);but the noreplace() pool seems to contain the same, single page (imaginationtastk1) twice.Other than that, no, this makes no guarantee that the distribution would be 50/50 across subjects, although just by virtue of random sampling theory it should be close to that given a large number of participants. Ultimately, it's like a coin toss: If you perform 100 coin tosses, there's no guarantee that it'll come up 50 times heads and 50 times tails -- each coin toss is an independent event.Sorry, I mistyped. It should have read:3=noreplace(imaginationtastk1, imaginationtastk2);Pertaining to the second part: That's why I wrote "approximately". I'm not that experienced with conducting experiments, but if I choose the 'grouping'-approach by / groups = (1 of 2)I myself would have to use some sort of randomization procedure that would not necessarily guarantee a 50/50 distribution, or is there some way to make sure there is? Eventually I want to use my experiment online, where I wouldn't be able to assign participants to groups myself. Could you explain to me, how randomization would work in this case with Inquisit 6? Or point me to resources, where this is explained?Thank you very much! By Dave - 7/29/2020 +x RobinW - 7/29/2020+x Dave - 7/29/2020+x RobinW - 7/29/2020+x Dave - 7/29/2020+x RobinW - 7/29/2020Hello,working through the tutorials I noticed that the tutorial for the simple IAT is not working for Inquisit 6 (apparently it was written for Inquisit 3). I managed to adapt it to inquisit 6, but I can't figure out how to make groups like is done in the tutorial: / group = (1 of 2) (block2=fi_practice, block3=incompatible, block4=if_practice, block5=compatible)/ group = (2 of 2) (block2=if_practice, block3=compatible, block4=fi_practice, block5=incompatible)How can I achieve this in Inquisit 6?Thanks for your help!RobinThe element does not exist anymore under Inquisit 6. The way to do this then is to define two elements with the respective block orders:/ groups = (1 of 2)/ blocks = [...; 2=fi_practice; 3=incompatible; 4=if_practice; 5=compatible; ...]/ groups = (2 of 2)/ blocks = [...; 2=if_practice; 3=compatible, 4=fi_practice, 5=incompatible; ...]Thanks four your quick response! In an experiment of mine I'm using a different approach to assigning subjects to one of two groups within a survey:/ pages = [1=questions1; 2=questions2; 3=noreplace(imaginationtastk1, imaginationtastk1); 4=questions3]/ responsefontstyle = ("Verdana", -12, false, false, false, false, 5, 0)/ itemfontstyle = ("Verdana", -13, false, false, false, false, 5, 0)/ itemspacing = 2%/ showpagenumbers = falseI just changed that to the 'group'-approach, but I just realized that I would have to enter a group for every subject in this case. Does the approach I take in the example above have any disadvantages? With this approach I wouldn't be able to control randomization but Inquisit would approximately assign 50% to one and 50% to the other condition, right?I don't see any between-subjects condition in that survey code you posted. I see that you are sampling one random page here3=noreplace(imaginationtastk1, imaginationtastk1);but the noreplace() pool seems to contain the same, single page (imaginationtastk1) twice.Other than that, no, this makes no guarantee that the distribution would be 50/50 across subjects, although just by virtue of random sampling theory it should be close to that given a large number of participants. Ultimately, it's like a coin toss: If you perform 100 coin tosses, there's no guarantee that it'll come up 50 times heads and 50 times tails -- each coin toss is an independent event.Sorry, I mistyped. It should have read:3=noreplace(imaginationtastk1, imaginationtastk2);Pertaining to the second part: That's why I wrote "approximately". I'm not that experienced with conducting experiments, but if I choose the 'grouping'-approach by / groups = (1 of 2)I myself would have to use some sort of randomization procedure that would not necessarily guarantee a 50/50 distribution, or is there some way to make sure there is? Eventually I want to use my experiment online, where I wouldn't be able to assign participants to groups myself. Could you explain to me, how randomization would work in this case with Inquisit 6? Or point me to resources, where this is explained?Thank you very much! When you upload your script to the web, you can select how group IDs should be generated. To achieve an approximately equal distribution across the two conditions, you should select either sequential or random (without replacement) group IDs and indicate the number of groups as 2.