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How many English words do you know?

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21 messages over 3 pages: 1 2 3  Next >>
kanewai
Triglot
Senior Member
United States
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Speaks: English*, French, Marshallese
Studies: Italian, Spanish

 
 Message 1 of 21
30 January 2014 at 12:03am | IP Logged 
Here's an interesting research project out of Ghent University's Center for Reading
Research:

How many English words do you know? With this test you get a valid estimate of your
English vocabulary size within 4 minutes and you help scientific research.


Go to the test

How many words and nonwords are included in the test?

The master list includes 60,469 words and 304,275 nonwords. From these you get a random
sample of 70 words and 30 nonwords. It's ok to take the test multiple times, as you'll
get a different words and a slight difference in your score.

And here is an interesting tidbit from the FAQ page:

How many words do people know?

This is one of the questions we’d like to answer with our test. However, on the basis
of our experiences with a similar test in Dutch and previous rating studies in English
we estimate that a proficient native speaker will know some 40,000 words of the list
(i.e., 67%). Older people know more words than younger people. The situation is
different for second language speakers. Here, our estimates range from 6,000 words
(10%) for a medium proficiency speaker to 20,000 words (33%) for a high-proficiency
speaker.

Edited by kanewai on 30 January 2014 at 12:40am

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Chav
Diglot
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Speaks: Lowland Scots, English*
Studies: Dutch, Polish

 
 Message 2 of 21
30 January 2014 at 12:51am | IP Logged 
93%

I noticed a fair number of science/technical terms in the set I got.
1 person has voted this message useful





Iversen
Super Polyglot
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Denmark
berejst.dk
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 Message 3 of 21
30 January 2014 at 1:30am | IP Logged 
Somewhat lower for me: 91% minus 7% (and a couple of 'no' answers to words I understood, but thought I just had guessed). This would be around 50.000 words, and based on my own estimates (based on dictionary samples) I would have counted on max. 40.000 so this new tests is somewhat more lenient - even though it caught me out on some non-words which looked liked something I know ("umpish"/"impish"). It would be interesting to see similar tests for other languages.

EDIT: Now I also did the Dutch test - 51% (but I pressed "no" to jonglerende, leescafé, codeerde and voltmeter, so I actually cheated myself, but still got a higher estimate here than I got from my own tests).



Edited by Iversen on 30 January 2014 at 2:01am

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patrickwilken
Senior Member
Germany
radiant-flux.net
Joined 4533 days ago

1546 posts - 3200 votes 
Studies: German

 
 Message 4 of 21
30 January 2014 at 1:55am | IP Logged 
Is there some simple correspondence between "words" and "word groups". The last test I did of this type estimated that I knew about 24.4k word groups.



Edited by patrickwilken on 30 January 2014 at 2:18am

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Iversen
Super Polyglot
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Studies: Afrikaans, Greek, Norwegian, Russian, Serbian, Icelandic, Latin, Irish, Lowland Scots, Indonesian, Polish, Croatian
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 Message 5 of 21
30 January 2014 at 2:08am | IP Logged 
According to the background information from the research institute the researchers have tried to give basic forms rather than derivations and compound words. I have seen this information somewhere in the English material, but it is unequivocally stated in Dutch here.

Edited by Iversen on 30 January 2014 at 2:11am

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patrickwilken
Senior Member
Germany
radiant-flux.net
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1546 posts - 3200 votes 
Studies: German

 
 Message 6 of 21
30 January 2014 at 2:18am | IP Logged 
You can't (obviously) just take your percentage correct as an estimate of the number of words you know.

To get a measure of sensitivity (how well you can distinguish words from non-words), you have to have two estimates: (1) how likely you are to correctly say yes to real words (hit rate - H); (2) and how how likely you are to say yes to non-real words (false alarm rate - FA).

It is tempting to simply subtract the FA from H to get a sensitivity number, but there are mathematical problems doing this so psychologists generally prefer a measure called d' (d prime).

My score was: Hit rate = 84%, and FA = 0%.

This gives a d' of 5.3, which is basically at ceiling. Normally you'd assume FA is at least 0.01, so my adjusted d' would be 3.4. Iversen your d' was 2.8 which was also very high.

The basic idea of d' prime is to somehow give you a measure of sensitivity that is independent of your response bias - a number that doesn't get affected by whether you guess a lot or not at all.

d's above 2 are an indication of being able to clearly distinguish between to stimuli (here words and non-words). d'=0 means you can't distinguish at all and are essentially just guessing. A d' < 0 would mean you were deliberately answering falsely.

There is a d' prime calculator here: http://memory.psych.mun.ca/models/dprime/

Edited by patrickwilken on 30 January 2014 at 2:22am

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Hekje
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842 posts - 1330 votes 
Speaks: English*, Dutch
Studies: French, Indonesian

 
 Message 7 of 21
30 January 2014 at 3:30am | IP Logged 
Wo-ow, I took the English test first and scored a 74%.

Then I took the Dutch test and scored 70%.

Either my Dutch is very good, or my English is bad and I should be sent home. :-P

Edited by Hekje on 30 January 2014 at 3:33am

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tastyonions
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Speaks: English*, French, Spanish
Studies: Italian

 
 Message 8 of 21
30 January 2014 at 9:51am | IP Logged 
Don't feel bad, I got 73%.

I said "no" to a number of words that were perfectly understandable but I have never in my life seen or heard. Stuff like "instable", for example: "unstable", yes, "instability", yes, but never "instable." Honestly if I saw "instable" on a forum I would probably assume that the writer was a non-native speaker, even though the word is apparently in at least some dictionaries... :-)

No false positives.

Edited by tastyonions on 30 January 2014 at 9:58am



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