MySQL: AVG in SELECT

hi, i have a select statement to give me the averages for various mysql columns. for one of the columns, i need to take the average for only those fields where the content !='6' (with 6 being a Flag telling me to exclude it). so i came up with the following select statement, which returns no mysql_error, but does not return the correct figure for the AVG: SELECT AVG(q1), AVG(q2), AVG(q3), AVG(q4), AVG(q5), AVG(q6), AVG(q7), AVG(q8), AVG(q11), AVG(q16), AVG(q18!='6') FROM hr_form2 WHERE Answer != 0 from this i should get a return of something like this, since NONE of my fields in column q18 contain a '6': AVG(q18) = 2.7143 instead, i get this: AVG(q18) = 1.0000 could someone please tell me--is there a logical error in my thinking here? i want to select the AVG from q18 ONLY for those fields where q18!=6 : for those fields where q18=6, I just want it left out of the AVG. there's definitely a better way to formulate my sql, right? muchos gracias, tom