test of association- chi squared

Question 1

why use chi?

Answer 1

-for difference or associaiton
-nominal data
-independant unrelated designs

Question 2

what may be an alternative hypothesis?

Answer 2

-more 8yr olds than 5yr olds are able to select a phtotgraph that represents a persepctive different from their own(directional one tailed)

Question 3

what may be a null hypothesis?

Answer 3

-there is no difference between the number of 5yr olds and 8 yr olds who can select a photograph that represents a persepctive different from their own

Question 4

what is step 1?

Answer 4

-draw 2 times 2 contigency table
-shows observed values
-calc totals for each row, column and overall total
(table in folder)

Question 5

what is step 2?

Answer 5

-the table of expected frequencies
-a frequency expected if there was no difference between the 2 grps
-calculated by multiply the total for the row by the total for the column divided by grand total

Question 6

step 3?

Answer 6

-work out value of X squared
-look at sum in revision folder

Question 7

what is step 4?

Answer 7

-the calculated and critical values
-to find critical value work out degrees of freedom
-(rows -1) * (columns-1) at contigency table
-one tail? DF? 0.05?
-if calc value is more than critical we can accept alternative H SIGNIFICANT.