Hallmon, David. (Summer 2008) C&I 557 Mid-Term. Two Task Analysis of the Chi-Square Method. Southern Illinois University-Carbondale.
Information Processing Analysis of the Chi-Square Method Start
A
End
Enter with previously unseen data set 2 & a Chi-Square x Table of statistical significance.
No
Locate the df value in the 2 appropriate column of the x Table.
Chi-Square Method not applicable.
Recall the definition of nominal categorical: Data is not represented in scale, order, or intervals. Data is only in observed categories with one variable.
Is the data sets level of measurement nominal categorical?
Locate the value closest to your 2 calculated x on that df row.
Move up the column and determine your probability p value.
Recall the Chi-Square Formula:
Yes
x2 = ∑ (fo - fe)2 / fe ,
Determine the null hypothesis by dividing the total sample numbers N by the total number of categories k to determine what the expected frequency fe would be if due to chance.
Calculate the difference between observed frequency fo of a category and the expected fe & square the amount.
fe = N / k
Is the p value for 2 the calculated x
2
is p
Recall part of the x 2 formula: (fo - fe) / fe
Reject your null hypothesis. The results are statistically significant. Other factors are involved!
Recall categories:
c1= (fo - fe)2 / fe c2= (fo - fe)2 / fe End
Yes
No Sum ∑ all categories in sample
c1 + c2 + c3 + …
Recall summation of categories formula:
x2 = ∑ c1 + c2
+…
Complete calculations to 4 significant digits. Round off answer to 3 significant digits.
Determine the degrees of freedom
df
Recall degrees of freedom formula: df = k -1
A
Yes
> 0.05?
No
2
Record as Category1 c1 or subsequent categories c2, c3, c4…
Additional categories in sample?
Recall determining the null hypothesis:
Recall part of the x 2 formula: (fo - fe)
Divide the calculated difference by the expected frequency fe.
Multiply the p value by 100 to determine the percentage % probability that any deviation from expected is only due to chance.
Accept your null hypothesis. The results are not statistically significant
End