A chi-square test for independence step by step explanation please?

Step 1. Calculate all row and column totals, as well as the grand total.

Example: Row 1 total is #50+120+65=235#
Example: Column 1 total is #50+25 = 75#
Example: Grand total is #300#

Step 2. Compute the expected value of each of the 6 cells as follows:

#E("row i, col j") = ("Row i total " xx " Col j total")/"Grand total"#

Example for row 1, column 1:

#E(1,1) = (235xx75)/300=58.75#

Step 3. Compute the following value for each of the 6 cells:

#("Observed " - " Expected")^2/"Expected"#

Example for row 1, column 1:

#(50-58.75)^2/58.75=1.3032#

Step 4. Add all 6 values together to get your #chi^2# (chi-squared) statistic.

Step 5. Compare this statistic to the test-value of #chi_("  "0.05, 2)^2# to see which one is larger.

Why 0.05? This is your level of significance.

Why 2? The 2 comes from
#("row count" - 1) xx ("column count" - 1)#
#=(2-1)xx(3-1)#
#= 1xx2#
#= 2#

Using a #chi^2# table, we see that #chi_("  "2, 0.05)^2=5.991.# If your #chi^2# statistic is #>=5.991,# then it is unlikely (at the 0.05 significance level) to have come from a #chi^2# distribution with 2 degrees of freedom, and thus we would reject the null hypothesis #(H_0)# and conclude that vacation preference and income level are not independent.

If your #chi^2# statistic is #<5.991# then we cannot reject #H_0.#

#-#

Unfortunately, I do not know how to compute a confidence interval for a single cell proportion.