Question

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

A travel company wishes to determine if the type of vacation purchased in its market area is independent of income level of purchasers. A random survey of purchasers gave the following results:

a. At the 0.05 level of significance, can it be concluded that vacation preference and income level are statistically
independent?

b. Construct a 94 percent confidence interval estimate for the proportion of medium income individuals who purchase a foreign vacation.

Answer 1

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.