## Question 535:

1

1. Q535_1_chiSquareQ535.xls

So we have categorical data setup here and want to know if there is any independence between the car types at each mall. There certainly are differences in the observed counts, but we need to take into account chance fluctuations in these differences. To do so we can use the Chi-Square test of Independence. The null hypothesis is that there is no difference in car-type by mall location and we will reject this hypothesis if p < .05.

The Chi-Square test uses the formula

Sum of       (O-E)2
---------
E

• O means the observed frequencies
• E means the expected frequencies

We will need to do this for each cell. I'll walk through the first cells example. We have 44 cars at the Somerset Mall. This is the observed count. The expected count is found by multiplying the row total and column totals together and then multiplying this by the grand total. We get (193*100)/400 = 48.25 as our expected value. Now we just plug and chug in the formula (44-48.25)2/48.25 = .37435.

We now repeat this step for all values and then add them all up. This gets us 24.5315, which is our test statistic. To see if this is significant, we can use the excel formula =CHIDIST()(24.5315,12) where the second parameter is the degrees of freedom, found as (row count -1)*(column count-1) = 4*3=12. We get the p-value .0172 which is less than our alpha of .05 so we would reject the null hypothesis and say that there is a difference in car types found at each mall.