See the attached file for the same text in a better format.

Type of Vehicle Used
Gender Car Truck SUV Van
Male 90 100 50 7
Female 110 75 65 3

Use the given contingency table to
(a) Find the expected frequencies of each cell in the table,

First, you want to calculate all of the row and column totals (in bold in the table below):

Type of Vehicle Used
Gender Car Truck SUV Van Total
Male 90 100 50 7 247
Female 110 75 65 3 253
Total 200 175 115 10 500

Then, you want to find the expected frequency for each cell (this is the number that would be expected for each category if there were no relationship between gender and the type of vehicle used).

You do this by, for every cell in the table, multiplying the row total by the column total, and dividing by the grand total.

Type of Vehicle Used
Gender Car Truck SUV Van Total
Male 247
Female 253
Total 200 175 115 10 500

So, for the first row and first column (male, car), you would calculate (247)(200)/500 to get 98.8, the expected number. If you do this for all of the cells, you should get this:

Type of Vehicle Used
Gender Car Truck SUV Van Total
Male 98.8 86.45 56.81 4.94 247
Female 101.2 88.55 58.19 5.06 253
Total 200 175 115 10 500

(b) Perform a chi-square test for independence, and

The chi-squared test tests the null hypothesis that there is no relationship between the variables (i.e. they are independent of one another) against the alternative hypothesis that they are related in some way (i.e. they are not independent).

The test statistic is calculated as:

X2 = (O - E)^2/E where O is the observed number and E is the expected.

So, in this case...

X2 = (90 - 98.8)^2/98.8 + (86.45 - 100)2/100 + ... + (5.06 - 3)^2/3

X2 = 9.057

Notice that the further apart the observed and expected values are, the larger X2 will be. To see if X2 is large enough to be statistically significant, you have to compare it to a X2 distribution with the appropriate degrees of freedom. The degrees of freedom are calculated as df = (number of rows - 1)(number of columns - 1). So, in this case, df = 3 * 1 = 3.

Compare X2 = 9.057 to a X2 distribution with 3 degrees of freedom to see that there a p-value of p < 0.05. This is because our observed X2 value of 9.057 is greater than the critical value of X2 at the 0.05 significance level. However, the observed X2 value of 9.057 is NOT greater than the critical value of X2 at the 0.01 significance level.

This table has several critical values of the X2 test with df = 3.

Significance levels: 0.20 0.10 0.05 0.025 0.01 0.001
Critical values: 4.64 6.25 7.82 9.35 11.34 16.27

(c) Comment on the relationship between the two variables. Assume the variables are independent.

Because the X2 test performed in part b had a p-value of p > 0.01, we cannot reject the null hypothesis that gender and the type of vehicle used are independent of one another.

If you were testing this using a significance level of alpha = 0.05 instead of 0.01, you would be able to reject the null hypothesis