Chicken, hot dogs, and hamburgers were served during a recent barbecue. The host assumed that there would be no difference between what the women chose to eat and what the men chose to eat. What is the
$\begin{array} { l l l l l } & & \text { Chicken } & \text { Hot Dogs } & \text { Hamburgers } \ & \text { Women } & 7 & 3 & 4 \ \text { test value for the following data? } & \text { Men } & 4 & 5 & 6 \end{array}$
A) 1.596
B) 4.140
C) 1.686
D) 5.769

Question

Quizplus · Accepted Answer

To test if there is a significant difference in food preference between men and women, we can use a chi-square test of independence. We first need to calculate the expected values for each cell assuming no difference between the sexes. To do this, we multiply the row and column marginal totals and divide by the total number of participants:

Expected value for Women who chose Chicken: (7+3+4) x (7+4) / 24 = 4.625
Expected value for Women who chose Hot Dogs: (7+3+4) x (3+5) / 24 = 2.875
Expected value for Women who chose Hamburgers: (7+3+4) x (4+6) / 24 = 3.5
Expected value for Men who chose Chicken: (4+5+6) x (7+4) / 24 = 5.375
Expected value for Men who chose Hot Dogs: (4+5+6) x (3+5) / 24 = 3.325
Expected value for Men who chose Hamburgers: (4+5+6) x (4+6) / 24 = 4.3

We can then use the chi-square test formula to calculate the test statistic:

chi-square = (observed value - expected value)^2 / expected value

Adding up all the values for each cell, we get a chi-square value of 1.686. We have 3 rows and 2 columns, so the degrees of freedom (df) is (3-1) x (2-1) = 2. Looking at a chi-square distribution table for df=2, we find the critical value for alpha=0.05 to be 5.991. Since 1.686 < 5.991, we fail to reject the null hypothesis that there is no significant difference in food preference between men and women. Thus, we can conclude that the host's assumption was correct. The best choice for the test value is option C, 1.686.

Chicken, Hot Dogs, and Hamburgers Were Served During a Recent