Question

We were told this question uses the Chi Square test for either Goodness of Fit or Independence in the following Question: According to the American Automobile Association, 32 million Americans travel over the Thanksgiving holiday. To determine whether to stay open or close, a national restaurant chain surveyed customers at 4 national locations on their Thanksgiving travel plans. At the 5% significance level, test the claim that the proportions of Americans who will travel over Thanksgiving are equal.

Philadelphia Will Travel: 48

Philadelphia Will not Travel: 32

LA will travel: 63

LA will not travel: 17

Austin will travel: 21

Austin will not travel: 59

St. Louis will travel: 25

St. Louis will not travel: 55

Please list the hypothesis statements, test statistic, critical value(s), anp-valueue. Test the claim and summarize your results with a statement.

Answer 1

Ans:

	Observed(fo)
	Philadelphia	LA	Austin	St. Louis	Total
travel	48	63	21	25	157
will not travel	32	17	59	55	163
Total	80	80	80	80	320
	Expected(fe)
	Philadelphia	LA	Austin	St. Louis	Total
travel	39.25	39.25	39.25	39.25	157
will not travel	40.75	40.75	40.75	40.75	163
Total	80	80	80	80	320
	Chi square=(fo-fe)^2/fe
	Philadelphia	LA	Austin	St. Louis	Total
travel	1.95	14.37	8.49	5.17	29.98
will not travel	1.88	13.84	8.17	4.98	28.88
Total	3.83	28.21	16.66	10.16	58.86

Test statistic:

Chi square=58.86

df=(2-1)*(4-1)=3

p-value=chidist(58.86,3)=0.0000

critical chi square=chiinv(0.05,3)=7.815

Reject the null hypothesis.

There is sufficient evidence to reject the claim that  the proportions of Americans who will travel over Thanksgiving are equal.