Please use Excel to explain answers. Thanks. A contingency analysis table has be
ID: 3153861 • Letter: P
Question
Please use Excel to explain answers. Thanks.
A contingency analysis table has been constructed from data obtained in a phone survey of customers in a market area in which respondents were asked to indicate whether they owned a domestic or foreign car and whether they were a member of a union or not. The following contingency table is provided. Use the chi-square approach to test whether type of car owned (domestic or foreign) is independent of union membership. Test using an alpha = 0.05 level. Calculate the p-value for this hypothesis test.Explanation / Answer
A)
Doing an Expected Value Chart,
123.1060606 501.8939394
71.89393939 293.1060606
Using chi^2 = Sum[(O - E)^2/E],
chi^2 = 27.90919379 [ANSWER, TEST STATISTIC]
With df = (a - 1)(b - 1), where a and b are the number of categories of each variable,
a = 2
b = 2
df = 1
Thus, the critical value is
significance level = 0.05
chi^2(critical) = 3.841458821
[type : =CHISQ.INV.RT(0.05, 1)]
As chi^2 > 3.841, we REJECT THE NULL HYPOTHESIS.
Thus, there is significant evidence that having a foreign and domestic car is not independent of being a member of a union or not at 0.05 level. [CONCLUSION]
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b)
Also, the p value is
P = 1.27144*10^-7 [ANSWER, P VALUE]
[type: =CHISQ.DIST.RT(27.90919379, 1) ]
As P < 0.05, we REJECT THE NULL HYPOTHESIS.
Thus, there is significant evidence that having a foreign and domestic car is not independent of being a member of a union or not at 0.05 level. [CONCLUSION]