Question

Define two events of your choosing. Speculate as to whether or not you believe these events will be independent and provide a brief explanation for your response. Then, conduct a test for independence and show your work. Do the results of this test support your speculation?

Subscriptions	Subscription Category Used
Renewal Status	1 ( Gift )	2 ( Prev Review )	3 ( Direct Mail )	4 (Sub Service)	5 (Catalog)	Grand Total
0 ( No Renewal )	180	1233	1090	742	43	3288
1 ( Renewed )	704	3907	1134	122	2	5869
Grand Total	884	5140	2224	864	45	9157

Answer 1

Consider two events

A: Subscription category used.

B: Renewal status.

We have to test to events

Subscription category used and renewal status are independent.

i.e. Null Hypothesis- Ho : Subscription category used and renewal status are independent.

against

Alternative Hypothesis-Ha:Subscription category used and renewal status are not independent.

We use chi-square independence test.

The value of chi-square test statistic

r=number of rows =2

c=number of columns =5

D.f. =(r-1)*(c-1)=4

Oi: Observed frequency and Ei: Expected frequency.

Expected frequency is calculated by

Ei = ( Row total * column total) / Total frequency, Here Total frequency = 9157

Oi	Ei	(Oi-Ei)^2/Ei
180	317.417	59.491
1233	1845.618	203.347
1090	798.571	106.354
742	310.236	600.897
43	16.158	44.590
704	566.583	33.329
3907	3294.382	113.921
1134	1425.429	59.583
122	553.764	336.642
2	28.842	24.981
9157	9157	1583.134

Value of chi-square test statistic is

Value of test statistic is 1583.134 and degrees of freedom are 4.

p-value is obtained by

using R

> pvalue=1-pchisq(1583.134,4)
> pvalue
[1] 0

Decision: since p-value is zero, we reject the null hypothesis Ho.

Conclusion: There is sufficient evidence to claim that subscription category used and renewal status are not independent.

i.e. Renewal status depends on subscription category used.