Ou#3(20 Marks) Online retailers have policies known as \"privacy disclaimers\" t

Question

Ou#3(20 Marks) Online retailers have policies known as "privacy disclaimers" that define the rules regarding their uses of information collected, the customer's right to refuse third-party promotional offers, and so on. You can access these policies through a Web link, found either on the Web site's home page, or on the order page (i.e., as you enter your credit card information), or on a client Web page, or on some other Web page. Location of the privacy disclaimer is considered to be a measure of the degree of consumer protection (the farther the link is from the home page, the less likely it is to be noticed). Marketing researchers did a survey of 291 Web sites in three countries (France, U.K., U.S.) and obtained the contingency table shown below. (You can make your own data for Qu#3) Nationalit US Location of Disclaimer Home Page Order Page Client Page Other Page Column Total93 France UK Row Total 159 56 19 6 68 19 10 35 28 16 32 34 291 106 92 Test if the Location of the Privacy Disclaimer independent of the Web site's nationality? a.

Explanation / Answer

(a)MiniTab output: (stat>table>chi-square test)

Expected counts are printed below observed counts
Chi-Square contributions are printed below expected counts

Fra us uk Total
1 56 68 35 159
50.81 57.92 50.27
0.529 1.755 4.637

2 19 19 28 66
21.09 24.04 20.87
0.208 1.057 2.439

3 6 10 16 32
10.23 11.66 10.12
1.747 0.235 3.421

4 12 9 13 34
10.87 12.38 10.75
0.118 0.925 0.471

Total 93 106 92 291

Chi-Sq = 17.544, DF = 6, P-Value = 0.007

So as you can see p-value < .05 . so we reject the null hypothsis. i.e The data is not independant, they are associated.

(b)one crucial condition is to check whether the cell frequncies are >5 or not! here all the frequencies are >5, so the test is good to go!

(c)Use goodness of fit. Just calculate the 3 proportions and use minitab Stat>Table>Goodness of fit

Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: C5

Using category names in C6

Test Contribution
Category Observed Proportion Expected to Chi-Sq
1 0.32 0.333333 0.333333 0.0005333
2 0.36 0.333333 0.333333 0.0021333
3 0.32 0.333333 0.333333 0.0005333

N DF Chi-Sq P-Value
1 2 0.0032 0.998

since p value>.05 we accept the null i.e they are equal.

(d)to test Ho: p1=4*p3 ; p2=2*p3; p3=p4

i.e H0: p1:p2:p3:p4= 4:2:1:1

so prepare a column for 4 observed proportions and another one for storing null proportions as 4/8,2/8,1/8and 1/8

then stat>table>goodness of fit>select specific proportions from their and select the null proportions column

Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: C8

Using category names in C7

Test Contribution
Category Observed Proportion Expected to Chi-Sq
1 0.546 0.500 0.495500 0.0051468
2 0.227 0.250 0.247750 0.0017379
3 0.101 0.125 0.123875 0.0042241
4 0.117 0.125 0.123875 0.0003816

N DF Chi-Sq P-Value
0.991 3 0.0114904 1.000

since p value is 1. so accept the null . i.e the given proportions are same.

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.