Online retailers have policies known as \"privacy disclaimers\" that define the

Question

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 Total 93 France UK Row Total 159 35 28 56 68 32 34 291 106 92 a. Test if the Location of the Privacy Disclaimer independent of the Web site's nationality?

Explanation / Answer

Solution:

(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.

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