4. Goodness of fit tests - Multinomial population Aa Aa Visitors to an e-commerc

Question

4. Goodness of fit tests - Multinomial population Aa Aa Visitors to an e-commerce website arrive at the site in one of four ways: by direct access (the visitor types the site's URL into a browser or clicks on a bookmark); via a link on another website, e-mail message, or banner ad; from a paid listing generated by a search engine; or via an organic (unpaid) listing generated by a search engine Engine Ready, an Internet marketing company, randomly selected 27 U.S. e-commerce companies from its 500 clients and studied a total of 18.7 million visits over a 2-year period. For each visit, data were collected on the traffic source and the size of the order placed by the visitor (in dollars). (If no order was placed, the order size is 0.) The relative frequency and average order size for each traffic-source type are shown in the following table Relative Frequency 0.20 0.14 0.41 0.25 Average Order (Dollars) 5.69 5.01 1.91 1.35 Traffic Source Direct Access in Paid Search Ad Search Result [Source: Engine Ready, "Examining the Role Traffic Source Plays in Visitor Purchase Behavior," 2008.] Suppose that SparkleMotion, an online jewelry retailer, conducts a test of the hypothesis that its traffic-source proportions are the same as the traffic-source proportions for the 18.7 million website visits in the Engine Ready study It selects a random sample of visits to its website and categorizes each visit according to the visit's traffic source. The resulting category counts are shown in the following table Traffic Source Observed Frequency Direct Access in Paid Search Ad Search Result 18 93

Explanation / Answer

following information is generated

null hypothesis H0: observed frequecny=expected frequency

alternate hypothesis Ha: observed frequecny is not =expected frequency

we use chi-square test of gooness of fit and chi-square=(sum(O-E)2/E)=8.35 with k-1=4-1=3 df

(1) Expected frequency of link category=28

(2) contribution of link categroty=3.57

(3)chi-square=8.35

(4)reject chi-squre if chi-squre>chi-square(0.05,3)=7.81

(5) using critical value approach, reject H0 as critical chi-square=7.81 is less than calcuated chi-square=8.35

(6)p-value=0.039,( using ms-excel command =chidist(8.35,3))

reject H0

(7)consider the average order size................websit is different the average.......study.

Observed (O) Relative frequency Expected frequency(E) (O-E) (O-E)2/E Direct Access 31 0.2 40 -9 2.025 Link 18 0.14 28 -10 3.571429 Paid search Ad 93 0.41 82 11 1.47561 Search result 58 0.25 50 8 1.28 sum= 200 1 200 0 8.352038

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