Question

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. The relative frequency for each traffic-source type are shown in the following table. [Source: Engine Ready, “Examining the Role Traffic Source Plays in Visitor Purchase Behavior,” 2008.]

Traffic Source	Relative Frequency
Direct Access	0.20
Link	0.14
Paid Search Ad	0.41
Search Result	0.25

Suppose that Paws-n-Claws, an online pet-products 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	50
Link	35
Paid Search Ad	72
Search Result	43

The Paws-n-Claws statistician confirms that the expected frequencies for each of the four categories is 5 or more and decides that the chi square goodness of fit test is appropriate to use.

Let p1 = the proportion of all of Paws-n-Claws’s traffic that arrives via direct access; p2 = the proportion that arrives via a link; p3 = the proportion that arrives via a paid search ad; and p4 = the proportion that arrives via a search result.

The null hypothesis is .

The expected frequency for the direct access category is .

Each of the four categories makes a contribution to the chi-square test statistic. The contribution of the direct access category is   .

The value of the test statistic is χ² =     .

Use the Distributions tool to help you answer the questions that follow.

Chi-Square Distribution

Degrees of Freedom = 6

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Use the critical value approach to perform the test for goodness of fit at a significance level of α = 0.10. The rejection rule is:

Reject H₀ if χ²       .

Using the critical value approach, the null hypothesis (that p₁ = 0.20, p₂ = 0.14, p₃ = 0.41, and p₄ = 0.25) is   .

The p-value is   .

At a significance level of α = 0.10, using the p-value approach, the null hypothesis (that p₁ = 0.20, p₂ = 0.14, p₃ = 0.41, and p₄ = 0.25) is   .

Answer 1

Following are the steps followed for Chi-Square Test for Goodness of Fit

Based on the above test following are answer for each part:

The null hypothesis is: p_1​ =0.20, p_2​ =0.14, p₃​ =0.41, p₄ ​= 0.25

The expected frequency for the direct access category is 200*0.20 = 40

The contribution of the direct access category is = (50-40)²/40 = 2.5

The value of the test statistic is χ² = 6.45

Degrees of freedom is df = 4 - 1 = 3

Reject H₀ χ² > 6.251 (Critical value was obtained using online calculator. Screenshot attached)

Since χ² = 6.45 > Critical value (6.251), we reject null hypothesis

Thus, Using the critical value approach, the null hypothesis (that p₁ = 0.20, p₂ = 0.14, p₃ = 0.41, and p₄ = 0.25) is rejected

The p-value is 0.0917 (P-value was obtained using online calculator. Screenshot attached)

Since p-value (0.0917) < α = 0.10, we reject null hypothesis.

Thus, at a significance level of α = 0.10, using the p-value approach, the null hypothesis (that p₁ = 0.20, p₂ = 0.14, p₃ = 0.41, and p₄ = 0.25) is rejected