Good website design can make Web navigation easier. The article "The Implications of Visualization Ability and Structure Preview Design for Web Information Search Tasks" (H. Zhang and G. Salvendy, International Journal of Human-Computer Interaction, 2001 :75-95) presents a comparison of item

Question

Good website design can make Web navigation easier. The article "The Implications of Visualization Ability and Structure Preview Design for Web Information Search Tasks" (H. Zhang and G. Salvendy, International Journal of Human-Computer Interaction, 2001 :75-95) presents a comparison of item recognition between two designs. A sample of 10 users using a conventional Web design averaged 32.3 items identified, with a standard deviation of 8.56. A sample of 10 users using a new structured Web design averaged 44.1 items identified, with a standard deviation of 10.09. Can we conclude that the mean number of items identified is greater with the new structured design?

Accepted Answer

Let [latex]\bar{X}=44.1[/latex] be the sample mean for the structured Web design. Then [latex]s_X[/latex] = 10.09 and [latex]n_X[/latex] = 10. Let [latex]\bar{Y}=32.3[/latex] be the sample mean for the conventional Web design. Then [latex]s_Y[/latex] = 8.56 and [latex]n_Y[/latex] = 10. Let [latex]μ_X[/latex] and [latex]μ_Y[/latex] denote the population mean measurements made by the structured and conventional methods, respectively. The null and alternate hypotheses are

[latex]H_0: \mu_X-\mu_Y \leq 0 \quad ext{versus}\quad H_1: \mu_X-\mu_Y>0[/latex]

The test statistic is

[latex]t=\frac{(\bar{X}-\bar{Y})-0}{\sqrt{s_X^2 / n_X+s_Y^2 / n_Y}}[/latex]

Substituting values for [latex]\bar{X}, \bar{Y}, s_X, s_Y, n_X[/latex] and [latex]n_Y[/latex], we compute the value of the test statistic to be t = 2.820. Under [latex]H_0[/latex] , this statistic has an approximate Student's t distribution, with the number of degrees of freedom given by

[latex]v=\frac{\left(\frac{10.09^2}{10}+\frac{8.56^2}{10} ight)^2}{\frac{\left(10.09^2 / 10 ight)^2}{9}+\frac{\left(8.56^2 / 10 ight)^2}{9}}=17.53 \approx 17[/latex]

Consulting the t table with 17 degrees of freedom, we find that the value cutting off 1% in the right-hand tail is 2.567, and the value cutting off 0.5% in the right-hand tail is 2.898. Therefore the area in the right-hand tail corresponding to values as extreme as or more extreme than the observed value of 2.820 is between 0.005 and 0.010. Therefore 0.005 < P < 0.01 (see Figure 7.4, page 288). There is strong evidence that the mean number of items identified is greater for the new design.

Question 7.7: Good website design can make Web navigation easier. The arti...

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