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

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

H_0: \mu_X-\mu_Y \leq 0 \quad \text{versus}\quad H_1: \mu_X-\mu_Y>0

The test statistic is

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

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

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

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.