Answer
There is not enough evidence to conclude that the proportion of sales between the two Web page designs is different.
John should have no preference between the two Web pages.
Work Step by Step
$N_1,n_1~and~p_1$ refer to Web page design I and $N_2,n_2~and~p_2$ refer to Web page design II.
$H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\ne p̂ _2$
$p̂ _1=\frac{x_1}{n_1}=\frac{54}{523}=0.1033$ and $p̂ _2=\frac{x_2}{n_2}=\frac{62}{512}=0.1211$
Requirements:
$n_1p̂ _1(1-p̂ _1)=523\times0.1033(1-0.1033)=48.4\geq10$
$n_2p̂ _2(1-p̂ _2)=512\times0.1211(1-0.1211)=54.5\geq10$
$n_1\leq0.05N_1$
$n_2\leq0.05N_2$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{54+62}{523+512}=0.1121$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.1033-0.1211}{\sqrt {0.1121(1-0.1121)}\sqrt {\frac{1}{523}+\frac{1}{512}}}=-0.91$
Two-tailed test:
Let's use $α=0.10$ level of significance.
$z_\frac{α}{2}=z_{0.05}$
If the area of the standard normal curve to the right of $z_{0.05}$ is 0.05, then the area of the standard normal curve to the left of $z_{0.05}$ is $1−0.05=0.95$
According to Table V, there are 2 z-scores which give the closest value to 0.95: 1.64 and 1.65. So, let's find the mean of these z-scores: $\frac{1.64+1.65}{2}=1.645$
Also, $-z_\frac{α}{2}=1.645$
Since $-z_\frac{α}{2}\lt z_0\lt z_{\frac{α}{2}}$, we do not reject the null hypothesis.
Notice that the null hypothesis would not be rejected for the $α=0.05$ or $α=0.01$ level of significance. They would provide a higher value of $z_\frac{α}{2}$ (and a lower value of $-z_\frac{α}{2}$).
