Answer
Do not reject the null hypothesis.
There is not enough evidence to conclude that Brand A is cleaner than brand D.
Work Step by Step
$n_1~and~p_1$ refer to brand A and $n_2~and~p_2$ refer to brand D.
$H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\lt p̂ _2$
$p̂ _1=0.04$ and $p̂ _2=0.07$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{8+14}{200+200}=0.055$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.04-0.07}{\sqrt {0.055(1-0.055)}\sqrt {\frac{1}{200}+\frac{1}{200}}}=-1.32$
Left-tailed test:
Level of significance: $α=0.05$
$z_α=z_{0.05}$
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, $-z_α=-z_{0.05}=-1.645$
Since $z_0\gt -z_α$, we do not reject the null hypothesis.
