Answer
There is not enough evidence to conclude that American adults have fewer guns at home.
Work Step by Step
$N_1,n_1~and~p_1$ refer to 2010 and $N_2,n_2~and~p_2$ refer to 2009.
$H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\lt p̂ _2$
$p̂ _1=\frac{x_1}{n_1}=\frac{441}{1134}=0.3889$ and $p̂ _2=\frac{x_2}{n_2}=\frac{458}{1134}=0.4039$
Requirements:
$n_1p̂ _1(1-p̂ _1)=1134\times0.3889(1-0.3889)=269.5\geq10$
$n_2p̂ _2(1-p̂ _2)=1134\times0.4039(1-0.4039)=273.0\geq10$
$n_1\leq0.05N_1$
$n_2\leq0.05N_2$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{441+458}{1134+1134}=0.3964$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.3889-0.4039}{\sqrt {0.3964(1-0.3964)}\sqrt {\frac{1}{1134}+\frac{1}{1134}}}=-0.73$
Left-tailed test:
Let's use $α=0.10$ level of significance.
$z_α=z_{0.10}$
If the area of the standard normal curve to the right of $z_{0.10}$ is 0.10, then the area of the standard normal curve to the left of $z_{0.10}$ is $1−0.10=0.90$
According to Table V, the z-score which gives the closest value to 0.90 is 1.28
So, $-z_\frac{α}{2}=-1.28$
Since $-z_α\lt z_0\lt z_α$, 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 $z_α$ and a lower value of $-z_α$.
