Answer
$-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$: null hypothesis is not rejected.
There is not enough evidence to conclude that $~p̂ _1\ne p̂ _2$.
Work Step by Step
- Proportion;
- Independent sampling.
$H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\ne p̂ _2$
$p̂ _1=\frac{x_1}{n_1}=\frac{43}{120}=0.3583$ and $p̂ _2=\frac{x_2}{n_2}=\frac{56}{130}=0.4308$
Requirements:
$n_1p̂ _1(1-p̂ _1)=120\times0.3583(1-0.3583)=24.5\geq10$
$n_2p̂ _2(1-p̂ _2)=130\times0.4308(1-0.4308)=31.9\geq10$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{43+56}{120+130}=0.396$
$z_0=\frac{p̂_2-p̂ _1}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.3583-0.4308}{\sqrt {0.396(1-0.396)}\sqrt {\frac{1}{120}+\frac{1}{130}}}=-1.17$
Two-tailed test:
$z_{\frac{α}{2}}=z_{0.005}$
If the area of the standard normal curve to the right of $z_{0.005}$ is 0.005, then the area of the standard normal curve to the left of $z_{0.005}$ is $1−0.005=0.995$
According to Table V, there are 2 z-scores which give the closest value to 0.995: 2.57 and 2.58. So, let's find the mean of these z-scores: $\frac{2.57+2.58}{2}=2.575$
Also, $-z_{\frac{α}{2}}=-2.575$
Since $-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$, we do not reject the null hypothesis.
