Answer:
There is no enough evidence that the proportions are different.
Step-by-step explanation:
We have to perform a hypothesis test on the difference of proportions.
In this case, the sample size is equal.
The null and alternative hypothesis are
[tex]H_0: \pi_1=\pi_2\\\\ H_1: \pi_1\neq\pi_2[/tex]
The significance level is assumed to be 0.05.
The weighted average p, as the sample sizes are the same, is the average of both proportions:
[tex]p=\frac{p_1+p_2}{2} =\frac{0.05+0.08}{2}=0.065[/tex]
The standard deviation is
[tex]s=\sqrt{\frac{2p(1-p)}{n} } =\sqrt{\frac{2*0.065(1-0.065)}{120}} =0.032[/tex]
The z-value for this sample is:
[tex]z=\frac{p_1-p_2}{s} =\frac{0.05-0.08}{0.032} =-0.9375[/tex]
The P-value for z=-0.9375 is P=0.3485.
The P-value (0.35) is greater than the significance level (0.05), so the null hypothesis is failed to reject.
There is no enough evidence that the proportions are different.
