Respuesta :
Answer:
Hypothesis
[tex]H_0: \pi=0.26\\\\H_a: \pi\neq 0.26[/tex]
P-value = 0.078
Conclusion: The null hypothesis is rejected. There is enough evidence to claim that the proportion of foreigners home buyers is different from 26%.
Step-by-step explanation:
We will perform a hypothesis test on a proportion.
The null hypothesis states that 26% of home buyers are foreigners.
The alternative hypothesis states that the proportion is different than 26%.
[tex]H_0: \pi=0.26\\\\H_a: \pi\neq 0.26[/tex]
The significance level is α=0.10.
The sample proportion is:
[tex]p=x/N=20/110=0.182[/tex]
The standard error is:
[tex]\sigma_p=\sqrt{\frac{\pi(1-\pi)}{n} }=\sqrt{\frac{0.26*0.74}{110}} =\sqrt{\frac{0.1924}{110} } =\sqrt{0.00175} \\\\\sigma_p=0.042[/tex]
Then, we can calculate the z-statistic as:
[tex]z=\frac{p-\pi+0.5/N}{\sigma_p} =\frac{0.182-0.26+0.004}{0.042}=- \frac{0.074}{0.042}= -1.762[/tex]
The P-value can be calculated as:
[tex]P-value=2*P(z<-1.762)=0.078\\\\P-value<\alpha[/tex]
The P-value is smaller than the significance level, so the effect is significant.
The null hypothesis is rejected. There is enough evidence to claim that the proportion of foreigners home buyers is different from 26%.
