Let p represents the population proportion.
From the given information, we have
[tex]H_0: p=0.06\\\\H_a: p>0.06[/tex], since alternative hypothesis is right tailed then the test is right tailed test.
Given : In a sample of 416 households that owned one or more vacation homes, 46 were minorities.
i.e. n= 416
Sample proportion : [tex]\hat{p}=\dfrac{46}{416}=0.110576923077\approx0.11[/tex]
Test statistic : [tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
i.e. [tex]z=\dfrac{0.11-0.06}{\sqrt{\dfrac{0.06(1-0.06)}{416}}}[/tex]
[tex]\Rightarrow\ z=4.29414907893\approx4.29[/tex]
P-value for right tailed test : [tex]P(z>4.29)=1-P(z\leq4.29)[/tex]
[tex]=1-0.999991=0.0000089[/tex]
Since the p-value is less than the significance level (0.01), so we reject the null hypothesis.
Thus , we conclude that we have enough evidence to support the claim that that the percentage of vacation-home owners who are minorities is larger than 6 percent.
