Answer:
Given:
Hypothesized Population Proportion $ (p)=0.30 $
Favorable Cases $ (x)=360 $
Sample Size $ (n)=1,024 $
Significance Level $ (\alpha) = 0.01$
Solution:
The null and alternative hypothesis:
$H_0:p=0.30$
$H_a:p>0.30$
The sample proportion:
$\hat{p} = \frac{x}{n} = \frac{360}{1,024} = 0.3516$
The test statistic:
$z = \frac{\hat{p} – p}{\sqrt{\frac{p(1-p)}{n}}} = \frac{0.3516 – 0.30}{\sqrt{\frac{0.30(1-0.30)}{1,024}}} = 3.601$
The p-value:
$\text{The p-value} = P(z > 3.601) = 0.0002$
The conclusion:
The p-value is less than the significance level. therefore, We reject the null hypothesis, suggesting that the fraction of supporters of party X is significantly greater than 30%.
Final answer:
The null and alternative hypothesis:
$H_0:p=0.30$
$H_a:p>0.30$
The test statistic $(z)=3.601$
The p-value $=0.0002$
The conclusion:
We reject the null hypothesis, suggesting that the fraction of supporters of party X is significantly greater than 30%.
