A researcher wishes to determine if the fraction of supporters of party X is equal to 30%, or more. In a sample of 1024 persons, 360 declared to be supporters. Verify the researcher’s hypothesis at a significance level of 0.01. What is the p-value of the resulting statistic?

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%.