Question

The town of Hayward (CA) has about 50,000 registered voters. A political research firm takes a simple random sample of 500 of these voters. In the sample, the breakdown by party affiliation is 115 Republicans, 331 Democrats, and 54 Independents. Calculate a 98% confidence interval for the true percentage of Independents among Haywards 50,000 registered voters.

Answer 1

Answer: (0.076, 0.140)

Step-by-step explanation:

Confidence interval for population proportion (p) is given by :-

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where $\hat{p}$ = sample proportion.

n= sample size.

$\alpha$ = significance level .

$z_{\alpha/2}$ = critical z-value (Two tailed)

As per given , we have

sample size : n= 500

The number of Independents.:  x= 54

Sample proportion of Independents$\hat{p}=\dfrac{x}{n}=\dfrac{54}{500}=0.108$

Significance level 98% confidence level : $\alpha=1-0.98=0.02$

By using z-table , Critical value : $z_{\alpha/2}=z_{0.01}=2.33$

The 98% confidence interval for the true percentage of Independents among Haywards 50,000 registered voters will be :-

$0.108\pm (2.33)\sqrt{\dfrac{0.108(1-0.108)}{500}}\\\\=0.108\pm2.33\times0.013880634\\\\=0.108\pm0.03234187722\\\\\approx0.108\pm0.032=(0.108-0.032,\ 0.108+0.032)=(0.076,\ 0.140)$

Hence, the 98% confidence interval for the true percentage of Independents among Haywards 50,000 registered voters.= (0.076, 0.140)