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 vote.

Answer 1

Answer:

(0.0757;0.1403)

Step-by-step explanation:

1) Data given and notation

Republicans =115

Democrats=331

Independents=54

Total= n= 115+331+54=500

$\hat p_{ind}=\frac{54}{500}=0.108$

Confidence=0.98=98%

2) Formula to use

The population proportion have the following distribution

$\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the population proportion is given by this formula

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

We have the proportion of independents calculated

$\hat p_{ind}=\frac{54}{500}=0.108$

We can calculate $\alpha=1-conf=1-0.98=0.02$

And we can find $\alpha/2 =0.02/2=0.01$, with this value we can find the critical value $z_{\alpha/2}$ using the normal distribution table, excel or a calculator.

On this case $z_{\alpha/2}=2.326$

3) Calculating the interval

And now we can calculate the interval:

$0.108 - 2.326\sqrt{\frac{0.108(1-0.108)}{500}}=0.0757$

$0.108 + 2.326\sqrt{\frac{0.108(1-0.108)}{500}}=0.1403$

So the 98% confidence interval for this case would be:

(0.0757;0.1403)