Question

your turn
for the given situation, find the specified confidence interval for the population proportion.

1. isabelle surveys a random sample of 80 voters in her large town and finds that 46 support raising property taxes in order to build a new library. find a 95% confidence interval for the proportion p of all voters in isabelles town who support raising property taxes in order to build a new library.

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 46$ (number of successes) and $n=80$ (sample size). So, $\hat{p}=\frac{46}{80}=0.575$.

Step2: Find the critical value

For a 95% confidence interval, the significance level $\alpha=1 - 0.95=0.05$. Then $\alpha/2=0.025$. The critical value $z_{\alpha/2}$ is the $z$-score such that the area to the right of $z_{\alpha/2}$ is $0.025$. From the standard normal table, $z_{\alpha/2}=1.96$.

Step3: Calculate the margin of error

The margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.575$, $n = 80$, and $z_{\alpha/2}=1.96$ into the formula:

$$
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$$

Step4: Calculate the confidence interval

The confidence interval for the population proportion $p$ is $\hat{p}-E$0.575-0.1084

Answer:

The 95% confidence interval for the proportion \(p\) of all voters in Isabelle's town who support raising property taxes is \((0.467,0.683)\) (rounded to three decimal places).