a survey of 300 union members in new york state reveals that 112 favor the republican candidate for…

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 112$ (number of successes) and $n=300$ (sample - size). So, $\hat{p}=\frac{112}{300}\approx0.3733$.

Step2: Determine the z - value

For a 99% confidence interval, the significance level $\alpha = 1 - 0.99=0.01$. Then $\alpha/2=0.005$. The z - value $z_{\alpha/2}=z_{0.005}$. Looking up in the standard normal table, $z_{0.005} = 2.576$.

Step3: Calculate the margin of error

The formula for the margin of error $E$ for a proportion is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.3733$, $n = 300$, and $z_{\alpha/2}=2.576$ into the formula. First, $1-\hat{p}=1 - 0.3733 = 0.6267$. Then $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.3733\times0.6267}{300}\approx\frac{0.2342}{300}\approx0.000781$. And $E = 2.576\sqrt{0.000781}\approx2.576\times0.028=0.0721$.

Step4: Construct the confidence interval

The confidence interval for the population proportion $p$ is $\hat{p}-E<p<\hat{p} + E$. Substitute $\hat{p}=0.3733$ and $E = 0.0721$ into the formula. We get $0.3733-0.0721 < p<0.3733 + 0.0721$, which simplifies to $0.3012<p<0.4454$.

Answer:

$(0.3012,0.4454)$