Question

a survey of 300 union members in new york state reveals that 112 favor the republican candidate for governor. construct the 99% confidence interval for the true population proportion of all new york state union members who favor the republican candidate
a. 0.301 < p < 0.445
b. 0.308 < p < 0.438
c. 0.304 < p < 0.442
d. 0.316 < p < 0.410

Explanation:

Step1: Calculate the sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 112$ (number of union members who favor the candidate) and $n=300$ (total number of union members surveyed). So, $\hat{p}=\frac{112}{300}\approx0.373$.

Step2: Find the critical value

For a 99% confidence interval, the critical value $z$ (using standard normal distribution tables or a calculator) is $z = 2.576$.

Step3: Calculate the margin of error

The margin of error $E=z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.373$, $n = 300$, and $z=2.576$ into the formula:

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Step4: Construct the confidence interval

The confidence interval is $\hat{p}-E$0.373-0.072

Answer:

A. $0.301 < p < 0.445$