Question

a survey of 300 union members in new york state reveals that 112 favor the republican candidate for governor. construct the 98% 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. 0.304 < p < 0.442
c. 0.308 < p < 0.438
d. 0.316 < p < 0.430

Explanation:

Step1: Calculate the sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 112$ (number of successes) and $n=300$ (sample size).
$\hat{p}=\frac{112}{300}\approx0.373$

Step2: Find the critical value

For a $98\%$ confidence interval, the significance level $\alpha=1 - 0.98=0.02$. Then $\alpha/2=0.01$.
Looking up in the standard normal distribution table, $z_{\alpha/2}=z_{0.01}\approx2.33$

Step3: Calculate the margin of error

The margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
Substitute $\hat{p}=0.373$, $n = 300$, and $z_{\alpha/2}=2.33$
$E=2.33\sqrt{\frac{0.373\times(1 - 0.373)}{300}}$
First calculate $0.373\times(1 - 0.373)=0.373\times0.627 = 0.234$
Then $\sqrt{\frac{0.234}{300}}=\sqrt{0.00078}\approx0.028$
$E=2.33\times0.028\approx0.065$

Step4: Calculate the confidence interval

The confidence interval is $\hat{p}-ESubstitute $\hat{p}=0.373$ and $E = 0.065$
$0.373-0.065 < p<0.373 + 0.065$
$0.308 < p<0.438$

Answer:

C. $0.308 < p<0.438$