In a survey of 1,000 people,420 are opposed to a particular tax increase.Construct a 95 percent confidence interval for the proportion of people in the population opposed to this tax increase.
A) $\left[ \begin{array} { l l l }
.394 & .446
\end{array} \right]$
B) $\left[ \begin{array} { l l }
.389 & .451
\end{array} \right]$
C) $[ .380 \quad .460 ]$
D) $[ .399 \quad .441 ]$

Question

In a survey of 1,000 people,420 are opposed to a particular tax increase.Construct a 95 percent confidence interval for the proportion of people in the population opposed to this tax increase.
A) $\left[ \begin{array} { l l l } 
.394 & .446
\end{array} \right]$
B) $\left[ \begin{array} { l l } 
.389 & .451
\end{array} \right]$
C) $[ .380 \quad .460 ]$
D) $[ .399 \quad .441 ]$

Quizplus · Accepted Answer

We can use the formula for a confidence interval for a proportion:
$$ \hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$
where $\hat{p}$ is the sample proportion, $n$ is the sample size, and $z_{\alpha/2}$ is the critical value from the standard normal distribution corresponding to the desired level of confidence.

Here, $\hat{p} = \frac{420}{1000} = 0.42$, $n=1000$, and using a 95% confidence level, $z_{\alpha/2} \approx 1.96$. So, the formula gives us:

$$0.42 \pm 1.96 \sqrt{\frac{(0.42)(0.58)}{1000}}$$

Simplifying this expression, we get:

$$\left[ \begin{array} { l l } 
0.389 & 0.451
\end{array} \right]$$

Therefore, the best choice is B, which is the closest to this interval.

In a Survey of 1,000 People,420 Are Opposed to a Particular