a u.s. travel data center survey conducted for better homes and gardens of 1500 adults found that 39% said that they would take more vacations this year than last year. find the 95% confidence interval for the true proportion of adults who said that they will travel more this year. write the confidence interval as an ordered pair to 3 decimal places with a space after the comma.

Question

Accepted Answer

# Explanation:
## Step1: Calculate $ \hat{p} $ and $ n $
Given $ \hat{p}=0.39 $, $ n = 1500 $
## Step2: Find $ z_{\alpha/2} $
For $ 95\% $ confidence interval, $ \alpha=1 - 0.95=0.05 $, $ \alpha/2=0.025 $. From standard normal table, $ z_{\alpha/2}=z_{0.025}=1.96 $
## Step3: Calculate the margin of error $ E $
$$
\begin{align*}
E&=z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\
&=1.96\sqrt{\frac{0.39\times(1 - 0.39)}{1500}}\
&=1.96\sqrt{\frac{0.39\times0.61}{1500}}\
&=1.96\sqrt{\frac{0.2379}{1500}}\
&=1.96\sqrt{0.0001586}\
&=1.96\times0.0126\
&=0.025
\end{align*}
$$
## Step4: Calculate the confidence interval
Lower limit $=\hat{p}-E=0.39 - 0.025=0.365$
Upper limit $=\hat{p}+E=0.39 + 0.025=0.415$

# Answer:
$ (0.365,0.415) $

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Question

Explanation:

Step1: Calculate \( \hat{p} \) and \( n \)

Step2: Find \( z_{\alpha/2} \)

Step3: Calculate the margin of error \( E \)

Step4: Calculate the confidence interval

Answer: