Question

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.

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 \( E \)

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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:

\( (0.365,0.415) \)