Question

graduation gifts ninety - three percent of students said that cash is their top choice for a graduation gift. if 100 students were surveyed, find the 99% confidence interval for the proportion. round your answers to at least three decimal places.

Explanation:

Step1: Identify the values

Given \(n = 100\), \(\hat{p}=0.93\) (since \(93\%=0.93\)), and for a \(99\%\) confidence interval, the critical value \(z_{\alpha/2}\) is \(z_{0.005}\). From the standard normal table, \(z_{0.005} = 2.576\).

Step2: Calculate the margin of error \(E\)

The formula for the margin of error for a proportion is \(E=z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\).
Substitute the values: \(\hat{p}=0.93\), \(1-\hat{p}=0.07\), \(n = 100\), \(z_{\alpha/2}=2.576\)

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Step3: Calculate the confidence interval

The confidence interval for a proportion is \(\hat{p}-ESubstitute \(\hat{p}=0.93\) and \(E = 0.0657\)
\(0.93-0.0657=0.8643\) and \(0.93 + 0.0657=0.9957\)

Answer:

\(0.864