Question

question 10 1 pts a 90% confidence interval is constructed in order to estimate the proportion of college students who are involved in at least one extracurricular activity. the interval is centered at 0.64, with a margin of error of 0.027. which one of the following intervals must be a 95% confidence interval constructed using the same sample of data? 0.620 to 0.662 0.608 to 0.672 0.611 to 0.669 0.604 to 0.676 0.615 to 0.665

Explanation:

Step1: Recall the relationship between confidence level and margin of error

As the confidence level increases, the margin of error also increases. Since \(95\%>90\%\), the margin of error for the \(95\%\) confidence interval will be larger than that for the \(90\%\) confidence interval.
The \(90\%\) confidence interval has a margin of error \(E = 0.027\). The \(95\%\) confidence interval will have \(E>0.027\).
The formula for a confidence interval for a proportion is \(\hat{p}-E

Step2: Calculate the lower and upper bounds for each option

• Option 1: \(\hat{p}-E=0.620\), so \(E = 0.64 - 0.620=0.02\); \(\hat{p}+E=0.662\), \(E = 0.662 - 0.64=0.022\) (too small)
• Option 2: \(\hat{p}-E=0.608\), \(E=0.64 - 0.608 = 0.032\); \(\hat{p}+E=0.672\), \(E=0.672 - 0.64=0.032\)
• Option 3: \(\hat{p}-E=0.611\), \(E=0.64 - 0.611=0.029\); \(\hat{p}+E=0.669\), \(E=0.669 - 0.64 = 0.029\) (margin of error is close to \(90\%\) margin of error, not likely for \(95\%\))
• Option 4: \(\hat{p}-E=0.604\), \(E=0.64 - 0.604=0.036\); \(\hat{p}+E=0.676\), \(E=0.676 - 0.64=0.036\) (margin of error is larger than \(90\%\) margin of error)
• Option 5: \(\hat{p}-E=0.615\), \(E=0.64 - 0.615=0.025\); \(\hat{p}+E=0.665\), \(E=0.665 - 0.64=0.025\) (too small)

Since the \(z -\)score for \(90\%\) confidence is \(z_{0.90}=1.645\) and for \(95\%\) confidence is \(z_{0.95} = 1.96\), and \(E=z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\), with the same \(\hat{p}\) and \(n\), the \(95\%\) margin of error should be larger. The margin of error for option 2: \(E = 0.032\) (calculated as \(0.64-0.608\) or \(0.672 - 0.64\)) is a reasonable increase from \(E = 0.027\) (for \(90\%\)) compared to option 4 which may be an over - estimate (if we assume a more moderate increase in \(z\) - value)

Answer:

0.608 to 0.672