We want to estimate with 99 percent confidence the percentage of buyers of cars who are under 30 years of age.A margin of error of 5 percentage points is desired.What sample size is needed? In an earlier sample,we found a 99 percent confidence interval of buyers under 30 years of age to be $\left[ \begin{array} { l l }
.18 & .27
\end{array} \right]$
)
A)104
B)664
C)392
D)523

Question

We want to estimate with 99 percent confidence the percentage of buyers of cars who are under 30 years of age.A margin of error of 5 percentage points is desired.What sample size is needed? In an earlier sample,we found a 99 percent confidence interval of buyers under 30 years of age to be $\left[ \begin{array} { l l } 
.18 & .27
\end{array} \right]$
)
A)104
B)664
C)392
D)523

Quizplus · Accepted Answer

To determine the sample size, we use the formula for sample size in estimating proportions: $ n = \left( \frac{Z^2 \cdot p \cdot (1-p)}{E^2} \right) $. Here, $ Z $ is the Z-value for 99% confidence (approximately 2.576), $ p $ is the estimated proportion (average of the interval, 0.225), and $ E $ is the margin of error (0.05). Plugging these values into the formula gives approximately 523.

We Want to Estimate with 99 Percent Confidence the Percentage [.18.27]\left[ \begin{array} { l l } .18 & .27\end{array} \right][.18​.27​]

We Want to Estimate with 99 Percent Confidence the Percentage $\left[ \begin{array} { l l } .18 & .27\end{array} \right]$