Questions: Obesity is defined as a body mass index (BMI) of 30 kg / m^2 or more. A 95% confidence interval for the percentage of U.S. women aged 18 to 29 who were obese was found to be 19.5% to 21.7%. What was the sample size? Round the intermediate calculations to four decimal places and round up your final answer to the next whole number. n=

Solution Steps

The confidence interval for the percentage of U.S. women aged 18 to 29 who were obese is given as \( 19.5\% \) to \( 21.7\% \). We can express these percentages as proportions:

\[ \text{Lower bound} = 0.195, \quad \text{Upper bound} = 0.217 \]

The sample proportion \( \hat{p} \) is calculated as the midpoint of the confidence interval:

\[ \hat{p} = \frac{0.195 + 0.217}{2} = 0.2060 \]

The margin of error \( E \) is half the width of the confidence interval:

\[ E = \frac{0.217 - 0.195}{2} = 0.0110 \]

For a \( 95\% \) confidence level, the Z-score \( Z \) is:

\[ Z = 1.96 \]

Using the formula for the margin of error in terms of sample size \( n \):

\[ E = Z \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]

We can rearrange this to solve for \( n \):

\[ n = \frac{Z^2 \cdot \hat{p} \cdot (1 - \hat{p})}{E^2} \]

Substituting the known values:

\[ n = \frac{(1.96)^2 \cdot 0.2060 \cdot (1 - 0.2060)}{(0.0110)^2} \]

Calculating this gives:

\[ n \approx 5192.5 \]

Rounding up to the next whole number, we find:

\[ n = 5193 \]

Final Answer