Questions: NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. Suppose a sample of 1026 people is drawn. Of these people, 523 passed out. Using the data, construct the 99% confidence interval for the population proportion of people who black out at G forces greater than 6. Round your answers to three decimal places.

Solution Steps

The sample proportion of people who blacked out at G forces greater than 6 is calculated as follows:

\[ \hat{p} = \frac{x}{n} = \frac{523}{1026} \approx 0.510 \]

For a 99% confidence interval, the significance level (\(\alpha\)) is given by:

\[ \alpha = 1 - 0.99 = 0.010 \]

The formula for the confidence interval for a single population proportion is:

\[ \hat{p} \pm z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]

Where:

• \(\hat{p} = 0.510\)
• \(n = 1026\)
• \(z\) is the z-score corresponding to the 99% confidence level, which is approximately \(2.576\).

Substituting the values into the formula:

\[ 0.510 \pm 2.576 \cdot \sqrt{\frac{0.510(1 - 0.510)}{1026}} \]

Calculating the margin of error:

\[ \sqrt{\frac{0.510(0.490)}{1026}} \approx 0.015 \]

Thus, the confidence interval becomes:

\[ 0.510 \pm 2.576 \cdot 0.015 \]

Calculating the bounds:

\[ 0.510 - 0.0387 \approx 0.4713 \quad \text{and} \quad 0.510 + 0.0387 \approx 0.5487 \]

Rounding to three decimal places, the confidence interval is:

\[ (0.471, 0.549) \]

Final Answer