NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6 . In an earlier study, the population proportion was
estimated to be 0.43 .
How large a sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the confidence level with an error of at most
0.03 ? Round your answer up to the next integer.
Answer
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Question

NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6 . In an earlier study, the population proportion was
estimated to be 0.43 .
How large a sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the

confidence level with an error of at most
0.03 ? Round your answer up to the next integer.
Answer
How to enter your answer (opens in new window)

UpStudy AI · Accepted Answer

To determine the sample size required to estimate the fraction of people who black out at 6 or more Gs at a 99% confidence level with an error of at most 0.03, we can use the formula for sample size calculation in proportion estimation:

$$ n = \left( \frac{{Z^2 \cdot p \cdot (1-p)}}{{E^2}} \right) $$

Where:
- $$ n $$ is the sample size
- $$ Z $$ is the Z-score corresponding to the desired confidence level (99% confidence level corresponds to a Z-score of 2.576)
- $$ p $$ is the estimated population proportion (0.43)
- $$ E $$ is the margin of error (0.03)

Substitute the values into the formula and calculate the sample size.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(2.576^{2}\times 0.43\left(1-0.43\right)\right)}{0.03^{2}}$$
- step1: Remove the parentheses:
  $$\frac{2.576^{2}\times 0.43\left(1-0.43\right)}{0.03^{2}}$$
- step2: Subtract the numbers:
  $$\frac{2.576^{2}\times 0.43\times 0.57}{0.03^{2}}$$
- step3: Convert the expressions:
  $$\frac{\left(\frac{322}{125}\right)^{2}\times 0.43\times 0.57}{0.03^{2}}$$
- step4: Convert the expressions:
  $$\frac{\left(\frac{322}{125}\right)^{2}\times 0.43\times 0.57}{\left(\frac{3}{100}\right)^{2}}$$
- step5: Multiply:
  $$\frac{\frac{63532371}{39062500}}{\left(\frac{3}{100}\right)^{2}}$$
- step6: Evaluate the power:
  $$\frac{\frac{63532371}{39062500}}{\frac{9}{10000}}$$
- step7: Multiply by the reciprocal:
  $$\frac{63532371}{39062500}\times \frac{10000}{9}$$
- step8: Reduce the numbers:
  $$\frac{21177457}{15625}\times \frac{4}{3}$$
- step9: Multiply the fractions:
  $$\frac{21177457\times 4}{15625\times 3}$$
- step10: Multiply:
  $$\frac{84709828}{46875}$$
The sample size required to estimate the fraction of people who black out at 6 or more Gs at a 99% confidence level with an error of at most 0.03 is approximately 1807.14.

Rounding up to the next integer, the required sample size is 1808.
The required sample size is 1808.

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