Question 1168930: You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately
σ=69.7.You would like to be 99.5% confident that your estimate is within 2 of the true population mean. How large of a sample size is required?
Do not round mid-calculation. However, use a critical value accurate to three decimal places — this is important for the system to be able to give hints for incorrect answers.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! To determine the required sample size, we can use the formula for the margin of error when the population standard deviation is known:
$E = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$
where:
$E$ is the desired margin of error (within 2)
$z_{\alpha/2}$ is the critical value from the standard normal distribution corresponding to the desired confidence level
$\sigma$ is the population standard deviation (69.7)
$n$ is the required sample size
We are given a 99.5% confidence level. The significance level $\alpha$ is:
$\alpha = 1 - 0.995 = 0.005$
The critical value $z_{\alpha/2}$ corresponds to the $1 - \alpha/2 = 1 - 0.005/2 = 1 - 0.0025 = 0.9975$ percentile of the standard normal distribution.
We need to find the z-score such that the area to the left is 0.9975. Using a standard normal distribution table or a calculator, we find the z-score:
$z_{0.9975} \approx 2.807$
Now, we can plug the values into the margin of error formula and solve for $n$:
$E = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$
$2 = 2.807 \times \frac{69.7}{\sqrt{n}}$
Now, solve for $\sqrt{n}$:
$\sqrt{n} = \frac{2.807 \times 69.7}{2}$
$\sqrt{n} = \frac{195.6479}{2}$
$\sqrt{n} = 97.82395$
Square both sides to find $n$:
$n = (97.82395)^2$
$n \approx 9569.52$
Since the sample size must be a whole number, we round up to the nearest integer to ensure the desired margin of error is met.
$n = 9570$
Final Answer: The final answer is $\boxed{9570}$
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