Country Insurance wants to estimate the cost of damage to cars due to accidents. The population standard deviation of the cost is assumed at $200. They want to estimate the mean cost using a 95% confidence interval with a maximum margin of error of $10. What is the minimum sample size n?
To determine the minimum sample size needed to estimate the mean cost with the given criteria, we can use the formula for margin of error:
Margin of Error = (Z-value) * (Standard Deviation / √n)
Here,
Margin of Error = $10
Standard Deviation = $200
To find the Z-value for a 95% confidence interval, we look up the corresponding value from the Z-table. For a 95% confidence level, the Z-value is approximately 1.96.
Plugging in the values into the formula, we have:
$10 = 1.96 * ($200 / √n)
Now, we can solve for n:
√n = (1.96 * $200) / $10
√n = 39.2
Squaring both sides gives:
n = 39.2^2 ≈ 1536
Therefore, the minimum sample size (n) needed to estimate the mean cost with a 95% confidence interval and a maximum margin of error of $10 is approximately 1536.