The Wall Street Journal reported that automobile crashes cost the United States $162 billion annually (The Wall Street Journal, March 5, 2008). The average cost per person for crashes in the Tampa, Florida, area was reported to be $1599. Suppose this average cost was based on a sample of 50 persons who had been involved in car crashes and that the population standard deviation is σ = $600.
What is the margin of error for a 95% confidence interval (to 2 decimals)?
Mean of error = Za/2* σ/√n
= 1.96* 600/√50
= $166.31
To find the margin of error for a 95% confidence interval, we need to use the formula:
Margin of Error = Z * (σ / √n)
Where:
Z = Z-score for the desired confidence level (95% confidence level corresponds to a Z-score of 1.96)
σ = population standard deviation ($600 in this case)
n = sample size (50 persons in this case)
Substituting the given values into the formula, we have:
Margin of Error = 1.96 * ($600 / √50)
Calculating this expression, we find:
Margin of Error ≈ 1.96 * ($600 / 7.071)
Therefore, the margin of error for a 95% confidence interval is approximately $83.97 when rounded to two decimal places.
To find the margin of error for a 95% confidence interval, you can use the formula:
Margin of Error = (Z * σ) / √n
Where:
Z = The Z-score corresponding to the desired confidence level (95% confidence level corresponds to a Z-score of 1.96)
σ = The population standard deviation
n = The sample size
Given the information provided:
Z = 1.96
σ = $600
n = 50
Substituting these values into the formula:
Margin of Error = (1.96 * 600) / √50
Calculating the square root of 50:
Margin of Error = 1176 / 7.071
Calculating the result:
Margin of Error ≈ 166.32
Therefore, the margin of error for a 95% confidence interval is approximately $166.32.