In a survey, 27 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $36 and standard deviation of $4. Find the margin of error at a 90% confidence level.

Answer 1

To find the margin of error at a 90% confidence level, we need to use the formula:

Margin of Error = z * (standard deviation / square root of sample size)

First, let's find the value of z for a 90% confidence level. Using a standard normal distribution table, we find that the z-score for a 90% confidence level is approximately 1.645.

Given:
Mean (μ) = $36
Standard deviation (σ) = $4
Sample size (n) = 27
Z-score (z) = 1.645

Now we can calculate the margin of error:

Margin of Error = 1.645 * ($4 / sqrt(27))

Using a calculator:

Margin of Error ≈ 1.645 * ($4 / 5.196)

Margin of Error ≈ 1.645 * $0.770

Margin of Error ≈ $1.266

Therefore, the margin of error at a 90% confidence level is approximately $1.266.