Question

find the margin of error given the standard error and the confidence level. round your answers to three decimal places, if necessary.
part: 0 / 4
part 1 of 4
(a) standard error=2.3, confidence level 95%
the margin of error is .

Explanation:

Step1: Recall the formula for margin of error

The margin of error (ME) for a confidence interval is calculated as \( ME = z_{\alpha/2} \times \text{Standard Error} \), where \( z_{\alpha/2} \) is the critical z - value corresponding to the given confidence level. For a 95% confidence level, the significance level \( \alpha=1 - 0.95 = 0.05 \), so \( \alpha/2=0.025 \). The critical z - value \( z_{0.025}\) (the z - score such that the area to the right of it is 0.025) is approximately 1.96.

Step2: Substitute the values into the formula

We are given that the standard error is 2.3 and \( z_{\alpha/2}=1.96 \). So we calculate the margin of error as \( ME=1.96\times2.3 \).
First, multiply 1.96 and 2.3: \( 1.96\times2.3 = 4.508 \)

Answer:

4.508