Question

find the margin of error given the standard error and the confidence level. round your answers to three decimal places, if necessary.
part 1 of 4
(a) standard error=2.3, confidence level 95%
the margin of error is 4.508.
part: 1 / 4
part 2 of 4
(b) standard error=0.1, confidence level 99%
the margin of error is .

Explanation:

Step1: Recall the formula for margin of error

The margin of error (ME) 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 99% confidence level, the significance level \( \alpha=1 - 0.99 = 0.01 \), so \( \alpha/2=0.005 \). The critical z - value \( z_{0.005} \) (the z - score such that the area to the right of it is 0.005) is approximately 2.576.

Step2: Substitute the values into the formula

We are given that the standard error is 0.1 and the critical z - value for 99% confidence level is 2.576. Using the formula \( ME=z_{\alpha/2}\times\text{Standard Error} \), we substitute \( z_{\alpha/2} = 2.576 \) and Standard Error \( = 0.1 \). So \( ME=2.576\times0.1 \).

Step3: Calculate the result

\( 2.576\times0.1 = 0.2576 \). Rounding to three decimal places, we get 0.258.

Answer:

0.258