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 2 of 4
(b) standard error=0.1, confidence level 99%
the margin of error is 0.258.
part: 2 / 4
part 3 of 4
(c) standard error=1.0, confidence level 90%
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 90% confidence level, the significance level \( \alpha=1 - 0.90 = 0.10 \), so \( \alpha/2=0.05 \).
We need to find the \( z \) - value such that the area to the right of \( z_{\alpha/2} \) is 0.05. Looking up in the standard normal distribution table (or using a calculator), the \( z \) - value for \( \alpha/2 = 0.05 \) is approximately \( z_{0.05}=1.645 \).

Step2: Calculate the margin of error

We are given that the standard error is 1.0 and the confidence level is 90%.
Using the formula \( ME=z_{\alpha/2}\times\text{Standard Error} \), substitute \( z_{\alpha/2} = 1.645 \) and standard error = 1.0.
\( ME=1.645\times1.0 = 1.645 \)

Answer:

1.645