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 3 of 4
(c) standard error = 1.0, confidence level 90%
the margin of error is 1.645.
part: 3 / 4
part 4 of 4
(d) standard error = 5.32, confidence level 98%
the margin of error is .

Explanation:

Step1: Find z - score for 98% confidence

For a 98% confidence level, the significance level $\alpha=1 - 0.98 = 0.02$. The critical value $z_{\alpha/2}$ is the z - score such that the area to the right of $z_{\alpha/2}$ is $\alpha/2=0.01$. Looking up in the standard normal table or using a calculator, $z_{0.01}\approx2.326$.

Step2: Calculate margin of error

The formula for margin of error (ME) when the standard error (SE) is known is $ME = z_{\alpha/2}\times SE$. Given $SE = 5.32$ and $z_{\alpha/2}=2.326$, we have $ME=2.326\times5.32$.
Calculating $2.326\times5.32$:
$2.326\times5.32=(2 + 0.3+0.02 + 0.006)\times5.32=2\times5.32+0.3\times5.32 + 0.02\times5.32+0.006\times5.32=10.64+1.596+0.1064 + 0.03192 = 12.37432\approx12.374$

Answer:

12.374