# Thread: Find the margin error E

1. ## Find the margin error E

Please help me solve this problem. I need to know what is the correct formula in order to solve the problem. Thanks

Use the given confidence level and sample data to find (a) the margin error E and (b) a confidence interval for estimation the population mean u.

Times between using a dishwasher: 90% confidence or 1.645 (critical value); n = 25, x = 5.24 sec, the population is normally distributed, and o is known to be 2.50 sec. Note: I was unable to put the minus sign over the x symbol.

Sin

2. Originally Posted by sinlee2010
Please help me solve this problem. I need to know what is the correct formula in order to solve the problem. Thanks

Use the given confidence level and sample data to find (a) the margin error E and (b) a confidence interval for estimation the population mean u.

Times between using a dishwasher: 90% confidence or 1.645 (critical value); n = 25, x = 5.24 sec, the population is normally distributed, and o is known to be 2.50 sec. Note: I was unable to put the minus sign over the x symbol.

Sin
For a sample of size $N=25$, we observe a mean $\bar x = 5.24 \ sec$.

We know that the population is normally distributed with known standard
deviation $\sigma=2.5\ sec$.

Then 90% margin of error for the mean is:

$E= \frac{1.645 \frac{\sigma}{\sqrt N}}{\bar x}100\ \%\ \sim 15.67\ \%$

The 90% confidence interval is:

$(\bar x-1.645 \frac{\sigma}{\sqrt N},\bar x+1.645 \frac{\sigma}{\sqrt N})=(4.4275,6.0725)$.

(Note that this retains more precision than is warranted)

RonL

3. ## Find the Margin of Error

Thanks for the help Captainblack......