# Find the margin error E

• Dec 19th 2005, 11:08 PM
sinlee2010
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
• Dec 20th 2005, 12:03 AM
CaptainBlack
Quote:

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 $\displaystyle N=25$, we observe a mean $\displaystyle \bar x = 5.24 \ sec$.

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

Then 90% margin of error for the mean is:

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

The 90% confidence interval is:

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

(Note that this retains more precision than is warranted)

RonL
• Dec 20th 2005, 06:25 AM
sinlee2010
Find the Margin of Error
Thanks for the help Captainblack...... :)