# Math Help - Stumped again!! Statistics... :(

1. ## Stumped again!! Statistics... :(

Hi guys!

THanks for all your help guys - esp Mr F!

I am stumped on these 2 questions! Any one have any idea on how to proceed?

Assuming the standard deviation for the cost of a bathroom remodel is known to be $545. How large a sample would need to be taken to estimate the mean cost of a bathroom remodel to within$136 with 99% confidence? I got 107 as my answer.

A researcher is interested in estimating the proportion of employees who are unhappy in their current. A prior study estimates this proportion at 0.13. How large a sample would need to be taken to estimate this proportion to within 4% with 95% accuracy? i got 71 as my answer

any help would be appreciated!

Cheers

Sunny

2. The first one is correct. The second one(proportion) should be 272 as the minimum sample size.

3. Originally Posted by galactus
The first one is correct. The second one(proportion) should be 272 as the minimum sample size.
How do you come up with 272?

Cheers

darren

4. I ran it through my Excel Stats sheet I use for this stuff.

But, here is the 'old-fashioned' way:

$n=pq(\frac{z}{E})^{2}$

p=.13, q=.87, z=1.96, E=.04

$n=(.13)(.87)(\frac{1.96}{.04})^{2}=271.55=272$

5. ## here is what i did!

he z value for the 95% interval is 1.96.

The interval for a proportion is:

mean - sqrt(p(1-p)/N) to mean + sqrt(p(1-p)/N)

We want the sqrt(p(1-p)/N) to be 0.04:

0.04 = sqrt(p(1-p)/N)

p = 0.13

0.04 = sqrt(0.13*0.87/N)

0.04 = sqrt(0.1131/N)

Square:

0.1131/N = 0.0016

Multiply:

0.0016N = 0.1131

Divide:

N = 0.1131/0.0016

N = 70.6875

Round up:

N = 71

6. The formula for sample size in proprotions is as I gave you:

$n=pq(\frac{z}{E})^{2}$

Run it through a calculator. I bet you get 272. I checked this with Excel and my TI-92

7. ok!

Thanks for the help! could u look at the other post that i just got going?

cheers

Darren