View Full Version : Margin of error. Confidence level
allabout grey
Mar 29, 2009, 02:29 PM
1.Use margin of error, confidence level, and standard deviation to find the minimum sample size required to estimate an unknown population mean.
Margin of error: $124, Confidence level; 99%, σ = $560
2.Find the margin of error for 95% confidence interval used to estimate the population propprtion.
In a clinical test with 2333 subjects, 50% showed improvement from the treament.
Thank, I'm having a hard time figure how to get the answers here.
galactus
Mar 29, 2009, 04:55 PM
1.Use margin of error, confidence level, and standard
deviation to find the minimum sample size required to estimate an unknown population mean.
Margin of error: $124, Confidence level; 99%, σ = $560
Do you have a stats book? If so, this formula should be in there:
n=\left(\frac{z\cdot\sigma}{E}\right)^{2}
You are given all the necessary info. E is the margin of error. You should know what sigma and z are. Look up z for a 99% CI in the z table.
2.Find the margin of error for 95% confidence interval used to estimate the population proportion.
In a clinical test with 2333 subjects, 50% showed improvement from the treatment.
Proportions has a minimum sample formula as well.
E=z\sqrt{\frac{pq}{n}}
You have all you need to find E.
allabout grey
Mar 30, 2009, 04:38 PM
Do you have a stats book?. If so, this formula should be in there:
n=\left(\frac{z\cdot\sigma}{E}\right)^{2}
You are given all the necessary info. E is the margin of error. You should know what sigma and z are. Look up z for a 99% CI in the z table.
Proportions has a minimum sample formula as well.
E=z\sqrt{\frac{pq}{n}}
You have all you need to find E.
Thank you, I figure the 1st one. I still confused about p and do i use 0.25 for q?
galactus
Mar 31, 2009, 06:14 AM
No, p+q=1. If p=.5, then what does q have to be?