# Statistics help.

• October 24th 2009, 08:32 PM
zerolx
Statistics help.
Need help with population mean and margin of error for homework.

Determine the point estimate of the population mean and margin of error for the Confidence interval.

Lower bound: 5, upper bound:23

I would be very grateful and appreciative If someone can help and explain it to me more clearly.
• October 24th 2009, 10:29 PM
mr fantastic
Quote:

Originally Posted by zerolx
Need help with population mean and margin of error for homework.

Determine the point estimate of the population mean and margin of error for the Confidence interval.

Lower bound: 5, upper bound:23

I would be very grateful and appreciative If someone can help and explain it to me more clearly.

More information is required. Please post the question eaxtly as it's written in wherever you got it from.
• October 24th 2009, 10:43 PM
zerolx
Quote:

Originally Posted by mr fantastic
More information is required. Please post the question eaxtly as it's written in wherever you got it from.

That's how it was written in the book. It's # 19
http://img20.imageshack.us/img20/985/68467805.th.jpg
I'm sorry if images are not allowed.
• October 24th 2009, 11:08 PM
matheagle
The point estimate is the sample mean which is the midpoint of the interval, 14.

The error, I assume is $z_{\alpha/2}s/\sqrt n$ which is 23-14=9
• October 25th 2009, 12:54 AM
mr fantastic
Quote:

Originally Posted by matheagle
The point estimate is the sample mean which is the midpoint of the interval, 14.

The error, I assume is $z_{\alpha/2}s/\sqrt n$ which is 23-14=9

Well they say eagles have sharp eyes. Because I just can't see where the values of $n$, $s$ and $\alpha$ have come from.