# slight discrepancy between my answer and textbook...

• Aug 28th 2007, 11:22 AM
earachefl
slight discrepancy between my answer and textbook...
OK, problem has sample n = 37, $\sum x = 277.875$, $\sum x^2 = 2093.109375$. Estimate with 99% confidence the mean.

I get $\mu = \frac{\sum x}{n} \Rightarrow \frac{277.875}{37} = 7.510$

and $\sigma = \sqrt{\sum x^2P(x) - \mu^2} = \sqrt {\frac{2093.109375}{37} - 7.51^2} \approx .413$

For a 99% confidence rating, $z_{.005} = 2.576$. Therefore, the confidence interval should be
$\mu \pm 2.576\frac {\sigma}{\sqrt n} \Rightarrow 7.51 \pm 2.576 \frac {.413}{\sqrt 37} \Rightarrow 7.51 \pm .175$.

The book gives answer $7.51 \pm .176$. Not much of a difference, I know; I'm just wondering whether the difference might be due to rounding assumptions or whether I've made a more fundamental mistake.
• Aug 28th 2007, 02:03 PM
CaptainBlack
Quote:

Originally Posted by earachefl
OK, problem has sample n = 37, $\sum x = 277.875$, $\sum x^2 = 2093.109375$. Estimate with 99% confidence the mean.

I get $\mu = \frac{\sum x}{n} \Rightarrow \frac{277.875}{37} = 7.510$

and $\sigma = \sqrt{\sum x^2P(x) - \mu^2} = \sqrt {\frac{2093.109375}{37} - 7.51^2} \approx .413$

For a 99% confidence rating, $z_{.005} = 2.576$. Therefore, the confidence interval should be
$\mu \pm 2.576\frac {\sigma}{\sqrt n} \Rightarrow 7.51 \pm 2.576 \frac {.413}{\sqrt 37} \Rightarrow 7.51 \pm .175$.

The book gives answer $7.51 \pm .176$. Not much of a difference, I know; I'm just wondering whether the difference might be due to rounding assumptions or whether I've made a more fundamental mistake.

I don't know that you are doing this but I suspect it is the case. The variance you should be using is the unbiased estimator of the population variance based on the sample, which is:

$
V=\frac{1}{n-1}\sum(x- \bar{x})^2 = \frac{1}{n-1}\sum x^2 - \frac{n}{n-1}\bar{x}^2
$

RonL
• Aug 28th 2007, 02:36 PM
earachefl
Thanks - that's not even mentioned in my book.... I'll check it out.
• Aug 28th 2007, 03:20 PM
earachefl
Actually, when I use that formula, I get the answer 7.51 $\pm .17723903$, which is also off from the book's answer but a bit higher. Still confused.
• Aug 28th 2007, 08:55 PM
earachefl
D'oh!! I have realized the error of my ways - I forgot how to do the standard deviation (hey, that was Chapter 2, I'm in Chapter 9 now :D )

so if I use $s = \sqrt
{ \frac {\sum x^2 - \frac { (\sum x)^2}{n}}{n-1}}$

then s = .416

and the answer magically becomes $7.51 \pm .176$ which is what the book says. Hoorah!

It's amazing how easy it is to become confused between the standard deviation of a population and the standard deviation of a sample... at least for me :(
• Aug 28th 2007, 09:01 PM
CaptainBlack
Quote:

Originally Posted by earachefl
Actually, when I use that formula, I get the answer 7.51 $\pm .17723903$, which is also off from the book's answer but a bit higher. Still confused.

Quote:

Originally Posted by earachefl
D'oh!! I have realized the error of my ways - I forgot how to do the standard deviation (hey, that was Chapter 2, I'm in Chapter 9 now :D )

so if I use $s = \sqrt
{ \frac {\sum x^2 - \frac { (\sum x)^2}{n}}{n-1}}$

then s = .416

and the answer magically becomes $7.51 \pm .176$ which is what the book says. Hoorah!

It's amazing how easy it is to become confused between the standard deviation of a population and the standard deviation of a sample... at least for me :(

These should give the same result as what I gave you was a rewrite (if I did it correctly) of:

$
V = \frac {\sum x^2 - \frac { (\sum x)^2}{n}}{n-1}
$

RonL