# p value and confidence interval

• Jul 7th 2006, 05:03 AM
bombo31
p value and confidence interval
In a random sample of ten cans of corn from supplier B, the average weight per can of corn was sample mean or x bar = 9.4 ounces with standard deviation s = 1.8 ounces. Does this sample contain sufficient evidence to indicate the mean weight is less than 10 ounces? Use the p-value approach. Find a 98% confidence interval for population variance
• Jul 7th 2006, 11:52 PM
CaptainBlack
Quote:

Originally Posted by bombo31
In a random sample of ten cans of corn from supplier B, the average weight per can of corn was sample mean or x bar = 9.4 ounces with standard deviation s = 1.8 ounces. Does this sample contain sufficient evidence to indicate the mean weight is less than 10 ounces? Use the p-value approach.

Assuming that the distribution of can weights is normal, the random variable:

$\displaystyle t=\sqrt{N}\frac{\bar{x}-\mu}{s}$

where $\displaystyle \bar{x}$ is the sample mean, $\displaystyle \mu$ is the population mean,
$\displaystyle s$ is the sample standard deviation (the one where division is by
$\displaystyle N-1$ rather than $\displaystyle N$ as we need the unbiased estimator of the
population variance here), and $\displaystyle N$ is the sample size, has a
Student's t-distribution with $\displaystyle N-1$ degrees of freedom.

The null hypothesis is that $\displaystyle \mu=10$, and then we have:

$\displaystyle t=\sqrt{10} \frac{9.4-10}{1.8}\approx -1.054$

Now looking this up in an appropriate table for $\displaystyle 9$ degrees of freedom gives:

$\displaystyle p(t \le -1.054)=0.1597$,

or $\displaystyle \sim 16 \%$, and with this level of probability we would not reject the
hypothesis that $\displaystyle \mu=10$.

Now we could have tried using the large sample assumption, which would then
justify treating:

$\displaystyle z=\sqrt{10} \frac{\bar{x}-\mu}{s}$

as being normally distributed, then looking this up in a normal table we would
find that:

$\displaystyle p(z \le -1.054)=0.1459$,

and so the same conclusion would be arrived at.
• Jul 8th 2006, 01:07 AM
CaptainBlack
Quote:

Originally Posted by bombo31
In a random sample of ten cans of corn from supplier B, the average weight per can of corn was sample mean or x bar = 9.4 ounces with standard deviation s = 1.8 ounces. ..

.. Find a 98% confidence interval for population variance

If we assume that the weight of cans has a normal distribution then:

$\displaystyle x^2=\frac{(N-1)s^2}{\sigma^2} \sim \chi^2_{N-1}$

That is $\displaystyle x^2$ has a chi-squared distribution with $\displaystyle N-1$ degrees of freedom.

Now we want a 98% confidence region so we need the critical values
of $\displaystyle \chi^2_{9}$ for 1% and 99% (as the area between will constitute 98%.
These critical values are: $\displaystyle 21.7$ and $\displaystyle 2.09$, so for
our confidence region:

$\displaystyle 2.09 \le x^2 \le 21.7$,

or:

$\displaystyle 2.09 \le \frac{(N-1)s^2}{\sigma^2} \le 21.7$,

so:

$\displaystyle \frac{(N-1) s^2}{2.09} \ge \sigma^2 \ge \frac{(N-1) s^2}{21.7}$

substituting the values for $\displaystyle s$ and $\displaystyle N$:

$\displaystyle 10.12 \ge \sigma^2 \ge 0.975$

Hence the 98% confidence interval for $\displaystyle \sigma^2$ (variance) is:
$\displaystyle [0.975,10.12]$.

RonL
• Jul 8th 2006, 02:49 AM
bombo31
Thanks
Thanks much.I think there is a miscalculation in the original statistic. I got
-1.053. Thanks a lot. You are the best.
• Jul 8th 2006, 03:01 AM
CaptainBlack
Quote:

Originally Posted by bombo31
Thanks much.I think there is a miscalculation in the original statistic. I got
-1.053. Thanks a lot. You are the best.

Typo that propagated through the working :mad:

I will correct it.

Note same error crept into your other question as well, both now corrected

Thanks

RonL
• Nov 8th 2015, 07:16 PM
BreeLee
Re: p value and confidence interval
"The data do not present sufficient evidence to indicate that the mean weight per can is less than 10 ounces..."

the confidence interval for 98% for population mean:

is it CI= 7.7943, 11.0057 OR CI= 8.2843, 10.5157???