p value and confidence interval

**bombo31** · July 7th 2006, 05:03 AM

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

**CaptainBlack** · July 7th 2006, 11:52 PM

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:

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

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

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

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

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

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

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

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

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

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

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

and so the same conclusion would be arrived at.

**CaptainBlack** · July 8th 2006, 01:07 AM

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:

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

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

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

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

or:

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

so:

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

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

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

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

RonL

**bombo31** · July 8th 2006, 02:49 AM

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

**CaptainBlack** · July 8th 2006, 03:01 AM

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

I will correct it.

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

Thanks

RonL

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