1
$\begingroup$

I am presented with:

Two machines are used to fill plastic bottles with dishwashing detergent. The standard deviations of fill volume are known to be σ1 = 0.10 and σ2 = 0.15 fluid ounces for the two machines, respectively. Two random samples of n1=12 bottles from machine 1 and n2=10 bottles from machine 2 are selected, and the sample mean fill volumes are =30.65 and =30.34 fluid ounces. Assume normality.

Find a 95% upper-confidence bound on the mean difference in fill volume.

I was able to answer all other parts of this question, but I can't figure this problem out... I know the formula should be:

U=$x_1-x_2+z_asqrt(\sigma^2_1/n_1+\sigma^2_2/n_2)$

I'm not sure what $z_a$ is and I'm sure that's part of my problem... I also do not have a strong physical concept of what an upper-confidence bound is. Can anyone help out?

$\endgroup$
  • $\begingroup$ you should take $z_{\alpha}=1.96$ (that is $z_{1-\alpha/2}$) $\endgroup$ – mert Oct 24 '13 at 16:34
  • $\begingroup$ How did you get that number? $\endgroup$ – rphello101 Oct 24 '13 at 16:53
  • $\begingroup$ When I plug that number in, I get .41883. I got that number when I did the 2-SampZInt on the calculator as well, but that answer is not correct in the context of this problem apparently - electronic submission will not accept it. $\endgroup$ – rphello101 Oct 24 '13 at 16:59
2
$\begingroup$

What you need to calculate here is the 95% confidence interval ($1-\alpha=0.95$) for the mean difference. For large samples, it is given by:

$$\left[\bar{X}_{1}-\bar{X}_{2}-z_{1-\frac{\alpha}{2}}\sqrt{\frac{\sigma_{1}^2}{n_{1}}+\frac{\sigma_{2}^2}{n_{2}}},\bar{X}_{1}-\bar{X}_{2}+z_{1-\frac{\alpha}{2}}\sqrt{\frac{\sigma_{1}^2}{n_{1}}+\frac{\sigma_{2}^2}{n_{2}}}\right].$$

Intuitively, it is the interval in which the true value of the mean of the difference of the fill volumes is located with 95% likelihood.

More precisely, you are interested in the upper bound, i.e. $\bar{X}_{1}-\bar{X}_{2}+z_{1-\frac{\alpha}{2}}\sqrt{\frac{\sigma_{1}^2}{n_{1}}+\frac{\sigma_{2}^2}{n_{2}}}$.

Now, you were wondering what $z_{1-\frac{\alpha}{2}}$. It such value of $z$, which satisifies $\Phi(z)=1-\frac{\alpha}{2}$ where $\Phi(\cdot)$ is the cumulative distribution function of the standard normal distribution. The value given by our colleague, $1.96$, was the correct value, retrieved from the tables for standard normal distribution.

However, as your values for both $n$ are rather low, it seems reasonable to use the t-distribution instead of normal distribution. You upper bound changes thus to:

$$\bar{X}_{1}-\bar{X}_{2}+t\left(\frac{\alpha}{2},\nu\right)\sqrt{\frac{\sigma_{1}^2}{n_{1}}+\frac{\sigma_{2}^2}{n_{2}}}$$

where $\alpha/2$ is the upper tail probability ($1-\alpha=0.95$ is the confidence level) and $\nu=n_1+n_2-2$ are the degrees of freedom. You may find in the tables of t-distribution that $t(0.025,20)=2.086$ (though you may find other notations, such as $t_{0.975}$ - it usually follows from the table which notation is used). Plug in the rest of your numbers, and you should obtain the correct result.

$\endgroup$
  • $\begingroup$ This is a nice exposition but it addresses a slightly different problem: the question asks for a UCL of the mean, not a symmetric CI for the mean. $\endgroup$ – whuber Oct 25 '13 at 1:55
  • $\begingroup$ Correct if I'm wrong, but if that's the case, by symmetry, we could use a similar formula with $t(\alpha,\nu)$ instead of $t(\alpha/2,\nu)$. $\endgroup$ – Johnny Westerling Oct 25 '13 at 8:33
  • $\begingroup$ That's right--and leave out the left endpoint of the interval altogether. $\endgroup$ – whuber Oct 25 '13 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.