I have the following question:
A soft-drink machine is regulated so that the amount of drink
dispensed averages $240$ milliliters with a standard deviation of $15$
milliliters. Periodically, the machine is checked by taking a sample
of $40$ drinks and computing the average content. If the mean of the
$40$ drinks is a value within the interval $\mu_{ \bar X} ± 2\sigma_{
\bar X }$, the machine is thought to be operating satisfactorily;
otherwise, adjustments are made. The com- pany official found the mean
of $40$ drinks to be $\bar x = 236$ milliliters and concluded that the
machine needed no adjustment. Was this a reasonable decision?
The reason I am posting this is that I solved it in a different way than the solution provided and I want someone to verify my solution, I did the following:
I applied the central limit theorem since the central limit theorem as I think the central limit theorem tells how many standard deviations away from the mean, I got a value of $-1.68$ which is between $-2$ and $2$ so the decision is correct.
In the solution given they find the sample standard deviation, multiply it by $2$ and add/subtract it from the population mean to see if $236$ falls in this range, the decision was correct too but I am not very sure about my reasoning of the central limit theorem so I want someone to correct me if I am wrong and the right decision answer was just by luck.
Best Answer
I don't see how it is possible to calculate a sample standard deviation from the information that is provided in the question, so either:
The way I would answer the question is as follows. For a sample of $n = 40$ randomly selected drinks, the observed mean volume is $\bar x = 236$ milliliters, which is $4$ milliliters less than the hypothesized mean $\mu = 240$ milliliters. The standard error of the sample mean is $$\sigma_{\bar x} = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{40}} \approx 2.37171,$$ hence the permissible margin of error is $2\sigma_{\bar x} \approx 4.74342$ milliliters. Since $4$ milliliters does not exceed this margin of error, the machine is not adjusted: $$|\bar x - \mu| \le 2\sigma_{\bar x}.$$ My suspicion is that you have mistaken $\sigma_{\bar x}$ as the sample standard deviation. This would be incorrect because you are given that $\sigma = 15$ milliliters, consequently the sampling distribution of the sample mean for $n = 40$ is approximately normally distributed with mean $\mu$ and standard deviation $\sigma_{\bar x}$. The latter is not a sample standard deviation because it is not a statistic estimated from the sample: it is a value inherited from the known standard deviation of individual drink volumes.
That said, your approach is not materially different because in both cases, an asymptotically normal distribution is assumed for the sampling distribution. You are calculating a $z$-score for the sampling distribution and comparing it against the critical value $2$. In other words, my solution is working on the scale of milliliters; your solution is working on the scale of $z$-scores, which is measured in standard deviations and is unitless.