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We have a manufacturing process in which the finished products have the following requirements: individual unit must have a weight within ± 10% of average weight (test with 10 random units).

One of the in-process quality control test is: Take 10 units, determine their average weight (X) without measuring individual units. Let's say this test is repeated 100 times throughout the day.

So each measurement X is the average of 10 units. From the data of X1, X2,...X100, we can calculate the mean of X and its standard deviation (standard error in this case since X is itself an average).

My question is: Can you estimate the standard deviation and range of the population (individual units - 1 million in total) and the probability of passing the required finished tests, from this data and how? Can you use the formula SD = SE * sqrt (n)? (in this case n=10)

If anyone can point me to documentation or guides i'm very thankful.

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I would personnaly use the following estimator of the variance for your problem:

$$ \hat{\sigma}^2 = \hat{\sigma}^2_L + \hat{\sigma}^2_E $$

  • Empirical mean: $\bar{X}:= \frac{1}{nL} \sum_{j = 1}^L \sum_{i = 1}^n X_{ij}$

  • Within location variance $\hat{\sigma}^2_E = \frac{1}{L(n-1)} \sum_{j = 1}^L \sum_{i = 1}^n \left(X_{ij} - \bar{X}_j \right)^2 = MS_E$

  • Between-location variance $\hat{\sigma}^2_L = \frac{1}{L-1} \sum_{j = 1}^L \left( \bar{X}_j - \bar{X} \right)^2 - \frac{1}{n}\hat{\sigma}^2_E = \frac{1}{n}(MS_L - MS_E)$

  • $MS_L = \frac{n}{L-1} \sum_{j = 1}^L \left( \bar{X}_j - \bar{X} \right)^2$

  • $\bar{X}_j = \frac{1}{n} \sum_{i = 1}^n X_{ij}$

You can use tolerance intervals for example to determine with a given confidence level what is the probability to pass your requirements.

You can find the previous definition in Anand M. Joglekar. Statistical Methods for Six Sigma. John Wiley and Sons Inc. 2003 on pages 191-193.

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  • $\begingroup$ I mean for in-process control test we don't measure individual units weight, just their average (by putting all 10 of them on the scale, take the weight and divide by 10). Only the finished test would measure them individually AND also their average. $\endgroup$
    – Gnad
    Commented Apr 4, 2023 at 13:18
  • $\begingroup$ An example: I have done the in-process control test 5 time and my result is 103, 101, 95, 106, 98 - each of these results is the average of 10 units not individual data (theoretical weight is 100). This result in Xbar of 100.6 and SumofSquares is 73.2. Can we say anything about population standard deviation (individual units) assuming it is normally distributed? $\endgroup$
    – Gnad
    Commented Apr 4, 2023 at 13:28
  • $\begingroup$ Let's say I take the standard error, which is 4.28 (sqrt (73.2/(5-1)). Then standard deviation of population is 4.28 * sqrt (10) = 13.5. But I'm not sure it is correct practice. $\endgroup$
    – Gnad
    Commented Apr 4, 2023 at 13:38
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    $\begingroup$ You have SE ($MS_E$ in my notation) and you have the sum of square $SS$. You know that $SS = MS_E + MS_L$ and then from it you can obtain $MS_L$ and so, everything needed for the previous estimator $\hat{\sigma}^2$ to be known. $\endgroup$
    – lulufofo
    Commented Apr 4, 2023 at 13:43

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