# If pilot data of 30 samples of size 100 items are tested for nonconformity, and a total of 95 of the 3,000 items in the pilot data are defective, find the upper and lower control limits for a p-chart.

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### 1 Answer

` `Out of the first 30 pilot samples of `n=100` items per population/batch tested for nonconformity, 95 out of the 30 x 100 = 3000 items were found to be defective.

From these pilot data then, the proportion defective in the population is estimated to be

`hat p = 95/n = 95/3000 = 0.032`

The point of doing this sampling and testing for nonconformity is to estimate the proportion defective in a batch or production run without having to inspect the whole batch. If the estimated proportion is unacceptably high, the batch will be rejected or marked as lower quality.

A (Shewart) p-chart is a control chart that plots the results from batch inspection as each batch comes in. An alarm is signalled if an estimated proportion falls outside the `3sigma` limits. These are derived from 'in control' or 'pilot data'. If the data from these 30 samples here provide a benchmark for the in control state of the process, then the limits for the resulting p-chart would be the two-sided 99.7% confidence limits about the estimated proportion from the pilot data `hat p`. Specifically, making a Normal approximation to the Binomial, the limits are given by

`hat p pm 3 sigma`

where `sigma = sqrt((hat p (1-hat p))/n)` and `n` is the size of the samples tested as batches come in, so that `n= 100` here.

Therefore,

`sigma = sqrt(((95/3000)(2905/3000))/100)` ` ``= 0.0175`

The limits then are `hat p pm 3 sigma = 95/3000 pm 0.0525 = [-0.021,0.084] `

Now, the Normal approximation has led to a negative lower limit for the proportion. In practice this would simply be set to zero **so that the limits would be**

**0 and 0.084**

(this is in fact a 96.3% confidence interval rather than the typical 99.7% - the upper limit could be raised to lower the chance of a false alarm)

**The limits for the chart would be 0, 0.084**