Small sample confidence interval question

The manager of a milk factory finds that the first batch of 10 milk containers produced that day contains 3 which have spoilt milk.  He is concerned that the other batches might also have spoilt containers.  Although they might too have 30% of the containers spoilt, he is concerned there could be a larger percentage.  He wants to know the probable maximum percentage of containers that might be spoilt.

Solution

The approach to this problem is to assume that there is a probability ‘s’ that any one container is spoilt.  We want to find the highest probability this ‘s’ can be while staying consistent with the data from the first batch.  Obviously s = 0.3 is consistent with the first batch, what about s = 0.4? We need to pick a significance level – let’s use the standard 5%.  We look up the binomial table for n = 10:

                                                        s = 0.4, f = 0.6

Now we want to find out the minimum number of successes whose probabilities add to more than 5%:

                                                P(0) = 0.6 % – too small

                              P(0) + P(1) = 0.6% + 4.03% = 4.63% – too small

            P(0) + P(1) + P(2) = 0.6% + 4.03% + 12.09% = 16.72% – large enough (larger than our 5% significance level).

So if the number of spoilt containers in the first batch is 2 or more, it is consistent with s = 0.4 – it is consistent.

Let’s try s = 0.5:

                                                        s = 0.5, f = 0.5

Again, find out the minimum number of successes whose probabilities add up to more than 5%:

                                                 P(0) = 0.1% – too small

                              P(0) + P(1) = 0.1% + 0.98% = 1.08% – too small

               P(0) + P(1) + P(2) = 0.1% + 0.98% + 4.39% = 5.47% – large enough.

So once again, if the number of spoilt containers in the first batch is larger or equal to 2, it is consistent with s = 0.5 – it is.

Try:

                                                        s = 0.6, f = 0.4

Again, find out the minimum number of successes whose probabilities add up to more than 5%:

                                                P(0) = 0.1% – too small.

                              P(0) + P(1) = 0.1% + 0.16% = 0.26% – too small.

                  P(0) + P(1) + P(2) = 0.1% + 0.16% + 1.06% = 1.32% – too small

            P(0) + P(1) + P(2) + P(3) = 0.1% + 0.16% + 1.06% + 4.25% = 5.57%  – large enough.

So if there are 3 or more spoilt containers in the first batch, it is consistent with s = 0.6 – there are, so it is consistent.

Try:

                                                        s = 0.7, f = 0.3

                                                   P(0) = 0 – too small.

                                 P(0) + P(1) = 0 + 0.01% = 0.01% – too small.

                     P(0) + P(1) + P(2) = 0 + 0.01% + 0.14% = 0.15% – too small.

         P(0) + P(1) + P(2) + P(3) = 0 + 0.01% + 0.14% + 0.9% = 1.05% – too small.

P(0) + P(1) + P(2) + P(3) + P(4) = 0 + 0.01% + 0.14% + 0.9% + 3.68% = 4.73% – too small.

P(0) + P(1) + P(2) + P(3) + P(4) + P(5) = 0 + 0.01% + 0.14% + 0.9% + 3.68% + 10.29% = 15.02% – large enough.

So if there are 5 or more spoilt containers in the first batch, it will be consistent with an ‘s’ value of 0.7.  There aren’t, so we know that the highest values of ‘s’ we can have that is still consistent with the first batch sample, is between 0.6 and 0.7.

So the manager can say with confidence that there may be up to over 60% of containers spoilt on average, but it is unlikely that there is a 70% or higher average spoiling rate.  Since we were trying to find the largest value of ‘s’, we call this the upper 95% confidence limit (95% because we used a 5% level of significance – 100% – 5% = 95%).  If he wanted to have even more assurance he could use a smaller level of significance – maybe 1% instead of 5%. 

What we have just done in effect is assume higher and higher average probabilities of a container being spoilt, and then found whether only 3 being spoilt could have occurred from random variation.